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769 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8 Systems of Equations and Inequalities
Section 8.1 1. 3 4 8
4 41
x xxx
+ = −==
The solution set is { }1 .
2. a. 3 4 12x y+ = x-intercept: ( )3 4 0 12
3 124
xxx
+ =
==
y-intercept: ( )3 0 4 124 12
3
yyy
+ =
==
b. 3 4 12
4 3 123 34
x yy x
y x
+ == − +
= − +
A parallel line would have slope 34
− .
3. inconsistent
4. consistent
5. False
6. True
7. 2 55 2 8
x yx y− =⎧
⎨ + =⎩
Substituting the values of the variables: 2(2) ( 1) 4 1 55(2) 2( 1) 10 2 8
− − = + =⎧⎨ + − = − =⎩
Each equation is satisfied, so 2, 1x y= = − , or (2, 1)− , is a solution of the system of equations.
8. 3 2 2
7 30x yx y+ =⎧
⎨ − = −⎩
Substituting the values of the variables: 3( 2) 2(4) 6 8 2( 2) 7(4) 2 28 30− + = − + =⎧
⎨ − − = − − = −⎩
Each equation is satisfied, so 2, 4x y= − = , or ( 2, 4)− , is a solution of the system of equations.
9. 3 4 41 132 2
x y
x y
− =⎧⎪⎨
− = −⎪⎩
Substituting the values of the variables: 13(2) 4 6 2 42
1 1 3 1(2) 3 12 2 2 2
⎧ ⎛ ⎞− = − =⎜ ⎟⎪⎪ ⎝ ⎠⎨
⎛ ⎞⎪ − = − = −⎜ ⎟⎪ ⎝ ⎠⎩
Each equation is satisfied, so 12, 2x y= = , or
( )12, 2 , is a solution of the system of equations.
10. 12 02
193 4 2
x y
x y
⎧ + =⎪⎨
− = −⎪⎩
Substituting the values of the variables, we obtain:
( )
( )
1 12 2 1 1 02 2
1 3 193 4 2 82 2 2
⎧ ⎛ ⎞− + = − + =⎜ ⎟⎪⎪ ⎝ ⎠⎨⎛ ⎞⎪ − − = − − = −⎜ ⎟⎪ ⎝ ⎠⎩
Each equation is satisfied, so 1 , 22x y= − = , or
( )1 , 22− , is a solution of the system of equations.
11. 3
1 32
x y
x y
− =⎧⎪⎨
+ =⎪⎩
Substituting the values of the variables, we obtain: 4 1 31 (4) 1 2 1 32
− =⎧⎪⎨
+ = + =⎪⎩
Each equation is satisfied, so 4, 1x y= = , or (4, 1) , is a solution of the system of equations.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
770 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12. 3
3 1x y
x y− =⎧
⎨− + =⎩
Substituting the values of the variables: ( ) ( )
( ) ( )2 5 2 5 3
3 2 5 6 5 1
− − − = − + =⎧⎪⎨− − + − = − =⎪⎩
Each equation is satisfied, so 2, 5x y= − = − , or ( 2, 5)− − , is a solution of the system of equations.
13. 3 3 2 4 0
2 3 8
x y zx y z
y z
+ + =⎧⎪ − − =⎨⎪ − = −⎩
Substituting the values of the variables: 3(1) 3( 1) 2(2) 3 3 4 4
1 ( 1) 2 1 1 2 0 2( 1) 3(2) 2 6 8
+ − + = − + =⎧⎪ − − − = + − =⎨⎪ − − = − − = −⎩
Each equation is satisfied, so 1, 1, 2x y z= = − = , or (1, 1, 2)− , is a solution of the system of equations.
14. 4 78 5 0
5 6
x zx y zx y z
− =⎧⎪ + − =⎨⎪− − + =⎩
Substituting the values of the variables: ( )( ) ( )
( ) ( )
4 2 1 8 1 7
8 2 5 3 1 16 15 1 0
2 3 5 1 2 3 5 6
− = − =⎧⎪
+ − − = − − =⎨⎪− − − + = − + + =⎩
Each equation is satisfied, so 2x = , 3y = − , 1z = , or (2, 3, 1)− , is a solution of the system of
equations.
15. 3 3 2 4
3 105 2 3 8
x y zx y z
x y z
+ + =⎧⎪ − + =⎨⎪ − − =⎩
Substituting the values of the variables: ( ) ( ) ( )
( )( ) ( ) ( )
3 2 3 2 2 2 6 6 4 4
2 3 2 2 2 6 2 10
5 2 2 2 3 2 10 4 6 8
+ − + = − + =⎧⎪
− − + = + + =⎨⎪ − − − = + − =⎩
Each equation is satisfied, so 2x = , 2y = − , 2z = , or (2, 2, 2)− is a solution of the system of
equations.
16. 4 5 6
5 176 5 24
x zy z
x y z
− =⎧⎪ − = −⎨⎪− − + =⎩
Substituting the values of the variables: ( ) ( )( ) ( )( ) ( ) ( )
4 4 5 2 16 10 6
5 3 2 15 2 17
4 6 3 5 2 4 18 10 24
− = − =⎧⎪
− − = − − = −⎨⎪− − − + = − + + =⎩
Each equation is satisfied, so 4x = , 3y = − , 2z = , or (4, 3, 2)− , is a solution of the system
of equations.
17. 84
x yx y+ =⎧
⎨ − =⎩
Solve the first equation for y, substitute into the second equation and solve:
84
y xx y= −⎧
⎨ − =⎩
(8 ) 48 4
2 126
x xx x
xx
− − =− + =
==
Since 6, 8 6 2x y= = − = . The solution of the system is 6, 2x y= = or using ordered pairs (6, 2) .
18. 2 7
3x yx y+ = −⎧
⎨ + = −⎩
Solve the first equation for x, substitute into the second equation and solve:
7 23
x yx y= − −⎧
⎨ + = −⎩
( 7 2 ) 37 3
4
y yy
y
− − + = −− − = −
− =
Since 4, 7 2( 4) 1y x= − = − − − = . The solution of the system is 1, 4x y= = − or using an ordered pair, (1, 4)− .
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.1: Systems of Linear Equations: Substitution and Elimination
771 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19. 5 212 3 12
x yx y− =⎧
⎨ + = −⎩
Multiply each side of the first equation by 3 and add the equations to eliminate y:
15 3 632 3 12x yx y− =⎧⎪
⎨ + = −⎪⎩
17 51
3x
x==
Substitute and solve for y: 5(3) 21
15 216
6
yyyy
− =− =− =
= −
The solution of the system is 3, 6x y= = − or
using ordered an pair ( )3, 6− .
20. 3 5
2 3 8x yx y+ =⎧
⎨ − = −⎩
Add the equations: 3 5
2 3 8x yx y+ =⎧⎪
⎨ − = −⎪⎩
3 31
xx= −= −
Substitute and solve for y: 1 3 5
3 62
yyy
− + ===
The solution of the system is 1, 2x y= − = or using ordered pairs ( 1, 2)− .
21. 3 24
2 0xx y
=⎧⎨ + =⎩
Solve the first equation for x and substitute into the second equation:
82 0
xx y
=⎧⎨ + =⎩
8 2 02 8
4
yyy
+ == −= −
The solution of the system is 8, 4x y= = − or using ordered pairs (8, 4)−
22. 4 5 3
2 8x y
y+ = −⎧
⎨ − = −⎩
Solve the second equation for y and substitute into the first equation:
4 5 34
x yy
+ = −⎧⎨ =⎩
4 5(4) 34 20 3
4 23234
xx
x
x
+ = −+ = −
= −
= −
The solution of the system is 23 , 44
x y= − = or
using ordered pairs 23 , 44
⎛ ⎞−⎜ ⎟⎝ ⎠
.
23. 3 6 25 4 1
x yx y− =⎧
⎨ + =⎩
Multiply each side of the first equation by 2 and each side of the second equation by 3, then add to eliminate y:
6 12 415 12 3
x yx y− =⎧⎪
⎨ + =⎪⎩
21 7
13
x
x
=
=
Substitute and solve for y: ( )3 1/ 3 6 2
1 6 26 1
16
yyy
y
− =
− =− =
= −
The solution of the system is 1 1,3 6
x y= = − or
using ordered pairs 1 1,3 6
⎛ ⎞−⎜ ⎟⎝ ⎠
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
772 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
24. 22 43
3 5 10
x y
x y
⎧ + =⎪⎨⎪ − = −⎩
Multiply each side of the first equation by 5 and each side of the second equation by 4, then add to eliminate y:
1010 203
12 20 40
x y
x y
⎧ + =⎪⎨⎪ − = −⎩
11022 353
x
x
=
= −
Substitute and solve for y: ( )3 5 / 3 5 10
5 5 105 5
1
yyyy
− − = −
− − = −− = −
=
The solution of the system is 5 , 13
x y= − = or
using ordered pairs 5 , 13
⎛ ⎞−⎜ ⎟⎝ ⎠
.
25. 2 14 2 3
x yx y+ =⎧
⎨ + =⎩
Solve the first equation for y, substitute into the second equation and solve:
1 24 2 3y xx y= −⎧
⎨ + =⎩
4 2(1 2 ) 3
4 2 4 30 1
x xx x+ − =+ − =
=
This equation is false, so the system is inconsistent.
26. 5
3 3 2x yx y− =⎧
⎨− + =⎩
Solve the first equation for x, substitute into the second equation and solve:
53 3 2
x yx y= +⎧
⎨− + =⎩
3( 5) 3 23 15 3 2
0 17
y yy y
− + + =− − + =
=
This equation is false, so the system is inconsistent.
27. 2 04 2 12
x yx y− =⎧
⎨ + =⎩
Solve the first equation for y, substitute into the second equation and solve:
24 2 12y xx y=⎧
⎨ + =⎩
4 2(2 ) 124 4 12
8 1232
x xx x
x
x
+ =+ =
=
=
Since 3 3, 2 32 2
x y ⎛ ⎞= = =⎜ ⎟⎝ ⎠
The solution of the system is 3 , 32
x y= = or
using ordered pairs 3 ,3 .2
⎛ ⎞⎜ ⎟⎝ ⎠
28. 3 3 1
843
x y
x y
+ = −⎧⎪⎨ + =⎪⎩
Solve the second equation for y, substitute into the first equation and solve:
3 3 18 43
x y
y x
+ = −⎧⎪⎨ = −⎪⎩
83 3 4 13
3 8 12 19 9
1
x x
x xxx
⎛ ⎞+ − = −⎜ ⎟⎝ ⎠+ − = −
− = −=
Since 8 8 41, 4(1) 4 .3 3 3
x y= = − = − = −
The solution of the system is 41,3
x y= = − or
using ordered pairs 41,3
⎛ ⎞−⎜ ⎟⎝ ⎠
.
29. 2 4
2 4 8x yx y+ =⎧
⎨ + =⎩
Solve the first equation for x, substitute into the second equation and solve:
4 22 4 8x y
x y= −⎧
⎨ + =⎩
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.1: Systems of Linear Equations: Substitution and Elimination
773 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2(4 2 ) 4 88 4 4 8
0 0
y yy y
− + =− + =
=
These equations are dependent. The solution of the system is either 4 2x y= − , where y is any real
number or 42
xy −= , where x is any real number.
Using ordered pairs, we write the solution as { }( , ) 4 2 , is any real numberx y x y y= − or as
4( , ) , is any real number2
xx y y x⎧ − ⎫=⎨ ⎬
⎩ ⎭.
30. 3 79 3 21
x yx y− =⎧
⎨ − =⎩
Solve the first equation for y, substitute into the second equation and solve:
3 79 3 21y xx y= −⎧
⎨ − =⎩
9 3(3 7) 219 9 21 21
0 0
x xx x− − =− + =
=
These equations are dependent. The solution of the system is either 3 7y x= − , where x is any real
number is 73
yx += , where y is any real number.
Using ordered pairs, we write the solution as { }( , ) 3 7, is any real numberx y y x x= − or as
7( , ) , is any real number3
yx y x y⎧ + ⎫=⎨ ⎬
⎩ ⎭.
31. 2 3 1
10 11x yx y− = −⎧
⎨ + =⎩
Multiply each side of the first equation by –5, and add the equations to eliminate x:
10 15 5 10 11
x yx y
− + =⎧⎪⎨ + =⎪⎩
16 16
1yy==
Substitute and solve for x: 2 3(1) 1
2 3 12 2
1
xx
xx
− = −− = −
==
The solution of the system is 1, 1x y= = or using ordered pairs (1, 1).
32. 3 2 0
5 10 4x y
x y− =⎧
⎨ + =⎩
Multiply each side of the first equation by 5, and add the equations to eliminate y:
15 10 05 10 4
x yx y− =⎧⎪
⎨ + =⎪⎩
20 4
15
x
x
=
=
Substitute and solve for y: ( )5 1/ 5 10 4
1 10 410 3
310
yyy
y
+ =
+ ==
=
The solution of the system is 1 3,5 10
x y= = or
using ordered pairs 1 3,5 10
⎛ ⎞⎜ ⎟⎝ ⎠
.
33. 2 3 6
12
x y
x y
+ =⎧⎪⎨
− =⎪⎩
Solve the second equation for x, substitute into the first equation and solve:
2 3 612
x y
x y
+ =⎧⎪⎨
= +⎪⎩
12 3 62
2 1 3 65 5
1
y y
y yyy
⎛ ⎞+ + =⎜ ⎟⎝ ⎠
+ + ===
Since 1 31, 12 2
y x= = + = . The solution of the
system is 3 , 12
x y= = or using ordered pairs
3 , 12
⎛ ⎞⎜ ⎟⎝ ⎠
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
774 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
34. 1 22
2 8
x y
x y
⎧ + = −⎪⎨⎪ − =⎩
Solve the second equation for x, substitute into the first equation and solve:
1 22
2 8
x y
x y
⎧ + = −⎪⎨⎪ = +⎩
1 (2 8) 22
4 22 6
3
y y
y yyy
+ + = −
+ + = −= −= −
Since 3, 2( 3) 8 6 8 2y x= − = − + = − + = . The solution of the system is 2, 3x y= = − or using ordered pairs (2, 3)− .
35.
1 1 32 31 2 14 3
x y
x y
⎧ + =⎪⎪⎨⎪ − = −⎪⎩
Multiply each side of the first equation by –6 and each side of the second equation by 12, then add to eliminate x:
3 2 183 8 12
x yx y
− − = −⎧⎪⎨ − = −⎪⎩
10 303
yy
− = −=
Substitute and solve for x: 1 1 (3) 32 3
1 1 32
1 22
4
x
x
x
x
+ =
+ =
=
=
The solution of the system is 4, 3x y= = or using ordered pairs (4, 3).
36.
1 3 53 23 1 114 3
x y
x y
⎧ − = −⎪⎪⎨⎪ + =⎪⎩
Multiply each side of the first equation by –54 and each side of the second equation by 24, then add to eliminate x:
18 81 27018 8 264
x yx y
− + =⎧⎪⎨ + =⎪⎩
89 5346
yy==
Substitute and solve for x: 3 1 (6) 114 3
3 2 114
3 94
12
x
x
x
x
+ =
+ =
=
=
The solution of the system is 12, 6x y= = or using ordered pairs (12, 6).
37. 3 5 3
15 5 21x yx y− =⎧
⎨ + =⎩
Add the equations to eliminate y: 3 5 3
15 5 21x yx y− =⎧⎪
⎨ + =⎪⎩
18 2443
x
x
=
=
Substitute and solve for y: ( )3 4 / 3 5 3
4 5 35 1
15
yyy
y
− =
− =− = −
=
The solution of the system is 4 1,3 5
x y= = or
using ordered pairs 4 1,3 5
⎛ ⎞⎜ ⎟⎝ ⎠
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.1: Systems of Linear Equations: Substitution and Elimination
775 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
38. 2 1
1 32 2
x y
x y
− = −⎧⎪⎨
+ =⎪⎩
Multiply each side of the second equation by 2, and add the equations to eliminate y:
2 12 3
x yx y− = −⎧⎪
⎨ + =⎪⎩
4 212
x
x
=
=
Substitute and solve for y: 12 12
1 12
2
y
yyy
⎛ ⎞ − = −⎜ ⎟⎝ ⎠
− = −− = −
=
The solution of the system is 1 , 22
x y= = or
using ordered pairs 1 , 22
⎛ ⎞⎜ ⎟⎝ ⎠
.
39.
1 1 8
3 5 0
x y
x y
⎧ + =⎪⎪⎨⎪ − =⎪⎩
Rewrite letting 1 1,u vx y
= = :
83 5 0
u vu v+ =⎧
⎨ − =⎩
Solve the first equation for u, substitute into the second equation and solve:
83 5 0u vu v= −⎧
⎨ − =⎩
3(8 ) 5 024 3 5 0
8 243
v vv v
vv
− − =− − =
− = −=
Since 3, 8 3 5v u= = − = . Thus, 1 15
xu
= = ,
1 13
yv
= = . The solution of the system is
1 1,5 3
x y= = or using ordered pairs 1 1,5 3
⎛ ⎞⎜ ⎟⎝ ⎠
.
40.
4 3 0
6 3 22
x y
x y
⎧ − =⎪⎪⎨⎪ + =⎪⎩
Rewrite letting 1 1,u vx y
= = :
4 3 036 22
u v
u v
− =⎧⎪⎨
+ =⎪⎩
Multiply each side of the second equation by 2, and add the equations to eliminate v:
4 3 012 3 4
u vu v− =⎧
⎨ + =⎩
16 4
4 116 4
u
u
=
= =
Substitute and solve for v: 14 3 04
1 3 03 1
13
v
vv
v
⎛ ⎞ − =⎜ ⎟⎝ ⎠
− =− = −
=
Thus, 1 14, 3x yu v
= = = = . The solution of the
system is 4, 3x y= = or using ordered pairs (4, 3).
41. 6
2 3 162 4
x yx zy z
− =⎧⎪ − =⎨⎪ + =⎩
Multiply each side of the first equation by –2 and add to the second equation to eliminate x:
2 2 122 3 16
2 3 4
x yx z
y z
− + = −− =
− =
Multiply each side of the result by –1 and add to the original third equation to eliminate y:
2 3 42 4
4 00
y zy z
zz
− + = −+ =
==
Substituting and solving for the other variables:
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
776 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 0 4
2 42
yyy
+ ===
2 3(0) 16
2 168
xxx
− ===
The solution is 8, 2, 0x y z= = = or using ordered triples (8, 2, 0).
42. 2 42 4 03 2 11
x yy zx z
+ = −⎧⎪− + =⎨⎪ − = −⎩
Multiply each side of the first equation by 2 and add to the second equation to eliminate y: 4 2 8
2 4 0
4 4 8
x yy z
x z
+ = −− + =
+ = −
Multiply each side of the result by 12
and add to
the original third equation to eliminate z: 2 2 43 2 115 15
3
x zx zx
x
+ = −− = −
= −= −
Substituting and solving for the other variables: 2( 3) 4
6 42
yyy
− + = −− + = −
=
3( 3) 2 119 2 11
2 21
zzzz
− − = −− − = −
− = −=
The solution is 3, 2, 1x y z= − = = or using ordered triples ( 3, 2, 1)− .
43. 2 3 7
2 43 2 2 10
x y zx y zx y z
− + =⎧⎪ + + =⎨⎪− + − = −⎩
Multiply each side of the first equation by –2 and add to the second equation to eliminate x; and multiply each side of the first equation by 3 and add to the third equation to eliminate x:
2 4 6 142 4
5 5 10
x y zx y z
y z
− + − = −+ + =
− = −
3 6 9 21
3 2 2 10
4 7 11
x y zx y z
y z
− + =− + − = −
− + =
Multiply each side of the first result by 45
and
add to the second result to eliminate y:
4 4 84 7 11
y zy z− = −
− + =
3 3
1zz==
Substituting and solving for the other variables:
1 2
1y
y− = −
= −
2( 1) 3(1) 72 3 7
2
xx
x
− − + =+ + =
=
The solution is 2, 1, 1x y z= = − = or using ordered triples (2, 1, 1)− .
44. 2 3 02 2 73 4 3 7
x y zx y zx y z
+ − =⎧⎪− + + = −⎨⎪ − − =⎩
Multiply each side of the first equation by –2 and add to the second equation to eliminate y; and multiply each side of the first equation by 4 and add to the third equation to eliminate y:
4 2 6 02 2 7
6 7 7
x y zx y z
x z
− − + =− + + = −
− + = −
8 4 12 03 4 3 7
11 15 7
x y zx y z
x z
+ − =− − =
− =
Multiply each side of the first result by 11 and multiply each side of the second result by 6 to eliminate x:
66 77 7766 90 42
x zx z
− + = −− =
13 353513
z
z
− = −
=
Substituting and solving for the other variables: 356 7 7132456 713
336613
5613
x
x
x
x
⎛ ⎞− + = −⎜ ⎟⎝ ⎠
− + = −
− = −
=
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.1: Systems of Linear Equations: Substitution and Elimination
777 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
56 352 3 013 13
112 105 013 13
713
y
y
y
⎛ ⎞ ⎛ ⎞+ − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+ − =
= −
The solution is 56 7 35, ,13 13 13
x y z= = − = or
using ordered triples 56 7 35, ,13 13 13⎛ ⎞−⎜ ⎟⎝ ⎠
.
45. 1
2 3 23 2 0
x y zx y zx y
− − =⎧⎪ + + =⎨⎪ + =⎩
Add the first and second equations to eliminate z: 12 3 2
3 2 3
x y zx y z
x y
− − =+ + =
+ =
Multiply each side of the result by –1 and add to the original third equation to eliminate y:
3 2 33 2 0
0 3
x yx y
− − = −+ =
= −
This equation is false, so the system is inconsistent.
46. 2 3 0
2 53 4 1
x y zx y zx y z
− − =⎧⎪− + + =⎨⎪ − − =⎩
Add the first and second equations to eliminate z; then add the second and third equations to eliminate z: 2 3 0
2 5
5
x y zx y z
x y
− − =− + + =
− =
2 5
3 4 1
2 2 6
x y zx y z
x y
− + + =− − =
− =
Multiply each side of the first result by –2 and add to the second result to eliminate y:
2 2 102 2 6
x yx y
− + = −− =
0 2= − This equation is false, so the system is inconsistent.
47. 1
2 3 43 2 7 0
x y zx y zx y z
− − =⎧⎪− + − = −⎨⎪ − − =⎩
Add the first and second equations to eliminate x; multiply the first equation by –3 and add to the third equation to eliminate x:
12 3 4
4 3
x y zx y z
y z
− − =− + − = −
− = −
3 3 3 33 2 7 0
4 3
x y zx y z
y z
− + + = −− − =
− = −
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 34 3
0 0
y zy z
− + =− = −
=
The system is dependent. If z is any real number, then 4 3y z= − . Solving for x in terms of z in the first equation:
(4 3) 14 3 1
5 3 15 2
x z zx z z
x zx z
− − − =− + − =
− + == −
The solution is {( , , )x y z 5 2,x z= − 4 3y z= − , z is any real number}.
48. 2 3 03 2 2 2
5 3 2
x y zx y zx y z
− − =⎧⎪ + + =⎨⎪ + + =⎩
Multiply the first equation by 2 and add to the second equation to eliminate z; multiply the first equation by 3 and add to the third equation to eliminate z: 4 6 2 03 2 2 2
7 4 2
x y zx y z
x y
− − =+ + =
− =
6 9 3 0
5 3 2
7 4 2
x y zx y z
x y
− − =+ + =
− =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
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7 4 27 4 2
0 0
x yx y
− + = −− =
=
The system is dependent. If y is any real
number, then 4 27 7
x y= + .
Solving for z in terms of x in the first equation: 2 3
4 22 37
8 4 217
13 47
z x yy y
y y
y
= −
+⎛ ⎞= −⎜ ⎟⎝ ⎠+ −
=
− +=
The solution is 4 2( , , )7 7
x y z x y⎧= +⎨
⎩,
13 4 , is any real number7 7
z y y ⎫= − + ⎬⎭
.
49. 2 2 3 64 3 2 02 3 7 1
x y zx y zx y z
− + =⎧⎪ − + =⎨⎪− + − =⎩
Multiply the first equation by –2 and add to the second equation to eliminate x; add the first and third equations to eliminate x:
4 4 6 124 3 2 0
4 12
x y zx y z
y z
− + − = −− + =
− = −
2 2 3 6
2 3 7 1
4 7
x y zx y z
y z
− + =− + − =
− =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 12 4 7
0 19
y zy z
− + =− =
=
This result is false, so the system is inconsistent.
50. 3 2 2 67 3 2 12 3 4 0
x y zx y zx y z
− + =⎧⎪ − + = −⎨⎪ − + =⎩
Multiply the first equation by –1 and add to the second equation to eliminate z; multiply the first equation by –2 and add to the third equation to eliminate z:
3 2 2 67 3 2 1
4 7
x y zx y z
x y
− + − = −− + = −
− = −
6 4 4 122 3 4 0
4 12
x y zx y z
x y
− + − = −− + =
− + = −
Add the first result to the second result to eliminate y:
4 74 12
0 19
x yx y− = −
− + = −
= −
This result is false, so the system is inconsistent.
51. 6
3 2 53 2 14
x y zx y zx y z
+ − =⎧⎪ − + = −⎨⎪ + − =⎩
Add the first and second equations to eliminate z; multiply the second equation by 2 and add to the third equation to eliminate z:
63 2 5
4 1
x y zx y z
x y
+ − =− + = −
− =
6 4 2 10
3 2 14
7 4
x y zx y z
x y
− + = −+ − =
− =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 17 4
x yx y
− + = −− =
3 31
xx==
Substituting and solving for the other variables: 4(1) 1
33
yyy
− =− = −
=
3(1) 2(3) 53 6 5
2
zzz
− + = −− + = −
= −
The solution is 1, 3, 2x y z= = = − or using ordered triplets (1, 3, 2)− .
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.1: Systems of Linear Equations: Substitution and Elimination
779 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
52. 4
2 3 4 155 2 12
x y zx y zx y z
− + = −⎧⎪ − + = −⎨⎪ + − =⎩
Multiply the first equation by –3 and add to the second equation to eliminate y; add the first and third equations to eliminate y:
3 3 3 122 3 4 15x y zx y z
− + − =− + = −
33
x zz x
− + = −= −
45 2 12
6 8
x y zx y z
x z
− + = −+ − =
− =
Substitute and solve: 6 ( 3) 8
6 3 85 5
1
x xx x
xx
− − =− + =
==
3 1 3 2z x= − = − = − 12 5 2 12 5(1) 2( 2) 3y x z= − + = − + − =
The solution is 1, 3, 2x y z= = = − or using ordered triplets (1, 3, 2)− .
53. 2 3
2 4 72 2 3 4
x y zx y zx y z
+ − = −⎧⎪ − + = −⎨⎪− + − =⎩
Add the first and second equations to eliminate z; multiply the second equation by 3 and add to the third equation to eliminate z: 2 32 4 73 2 10
x y zx y zx y
+ − = −− + = −− = −
6 12 3 212 2 3 44 10 17
x y zx y zx y
− + = −− + − =
− = −
Multiply each side of the first result by –5 and add to the second result to eliminate y:
15 10 504 10 17
x yx y
− + =− = −
11 333
xx
− == −
Substituting and solving for the other variables: 3( 3) 2 10
9 2 102 1
12
yyy
y
− − = −− − = −
− = −
=
13 2 32
3 1 31
1
z
zzz
⎛ ⎞− + − = −⎜ ⎟⎝ ⎠
− + − = −− = −
=
The solution is 13, , 12
x y z= − = = or using
ordered triplets 13, , 12
⎛ ⎞−⎜ ⎟⎝ ⎠
.
54. 4 3 8
3 3 126 1
x y zx y zx y z
+ − = −⎧⎪ − + =⎨⎪ + + =⎩
Add the first and second equations to eliminate z; multiply the first equation by 2 and add to the third equation to eliminate z:
4 3 83 3 12
4 3 4
x y zx y z
x y
+ − = −− + =
+ =
2 8 6 16
6 1
3 9 15
x y zx y z
x y
+ − = −+ + =
+ = −
Multiply each side of the second result by 1/ 3− and add to the first result to eliminate y: 4 3 4
3 5x yx y+ =
− − =
3 93
xx==
Substituting and solving for the other variables: 3 3 5
3 883
yy
y
+ = −= −
= −
83 6 13
26319
z
z
z
⎛ ⎞+ − + =⎜ ⎟⎝ ⎠
=
=
The solution is 8 13, ,3 9
x y z= = − = or using
ordered triplets 8 13, ,3 9
⎛ ⎞−⎜ ⎟⎝ ⎠
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
780 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
55. Let l be the length of the rectangle and w be the width of the rectangle. Then:
2l w= and 2 2 90l w+ = Solve by substitution:
2(2 ) 2 904 2 90
6 9015 feet2(15) 30 feet
w ww w
wwl
+ =+ =
=== =
The floor is 15 feet by 30 feet.
56. Let l be the length of the rectangle and w be the width of the rectangle. Then:
50l w= + and 2 2 3000l w+ = Solve by substitution:
2( 50) 2 30002 100 2 3000
4 2900725 meters725 50 775 meters
w ww w
wwl
+ + =+ + =
=== + =
The dimensions of the field are 775 meters by 725 meters.
57. Let x = the number of commercial launches and y = the number of noncommercial launches.
Then: 55x y+ = and 2 1y x= + Solve by substitution:
(2 1) 553 54
18
x xxx
+ + ===
2(18) 136 137
yyy
= += +=
In 2005 there were 18 commercial launches and 37 noncommercial launches.
58. Let x = the number of adult tickets sold and y = the number of senior tickets sold. Then:
3259 7 2495
x yx y+ =⎧
⎨ + =⎩
Solve the first equation for y: 325y x= − Solve by substitution: 9 7(325 ) 24959 2275 7 2495
2 220110
x xx x
xx
+ − =+ − =
==
325 110 215y = − = There were 110 adult tickets sold and 215 senior citizen tickets sold.
59. Let x = the number of pounds of cashews. Let y = is the number of pounds in the mixture. The value of the cashews is 5x . The value of the peanuts is 1.50(30) = 45. The value of the mixture is 3y . Then 30x y+ = represents the amount of mixture. 5 45 3x y+ = represents the value of the mixture.
Solve by substitution: 5 45 3( 30)
2 4522.5
x xxx
+ = +==
So, 22.5 pounds of cashews should be used in the mixture.
60. Let x = the amount invested in AA bonds. Let y = the amount invested in the Bank Certificate.
a. Then 150,000x y+ = represents the total investment. 0.10 0.05 12,000x y+ = represents the earnings on the investment.
Solve by substitution: 0.10(150,000 ) 0.05 12,000
15,000 0.10 0.05 12,0000.05 3000
60,000
y yy y
yy
− + =− + =
− = −=
150,000 60,000 90,000x = − = Thus, $90,000 should be invested in AA Bonds and $60,000 in a Bank Certificate.
b. Then 150,000x y+ = represents the total investment. 0.10 0.05 14,000x y+ = represents the earnings on the investment.
Solve by substitution: 0.10(150,000 ) 0.05 14,000
15,000 0.10 0.05 14,0000.05 1000
20,000
y yy y
yy
− + =− + =
− = −=
150,000 20,000 130,000x = − = Thus, $130,000 should be invested in AA Bonds and $20,000 in a Bank Certificate.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.1: Systems of Linear Equations: Substitution and Elimination
781 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
61. Let x = the plane’s average airspeed and y = the average wind speed.
Rate Time DistanceWith Wind 3 600Against 4 600
x yx y+−
( )(3) 600( )(4) 600x yx y+ =⎧
⎨ − =⎩
Multiply each side of the first equation by 13
,
multiply each side of the second equation by 14
,
and add the result to eliminate y 200150
x yx y+ =− =
2 350175
xx==
175 20025
yy
+ ==
The average airspeed of the plane is 175 mph, and the average wind speed is 25 mph.
62. Let x = the average wind speed and y = the distance.
Rate Time Distance
With Wind 150 2Against 150 3
x yx y
+−
(150 )(2)(150 )(3)
x yx y
+ =⎧⎨ − =⎩
Solve by substitution: (150 )(2) (150 )(3)
300 2 450 35 150
30
x xx xxx
+ = −+ = −
==
Thus, the average wind speed is 30 mph.
63. Let x = the number of $25-design. Let y = the number of $45-design. Then x y+ = the total number of sets of dishes. 25 45x y+ = the cost of the dishes.
Setting up the equations and solving by substitution:
20025 45 7400
x yx y+ =⎧
⎨ + =⎩
Solve the first equation for y, the solve by substitution: 200y x= −
25 45(200 ) 740025 9000 45 7400
20 160080
x xx x
xx
+ − =+ − =
− = −=
200 80 120y = − = Thus, 80 sets of the $25 dishes and 120 sets of the $45 dishes should be ordered.
64. Let x = the cost of a hot dog. Let y = the cost of a soft drink. Setting up the equations and solving by substitution:
10 5 35.007 4 25.25
x yx y+ =⎧
⎨ + =⎩
10 5 35.00
2 77 2
x yx y
y x
+ =+ =
= −
7 4(7 2 ) 25.257 28 8 25.25
2.752.75
x xx x
xx
+ − =+ − =
− = −=
7 2(2.75) 1.50y = − = A single hot dog costs $2.75 and a single soft drink costs $1.50.
65. Let x = the cost per package of bacon. Let y = the cost of a carton of eggs. Set up a system of equations for the problem:
3 2 13.452 3 11.45x yx y+ =⎧
⎨ + =⎩
Multiply each side of the first equation by 3 and each side of the second equation by –2 and solve by elimination:
9 6 40.354 6 22.90
x yx y+ =
− − = −
5 17.453.49
xx==
Substitute and solve for y: 3(3.49) 2 13.45
10.47 2 13.452 2.98
1.49
yyyy
+ =+ =
==
A package of bacon costs $3.49 and a carton of eggs cost $1.49. The refund for 2 packages of bacon and 2 cartons of eggs will be 2($3.49) + 2($1.49) = $9.96.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
782 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
66. Let x = Pamela’s average speed in still water. Let y = the speed of the current.
Rate Time Distance
Downstream 3 15Upstream 5 15
x yx y+−
Set up a system of equations for the problem:
3( ) 155( ) 15
x yx y+ =⎧
⎨ − =⎩
Multiply each side of the first equation by 13
,
multiply each side of the second equation by 15
,
and add the result to eliminate y:
53
x yx y+ =− =
2 84
xx==
4 51
yy
+ ==
Pamela's average speed is 4 miles per hour and the speed of the current is 1 mile per hour.
67. Let x = the # of mg of compound 1. Let y = the # of mg of compound 2. Setting up the equations and solving by substitution:
0.2 0.4 40 vitamin C0.3 0.2 30 vitamin D
x yx y+ =⎧
⎨ + =⎩
Multiplying each equation by 10 yields 2 4 4006 4 600
x yx y+ =⎧
⎨ + =⎩
Subtracting the bottom equation from the top equation yields
( )2 4 6 4 400 6002 6 200
4 20050
x y x yx x
xx
+ − + = −
− = −− = −
=
( )2 50 4 400100 4 400
4 300300 75
4
yyy
y
+ =
+ ==
= =
So 50 mg of compound 1 should be mixed with 75 mg of compound 2.
68. Let x = the # of units of powder 1. Let y = the # of units of powder 2. Setting up the equations and solving by substitution:
120.2 0.4 12 vitamin B0.3 0.2 12 vitamin E
x yx y+ =⎧
⎨ + =⎩
Multiplying each equation by 10 yields
2 4 1206 4 240
x yx y+ =⎧
⎨ + =⎩
Subtracting the bottom equation from the top equation yields
( )2 4 6 4 120 2404 120
30
x y x yxx
+ − + = −
− = −=
( )2 30 4 12060 4 120
4 6060 154
yyy
y
+ =
+ ==
= =
So 30 units of powder 1 should be mixed with 15 units of powder 2.
69. 2y ax bx c= + + At (–1, 4) the equation becomes:
24 (–1) ( 1)4
a b ca b c
= + − += − +
At (2, 3) the equation becomes: 23 (2) (2)
3 4 2a b c
a b c= + += + +
At (0, 1) the equation becomes: 21 (0) (0)
1a b cc
= + +=
The system of equations is: 4
4 2 31
a b ca b c
c
− + =⎧⎪ + + =⎨⎪ =⎩
Substitute 1c = into the first and second equations and simplify:
1 43
3
a ba b
a b
− + =− =
= +
4 2 1 34 2 2
a ba b+ + =
+ =
Solve the first result for a, substitute into the second result and solve:
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.1: Systems of Linear Equations: Substitution and Elimination
783 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4( 3) 2 24 12 2 2
6 1053
b bb b
b
b
+ + =+ + =
= −
= −
5 433 3
a = − + =
The solution is 4 5, , 13 3
a b c= = − = . The
equation is 24 5 13 3
y x x= − + .
70. 2y ax bx c= + + At (–1, –2) the equation becomes:
22 ( 1) ( 1)2
a b ca b c
− = − + − +− + = −
At (1, –4) the equation becomes: 24 (1) (1)
4a b c
a b c− = + +
+ + = −
At (2, 4) the equation becomes: 24 (2) (2)
4 2 4a b c
a b c= + +
+ + =
The system of equations is: 2
– 44 2 4
a b ca b ca b c
− + = −⎧⎪ + + =⎨⎪ + + =⎩
Multiply the first equation by –1 and add to the second equation; multiply the first equation by –1 and add to the third equation to eliminate c:
2
– 4a b ca b c
− + − =⎧⎪⎨ + + =⎪⎩
2
4 2 4a b ca b c− + − =
+ + =
2 21
bb= −= −
3 3 6
2a b
a b+ =
+ =
Substitute and solve: ( 1) 2
3a
a+ − =
= 4
3 ( 1) 46
c a b= − − −= − − − −= −
The solution is 3, 1, 6a b c= = − = − . The
equation is 23 6y x x= − −
71. 0.06 5000 2400.06 6000 900
Y rY r− =⎧
⎨ + =⎩
Multiply the first equation by 1− , the add the result to the second equation to eliminate Y.
0.06 5000 240 0.06 6000 900
Y rY r
− + = −+ =
11000 6600.06
rr==
Substitute this result into the first equation to find Y. 0.06 5000(0.06) 240
0.06 300 2400.06 540
9000
YY
YY
− =− =
==
The equilibrium level of income and interest rates is $9000 million and 6%.
72. 0.05 1000 100.05 800 100
Y rY r− =⎧
⎨ + =⎩
Multiply the first equation by 1− , the add the result to the second equation to eliminate Y.
0.05 1000 10 0.05 800 100
Y rY r
− + = −+ =
1800 900.05
rr==
Substitute this result into the first equation to find Y. 0.05 1000(0.05) 10
0.05 50 100.05 60
1200
YY
YY
− =− =
==
The equilibrium level of income and interest rates is $1200 million and 5%.
73. 2 1 3
1 2
2 3
5 3 5 010 5 7 0
I I II II I
= +⎧⎪ − − =⎨⎪ − − =⎩
Substitute the expression for 2I into the second and third equations and simplify:
1 1 3
1 3
5 3 5( ) 08 5 5
I I II I
− − + =
− − = −
1 3 3
1 3
10 5( ) 7 05 12 10
I I II I
− + − =
− − = −
Multiply both sides of the first result by 5 and multiply both sides of the second result by –8 to eliminate 1I :
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
784 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1 3
1 3
40 25 2540 96 80
I II I
− − = −
+ =
3
3
71 555571
I
I
=
=
Substituting and solving for the other variables:
1
1
1
1
558 5 5712758 571
80871
1071
I
I
I
I
⎛ ⎞− − = −⎜ ⎟⎝ ⎠
− − = −
− = −
=
210 55 6571 71 71
I ⎛ ⎞= + =⎜ ⎟⎝ ⎠
The solution is 1 2 310 65 55, ,71 71 71
I I I= = = .
74. 3 1 2
3 2
1 2
8 4 68 4 6
I I II I
I I
= +⎧⎪ = +⎨⎪ = +⎩
Substitute the expression for 3I into the second equation and simplify:
1 2 2
1 2
1 2
8 4( ) 68 4 104 10 8
I I II I
I I
= + += ++ =
1 2
1 2
8 4 68 6 4
I II I= +− =
Multiply both sides of the first result by –2 and add to the second result to eliminate 1I :
1 2
1 2
8 20 16 8 6 4
I II I
− − = −− =
2
2
26 1212 626 13
I
I
− = −−
= =−
Substituting and solving for the other variables:
1
1
1
1
64 10 813604 813
444131113
I
I
I
I
⎛ ⎞+ =⎜ ⎟⎝ ⎠
+ =
=
=
3 1 211 6 1713 13 13
I I I= + = + =
The solution is 1 2 311 6 17, ,13 13 13
I I I= = = .
75. Let x = the number of orchestra seats. Let y = the number of main seats. Let z = the number of balcony seats. Since the total number of seats is 500,
500x y z+ + = . Since the total revenue is $17,100 if all seats are sold, 50 35 25 17,100x y z+ + = . If only half of the orchestra seats are sold, the revenue is $14,600.
So, 150 35 25 14,6002
x y z⎛ ⎞ + + =⎜ ⎟⎝ ⎠
.
Thus, we have the following system: 500
50 35 25 17,10025 35 25 14,600
x y zx y zx y z
+ + =⎧⎪ + + =⎨⎪ + + =⎩
Multiply each side of the first equation by –25 and add to the second equation to eliminate z; multiply each side of the third equation by –1 and add to the second equation to eliminate z:
25 25 25 12,50050 35 25 17,100
25 10 4600
x y zx y z
x y
− − − = −+ + =
+ =
50 35 25 17,100
25 35 25 14,600x y zx y z+ + =
− − − = −
25 2500
100x
x==
Substituting and solving for the other variables: 25(100) 10 4600
2500 10 460010 2100
210
yyyy
+ =+ =
==
100 210 500310 500
190
zzz
+ + =+ =
=
There are 100 orchestra seats, 210 main seats, and 190 balcony seats.
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Section 8.1: Systems of Linear Equations: Substitution and Elimination
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76. Let x = the number of adult tickets. Let y = the number of child tickets. Let z = the number of senior citizen tickets. Since the total number of tickets is 405,
405x y z+ + = . Since the total revenue is $2320, 8 4.50 6 2320x y z+ + = . Twice as many children's tickets as adult tickets are sold. So, 2y x= . Thus, we have the following system:
4058 4.50 6 2320
2
x y zx y z
y x
+ + =⎧⎪ + + =⎨⎪ =⎩
Substitute for y in the first two equations and simplify:
(2 ) 4053 405
x x zx z
+ + =+ =
8 4.50(2 ) 6 232017 6 2320
x x zx z
+ + =+ =
Multiply the first result by –6 and add to the second result to eliminate z:
18 6 243017 6 2320
x zx z
− − = −⎧⎪⎨ + =⎪⎩
110
110x
x− = −
=
22(110)220
y x===
3 4053(110) 405
330 40575
x zzzz
+ =+ =+ =
=
There were 110 adults, 220 children, and 75 senior citizens that bought tickets.
77. Let x = the number of servings of chicken. Let y = the number of servings of corn. Let z = the number of servings of 2% milk.
Protein equation: 30 3 9 66x y z+ + = Carbohydrate equation: 35 16 13 94.5x y z+ + = Calcium equation: 200 10 300 910x y z+ + =
Multiply each side of the first equation by –16 and multiply each side of the second equation by 3 and add them to eliminate y; multiply each side of the second equation by –5 and multiply each side of the third equation by 8 and add to eliminate y:
480 48 144 1056 105 48 39 283.5
375 105 772.5
x y zx y z
x z
− − − = −+ + =
− − = −
175 80 65 472.5
1600 80 2400 7280
1425 2335 6807.5
x y zx y z
x z
− − − = −+ + =
+ =
Multiply each side of the first result by 19 and multiply each side of the second result by 5 to eliminate x:
7125 1995 14,677.57125 11,675 34,037.5
x zx z
− − = −+ =
9680 19,3602
zz==
Substituting and solving for the other variables: 375 105(2) 772.5
375 210 772.5375 562.5
1.5
xx
xx
− − = −− − = −
− = −=
30(1.5) 3 9(2) 6645 3 18 66
3 31
yy
yy
+ + =+ + =
==
The dietitian should serve 1.5 servings of chicken, 1 serving of corn, and 2 servings of 2% milk.
78. Let x = the amount in Treasury bills. Let y = the amount in Treasury bonds. Let z = the amount in corporate bonds.
Since the total investment is $20,000, 20,000x y z+ + =
Since the total income is to be $1390, 0.05 0.07 0.10 1390x y z+ + =
The investment in Treasury bills is to be $3000 more than the investment in corporate bonds. So, 3000x z= +
Substitute for x in the first two equations and simplify: (3000 ) 20,000
2 17,000z y z
y z+ + + =
+ =
5(3000 ) 7 10 139,0007 15 124,000
z y zy z
+ + + =+ =
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Chapter 8: Systems of Equations and Inequalities
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Multiply each side of the first result by –7 and add to the second result to eliminate y:
7 14 119,0007 15 124,000
5,000
y zy z
z
− − = −+ =
=
3000 3000 5000 8000x z= + = + = 2 17,000
2(5000) 17,00010,000 17,000
7000
y zyy
y
+ =+ =+ =
=
Kelly should invest $8000 in Treasury bills, $7000 in Treasury bonds, and $5000 in corporate bonds.
79. Let x = the price of 1 hamburger. Let y = the price of 1 order of fries. Let z = the price of 1 drink.
We can construct the system 8 6 6 26.1010 6 8 31.60
x y zx y z+ + =⎧
⎨ + + =⎩
A system involving only 2 equations that contain 3 or more unknowns cannot be solved uniquely.
Multiply the first equation by 12
− and the
second equation by 12
, then add to eliminate y:
4 3 3 13.05 5 3 4 15.80
x y zx y z
− − − = −+ + =
2.75
2.75x z
x z+ =
= −
Substitute and solve for y in terms of z: ( )5 2.75 3 4 15.80
13.75 3 15.803 2.05
1 413 60
z y zy z
y z
y z
− + + =
+ − == +
= +
Solutions of the system are: 2.75x z= − , 1 413 60
y z= + .
Since we are given that 0.60 0.90z≤ ≤ , we choose values of z that give two-decimal-place values of x and y with 1.75 2.25x≤ ≤ and 0.75 1.00y≤ ≤ . The possible values of x, y, and z are shown in the table.
x y z 2.13 0.89 0.62 2.10 0.90 0.65 2.07 0.91 0.68 2.04 0.92 0.71 2.01 0.93 0.74 1.98 0.94 0.77 1.95 0.95 0.80 1.92 0.96 0.83 1.89 0.97 0.86 1.86 0.98 0.89
80. Let x = the price of 1 hamburger.
Let y = the price of 1 order of fries. Let z = the price of 1 drink
We can construct the system 8 6 6 26.1010 6 8 31.603 2 4 10.95
x y zx y zx y z
+ + =⎧⎪ + + =⎨⎪ + + =⎩
Subtract the second equation from the first equation to eliminate y: 8 6 6 26.1010 6 8 31.60
2 2 5.5
x y zx y z
x z
+ + =+ + =− − = −
Multiply the third equation by –3 and add it to the second equation to eliminate y: 10 6 8 31.609 6 12 32.85
4 1.25
x y zx y zx z
+ + =− − − = −
− = −
Multiply the second result by 2 and add it to the first result to eliminate x:
2 2 5.52 8 2.5x zx z
− − = −− = −
10 80.8
zz
− = −=
Substitute for z to find the other variables: 4(0.8) 1.25
3.2 1.251.95
xx
x
− = −− = −
=
3(1.95) 2 4(0.8) 10.955.85 2 3.2 1.095
2 1.90.95
yy
yy
+ + =+ + =
==
Therefore, one hamburger costs $1.95, one order of fries costs $0.95, and one drink costs $0.80.
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Section 8.1: Systems of Linear Equations: Substitution and Elimination
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81. Let x = Beth’s time working alone. Let y = Bill’s time working alone. Let z = Edie’s time working alone.
We can use the following tables to organize our work:
Beth Bill EdieHours to do jobPart of job done 1 1 1in 1 hour
x y z
x y z
In 10 hours they complete 1 entire job, so 1 1 110 1
1 1 1 110
x y z
x y z
⎛ ⎞+ + =⎜ ⎟
⎝ ⎠
+ + =
Bill EdieHours to do jobPart of job done 1 1in 1 hour
y z
y z
In 15 hours they complete 1 entire job, so 1 115 1
1 1 115
y z
y z
⎛ ⎞+ =⎜ ⎟
⎝ ⎠
+ =
.
Beth Bill EdieHours to do jobPart of job done 1 1 1in 1 hour
x y z
x y z
With all 3 working for 4 hours and Beth and Bill working for an additional 8 hours, they complete
1 entire job, so 1 1 1 1 14 8 1
12 12 4 1
x y z x y
x y z
⎛ ⎞ ⎛ ⎞+ + + + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+ + =
We have the system 1 1 1 1
101 1 1
1512 12 4 1
x y z
y z
x y z
⎧+ + =⎪
⎪⎪
+ =⎨⎪⎪
+ + =⎪⎩
Subtract the second equation from the first equation:
1 1 1 1 10
1 1 1 15
x y z
y z
+ + =
+ =
1 1
3030
xx
=
=
Substitute x = 30 into the third equation: 12 12 4 130
12 4 35
y z
y z
+ + =
+ =.
Now consider the system consisting of the last result and the second original equation. Multiply the second original equation by –12 and add it to the last result to eliminate y:
12 12 1215
12 4 3 5
y z
y z
− − −+ =
+ =
8 315
40zz
− = −
=
Plugging z = 40 to find y: 12 4 3
512 4 3
40 512 1
224
y z
y
yy
+ =
+ =
=
=
Working alone, it would take Beth 30 hours, Bill 24 hours, and Edie 40 hours to complete the job.
82. – 84. Answers will vary.
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Chapter 8: Systems of Equations and Inequalities
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Section 8.2
1. matrix
2. augmented
3. True
4. True
5. Writing the augmented matrix for the system of equations:
5 5 5 514 3 6 3 64
x yx y− =⎧ ⎡ ⎤−→⎨ ⎢ ⎥+ =⎩ ⎣ ⎦
6. Writing the augmented matrix for the system of equations:
3 4 7 3 744 2 5 54 2
x yx y+ =⎧ ⎡ ⎤
→⎨ ⎢ ⎥− =⎩ ⎣ ⎦−
7. 2 3 6 04 6 2 0
x yx y+ − =⎧
⎨ − + =⎩
Write the system in standard form and then write the augmented matrix for the system of equations:
2 3 6 3 624 6 2 4 6 2
x yx y+ =⎧ ⎡ ⎤
→⎨ ⎢ ⎥− = −⎩ ⎣ ⎦− −
8. 9 0
3 4 0x y
x y− =⎧
⎨ − − =⎩
Write the system in standard form and then write the augmented matrix for the system of equations:
9 0 9 013 4 3 1 4
x yx y− =⎧ ⎡ ⎤−→⎨ ⎢ ⎥− =⎩ ⎣ − ⎦
9. Writing the augmented matrix for the system of equations:
0.01 0.03 0.06 0.01 0.060.030.13 0.10 0.20 0.13 0.10 0.20
x yx y− =⎧ ⎡ ⎤−→⎨ ⎢ ⎥+ =⎩ ⎣ ⎦
10. Writing the augmented matrix for the system of equations:
4 3 3 4 3 33 2 4 3 2 41 1 2 1 214 3 3 3 34
x y
x y
⎧ ⎡ ⎤− = −⎪ ⎢ ⎥⎪ → ⎢ ⎥⎨⎢ ⎥⎪− + = −⎢ ⎥⎪⎩ ⎣ ⎦
11. Writing the augmented matrix for the system of equations:
10 1 1 1 103 3 5 3 3 0 5
2 2 1 1 2 2
x y zx y
x y z
− + = −⎧ ⎡ ⎤⎪ ⎢ ⎥+ = →⎨ ⎢ ⎥⎪ ⎢ ⎥+ + =⎩ ⎣ ⎦
12. Writing the augmented matrix for the system of equations:
5 0 5 1 1 05 1 1 0 5
2 3 2 2 0 3 2
x y zx yx z
− − = − −⎧ ⎡ ⎤⎪ ⎢ ⎥+ = →⎨ ⎢ ⎥⎪ ⎢ ⎥− = −⎩ ⎣ ⎦
13. Writing the augmented matrix for the system of equations:
2 1 1 1 23 2 2 3 2 0 25 3 1 5 3 1 1
x y zx yx y z
+ − = −⎧ ⎡ ⎤⎪ ⎢ ⎥− = → −⎨ ⎢ ⎥⎪ ⎢ ⎥+ − = −⎩ ⎣ ⎦
14. 2 3 4 0
5 2 02 3 2
x y zx zx y z
+ − =⎧⎪ − + =⎨⎪ + − = −⎩
Write the system in standard form and then write the augmented matrix for the system of equations:
2 3 4 0 2 3 4 0 5 2 1 0 5 2
2 3 2 1 2 3 2
x y zx z
x y z
+ − = −⎧ ⎡ ⎤⎪ ⎢ ⎥− = − → − −⎨ ⎢ ⎥⎪ ⎢ ⎥+ − = − − −⎩ ⎣ ⎦
15. Writing the augmented matrix for the system of equations:
10 1 1 1 102 2 1 2 1 2 1
3 4 5 3 4 0 5
4 5 0 4 5 1 0
x y zx y z
x yx y z
− − = ⎡ − − ⎤⎧⎢ ⎥⎪ + + = − −⎪ ⎢ ⎥→⎨ ⎢ ⎥− + = −⎪ ⎢ ⎥⎪ − + = −⎢ ⎥⎩ ⎣ ⎦
16. Writing the augmented matrix for the system of equations:
2 5 1 1 2 1 53 4 2 2 1 3 4 2 2
3 5 1 3 1 5 1 1
x y z wx y z w
x y z w
− + − = ⎡ − − ⎤⎧⎪ ⎢ ⎥+ − + = → −⎨ ⎢ ⎥⎪ ⎢ ⎥− − − = − − − − −⎩ ⎣ ⎦
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Section 8.2: Systems of Linear Equations: Matrices
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17. 1 3 2 3 22 5 5 2 5 5
x yx y
⎡ − − ⎤ − = −⎧→ ⎨⎢ ⎥− − =⎩⎣ ⎦
2 1 22R r r= − +
1 3 2 1 3 22 5 5 2(1) 2 2( 3) 5 2( 2) 5
1 3 2
0 1 9
⎡ − − ⎤ ⎡ − − ⎤→⎢ ⎥ ⎢ ⎥− − + − − − − +⎣ ⎦ ⎣ ⎦
⎡ − − ⎤→ ⎢ ⎥
⎣ ⎦
18. 1 3 3 3 32 5 4 2 5 4
x yx y
⎡ − − ⎤ − = −⎧→ ⎨⎢ ⎥− − − = −⎩⎣ ⎦
2 1 22R r r= − +
1 3 3 1 3 32 5 4 2(1) 2 2( 3) 5 2( 3) 4
1 3 30 1 2
⎡ − − ⎤ ⎡ − − ⎤→⎢ ⎥ ⎢ ⎥− − − + − − − − − −⎣ ⎦ ⎣ ⎦
⎡ − − ⎤→ ⎢ ⎥
⎣ ⎦
19. 1 3 4 3 3 4 33 5 6 6 3 5 6 65 3 4 6 5 3 4 6
x y zx y zx y z
− − + =⎡ ⎤ ⎧⎪⎢ ⎥− → − + =⎨⎢ ⎥⎪⎢ ⎥− − + + =⎣ ⎦ ⎩
a. 2 1 23R r r= − +
1 3 4 33 5 6 65 3 4 6
1 3 4 33(1) 3 3( 3) 5 3(4) 6 3(3) 6
5 3 4 61 3 4 30 4 6 35 3 4 6
−⎡ ⎤⎢ ⎥−⎢ ⎥−⎢ ⎥⎣ ⎦
−⎡ ⎤⎢ ⎥→ − + − − − − + − +⎢ ⎥
−⎢ ⎥⎣ ⎦−⎡ ⎤
⎢ ⎥→ − −⎢ ⎥−⎢ ⎥⎣ ⎦
b. 3 1 35R r r= +
1 3 4 33 5 6 65 3 4 6
1 3 4 33 5 6 6
5(1) 5 5( 3) 3 5(4) 4 5(3) 61 3 4 33 5 6 60 12 24 21
−⎡ ⎤⎢ ⎥−⎢ ⎥−⎢ ⎥⎣ ⎦
−⎡ ⎤⎢ ⎥→ −⎢ ⎥
− − + + +⎢ ⎥⎣ ⎦−⎡ ⎤
⎢ ⎥→ −⎢ ⎥
−⎢ ⎥⎣ ⎦
20. 1 3 3 5 3 3 54 5 3 5 4 5 3 53 2 4 6 3 2 4 6
x y zx y zx y z
− − − + = −⎡ ⎤ ⎧⎪⎢ ⎥− − − − → − − − = −⎨⎢ ⎥⎪⎢ ⎥− − − − + =⎣ ⎦ ⎩
a. 2 1 24R r r= +
1 3 3 54 5 3 53 2 4 6
1 3 3 54(1) 4 4( 3) 5 4(3) 3 4( 5) 5
3 2 4 61 3 3 50 17 9 253 2 4 6
− −⎡ ⎤⎢ ⎥− − − −⎢ ⎥− −⎢ ⎥⎣ ⎦
− −⎡ ⎤⎢ ⎥→ − − − − − −⎢ ⎥
− −⎢ ⎥⎣ ⎦− −⎡ ⎤
⎢ ⎥→ − −⎢ ⎥− −⎢ ⎥⎣ ⎦
b. 3 1 33R r r= +
1 3 3 54 5 3 53 2 4 6
1 3 3 54 5 3 5
3(1) 3 3( 3) 2 3(3) 4 3( 5) 61 3 3 54 5 3 5
0 11 13 9
− −⎡ ⎤⎢ ⎥− − − −⎢ ⎥− −⎢ ⎥⎣ ⎦
− −⎡ ⎤⎢ ⎥→ − − − −⎢ ⎥
− − − + − +⎢ ⎥⎣ ⎦− −⎡ ⎤
⎢ ⎥→ − − − −⎢ ⎥
− −⎢ ⎥⎣ ⎦
21. 1 3 2 6 3 2 62 5 3 4 2 5 3 43 6 4 6 3 6 4 6
x y zx y zx y z
− − − + = −⎡ ⎤ ⎧⎪⎢ ⎥− − → − + = −⎨⎢ ⎥⎪⎢ ⎥− − − − + =⎣ ⎦ ⎩
a. 2 1 22R r r= − +
1 3 2 62 5 3 43 6 4 6
1 3 2 62(1) 2 2( 3) 5 2(2) 3 2( 6) 4
3 6 4 61 3 2 60 1 1 83 6 4 6
− −⎡ ⎤⎢ ⎥− −⎢ ⎥− −⎢ ⎥⎣ ⎦
− −⎡ ⎤⎢ ⎥→ − + − − − − + − − −⎢ ⎥
− −⎢ ⎥⎣ ⎦− −⎡ ⎤
⎢ ⎥→ −⎢ ⎥− −⎢ ⎥⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
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b. 3 1 33R r r= + 1 3 2 62 5 3 43 6 4 6
1 3 2 62 5 3 4
3(1) 3 3( 3) 6 3(2) 4 3( 6) 61 3 2 62 5 3 40 15 10 12
− −⎡ ⎤⎢ ⎥− −⎢ ⎥− −⎢ ⎥⎣ ⎦
− −⎡ ⎤⎢ ⎥→ − −⎢ ⎥
− − − + − +⎢ ⎥⎣ ⎦− −⎡ ⎤
⎢ ⎥→ − −⎢ ⎥
− −⎢ ⎥⎣ ⎦
22. 1 3 4 6 3 4 66 5 6 6 6 5 6 61 1 4 6 4 6
x y zx y zx y z
− − − − − = −⎡ ⎤ ⎧⎪⎢ ⎥− − → − + = −⎨⎢ ⎥⎪⎢ ⎥− − + + =⎣ ⎦ ⎩
a. 2 1 26R r r= − +
1 3 4 66 5 6 61 1 4 6
1 3 4 66(1) 6 6( 3) 5 6( 4) 6 6( 6) 6
1 1 4 61 3 4 60 13 30 301 1 4 6
− − −⎡ ⎤⎢ ⎥− −⎢ ⎥−⎢ ⎥⎣ ⎦
− − −⎡ ⎤⎢ ⎥→ − + − − − − − + − − −⎢ ⎥
−⎢ ⎥⎣ ⎦− − −⎡ ⎤
⎢ ⎥→⎢ ⎥−⎢ ⎥⎣ ⎦
b. 3 1 3R r r= +
1 3 4 6 1 3 4 66 5 6 6 6 5 6 61 1 4 6 1 1 3 1 4 4 6 6
1 3 4 66 5 6 60 2 0 0
− − − − − −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− − → − −⎢ ⎥ ⎢ ⎥− − − + − + − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
− − −⎡ ⎤⎢ ⎥→ − −⎢ ⎥
−⎢ ⎥⎣ ⎦
23. 5 3 1 2 5 3 22 5 6 2 2 5 6 24 1 4 6 4 4 6
x y zx y zx y z
− − − + = −⎡ ⎤ ⎧⎪⎢ ⎥− − → − + = −⎨⎢ ⎥⎪⎢ ⎥− − + + =⎣ ⎦ ⎩
a. 1 2 12R r r= − +
5 3 1 22 5 6 24 1 4 6
2(2) 5 2( 5) 3 2(6) 1 2( 2) 22 5 6 24 1 4 6
1 7 11 22 5 6 24 1 4 6
− −⎡ ⎤⎢ ⎥− −⎢ ⎥−⎢ ⎥⎣ ⎦
− + − − − − + − − −⎡ ⎤⎢ ⎥→ − −⎢ ⎥
−⎢ ⎥⎣ ⎦−⎡ ⎤
⎢ ⎥→ − −⎢ ⎥−⎢ ⎥⎣ ⎦
b. 1 3 1R r r= +
5 3 1 22 5 6 24 1 4 6
4 5 1 3 4 1 6 22 5 6 24 1 4 6
1 2 5 42 5 6 24 1 4 6
− −⎡ ⎤⎢ ⎥− −⎢ ⎥−⎢ ⎥⎣ ⎦
− + − + −⎡ ⎤⎢ ⎥→ − −⎢ ⎥
−⎢ ⎥⎣ ⎦−⎡ ⎤
⎢ ⎥→ − −⎢ ⎥−⎢ ⎥⎣ ⎦
24. 4 3 1 2 4 3 23 5 2 6 3 5 2 63 6 4 6 3 6 4 6
x y zx y zx y z
− − − − =⎡ ⎤ ⎧⎪⎢ ⎥− → − + =⎨⎢ ⎥ ⎪− − − − + =⎢ ⎥⎣ ⎦ ⎩
a. 1 2 1R r r= − +
4 3 1 23 5 2 63 6 4 6
(3) 4 ( 5) 3 (2) 1 (6) 23 5 2 63 6 4 6
1 2 3 43 5 2 63 6 4 6
− −⎡ ⎤⎢ ⎥−⎢ ⎥− −⎢ ⎥⎣ ⎦
− + − − − − − − +⎡ ⎤⎢ ⎥→ −⎢ ⎥
− −⎢ ⎥⎣ ⎦− −⎡ ⎤
⎢ ⎥→ −⎢ ⎥− −⎢ ⎥⎣ ⎦
b. 1 3 1R r r= +
4 3 1 23 5 2 63 6 4 6
3 4 6 3 4 1 6 23 5 2 63 6 4 6
1 9 3 83 5 2 63 6 4 6
− −⎡ ⎤⎢ ⎥−⎢ ⎥− −⎢ ⎥⎣ ⎦
− + − − − +⎡ ⎤⎢ ⎥→ −⎢ ⎥
− −⎢ ⎥⎣ ⎦−⎡ ⎤
⎢ ⎥→ −⎢ ⎥− −⎢ ⎥⎣ ⎦
25. 5
1xy=⎧
⎨ = −⎩
Consistent; 5, 1,x y= = − or using ordered pairs (5, 1)− .
26. 4
0xy= −⎧
⎨ =⎩
Consistent; 4, 0,x y= − = or using ordered pairs ( 4, 0)− .
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Section 8.2: Systems of Linear Equations: Matrices
791 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
27. 12
0 3
xy=⎧
⎪ =⎨⎪ =⎩
Inconsistent
28. 00
0 2
xy=⎧
⎪ =⎨⎪ =⎩
Inconsistent
29. 2 14 20 0
x zy z+ = −⎧
⎪ − = −⎨⎪ =⎩
Consistent; 1 22 4
is any real number
x zy zz
= − −⎧⎪ = − +⎨⎪⎩
or {( , , ) | 1 2 , 2 4 ,x y z x z y z= − − = − + z is any real number}
30. 4 43 2 0 0
x zy z+ =⎧
⎪ + =⎨⎪ =⎩
Consistent; 4 42 3
is any real number
x zy zz
= −⎧⎪ = −⎨⎪⎩
or {( , , ) | 4 4 , 2 3 ,x y z x z y z= − = − z is any real number}
31. 1
2 4
3 4
12
2 3
xx x
x x
=⎧⎪ + =⎨⎪ + =⎩
Consistent; 1
2 4
3 4
4
123 2
is any real number
xx xx xx
=⎧⎪ = −⎪⎨ = −⎪⎪⎩
or 1 2 3 4 1 2 4{( , , , ) | 1, 2 ,x x x x x x x= = −
3 4 43 2 , is any real number}x x x= −
32. 1
2 4
3 4
12 23 0
xx xx x
=⎧⎪ + =⎨⎪ + =⎩
Consistent; 1
2 4
3 4
4
12 2
3is any real number
xx xx xx
=⎧⎪ = −⎪⎨ = −⎪⎪⎩
or 1 2 3 4 1 2 4 3 4{( , , , ) | 1, 2 2 , 3 ,x x x x x x x x x= = − = −
4 is any real number}x
33. 1 4
2 3 4
4 2 3 3
0 0
x xx x x
+ =⎧⎪ + + =⎨⎪ =⎩
Consistent; 1 4
2 3 4
3 4
2 43 3
, are any real numbers
x xx x xx x
= −⎧⎪ = − −⎨⎪⎩
or 1 2 3 4 1 4 2 3 4{( , , , ) | 2 4 , 3 3 ,x x x x x x x x x= − = − −
3 4and are any real numbers}x x
34. 1
2
3 4
12
2 3
xx
x x
=⎧⎪ =⎨⎪ + =⎩
Consistent; 1
2
3 4
4
123 2
is any real number
xxx xx
=⎧⎪ =⎪⎨ = −⎪⎪⎩
or 1 2 3 4 1 2 3 4{( , , , ) | 1, 2, 3 2 ,x x x x x x x x= = = −
4 is any real number}x
35.
1 4
2 4
3 4
22 2
00 0
x xx x
x x
+ = −⎧⎪ + =⎪⎨ − =⎪⎪ =⎩
Consistent; 1 4
2 4
3 4
4
22 2
is any real number
x xx xx xx
= − −⎧⎪ = −⎪⎨ =⎪⎪⎩
or 1 2 3 4 1 4 2 4{( , , , ) | 2 , 2 2 ,x x x x x x x x= − − = −
3 4 4, is any real number}x x x=
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
792 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
36.
1
2
3
4
1230
xxxx
=⎧⎪ =⎪⎨ =⎪⎪ =⎩
Consistent; 1
2
3
4
1230
xxxx
=⎧⎪ =⎪⎨ =⎪⎪ =⎩
or (1,2,3,0)
37. 84
x yx y+ =⎧
⎨ − =⎩
Write the augmented matrix:
( )
( )
( )
2 1 2
12 22
1 2 1
8 81 1 1 1 01 1 4 2 4
81 1 0 1 2
0 61 0 1 2
R r r
R r
R r r
⎡ ⎤ ⎡ ⎤→ = − +⎢ ⎥ ⎢ ⎥
⎣ − ⎦ ⎣ ⎦− −⎡ ⎤
→ = −⎢ ⎥⎣ ⎦⎡ ⎤
→ = − +⎢ ⎥⎣ ⎦
The solution is 6, 2x y= = or using ordered pairs (6, 2).
38. 2 5
3x yx y+ =⎧
⎨ + =⎩
Write the augmented matrix:
( )
( )
( )
2 1 2
2 2
1 2 1
5 51 2 1 2 3 01 1 1 2
51 2 0 1 2
01 1 20 1 2
R r r
R r
R r r
⎡ ⎤ ⎡ ⎤→ = − +⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ − ⎦−⎡ ⎤
→ = −⎢ ⎥⎣ ⎦⎡ ⎤
→ = − +⎢ ⎥⎣ ⎦
The solution is 1, 2x y= = or using ordered pairs (1, 2).
39. 2 4 23 2 3
x yx y− = −⎧
⎨ + =⎩
Write the augmented matrix:
( )
( )
( )
( )
11 12
2 1 2
12 283
4
12
1 2 134
2 1 14 2 2 3 3 3 32 2
1 12 3 0 8 6
1 12 0 1
01 2 0 1
R r
R r r
R r
R r r
⎡ ⎤ ⎡ ⎤−− − − =→⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
⎡ ⎤−− = − +→ ⎢ ⎥⎣ ⎦⎡ ⎤−− =→ ⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
= +⎢ ⎥→⎢ ⎥⎣ ⎦
The solution is 1 3,2 4
x y= = or using ordered pairs
1 3,2 4
⎛ ⎞⎜ ⎟⎝ ⎠
.
40. 3 3 3
84 23
x y
x y
+ =⎧⎪⎨
+ =⎪⎩
Write the augmented matrix:
( )11 18 38
3 3
3 3 3 1 1 1
4 2 4 2R r⎡ ⎤ ⎡ ⎤
=→⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
( )
( )
( )
2 1 243
12 22
23
13 1 2 123
1 1 1 4 0 21 1 1 0 1
01 0 1
R r r
R r
R r r
⎡ ⎤ = − +→ ⎢ ⎥⎢ ⎥−⎣ ⎦−⎡ ⎤ = −→ ⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
= − +⎢ ⎥→⎢ ⎥⎣ ⎦
The solution is 1 2,3 3
x y= = or using ordered
pairs 1 2,3 3
⎛ ⎞⎜ ⎟⎝ ⎠
.
41. 2 4
2 4 8x yx y+ =⎧
⎨ + =⎩
Write the augmented matrix:
( )2 1 21 2 4 1 2 4 2
8 0 0 02 4R r r⎡ ⎤ ⎡ ⎤ = − +→⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
This is a dependent system. 2 4
4 2x y
x y+ =
= −
The solution is 4 2 , is any real numberx y y= − or {( , ) | 4 2 , is any real number}x y x y y= −
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.2: Systems of Linear Equations: Matrices
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42. 3 79 3 21
x yx y− =⎧
⎨ − =⎩
Write the augmented matrix:
( )
( )
713
1 13
713 3
2 1 2
113 71 39 3 21 9 3 21
-1 90 0 0
R r
R r r
⎡ ⎤−⎡ ⎤− ⎢ ⎥→ =⎢ ⎥ ⎢ ⎥−⎣ ⎦ −⎢ ⎥⎣ ⎦⎡ ⎤
= − +→ ⎢ ⎥⎢ ⎥⎣ ⎦
This is a dependent system. 3 73 7x yx y− =− =
The solution is 3 7, is any real numbery x x= − or {( , ) | 3 7, is any real number}x y y x x= −
43. 2 3 6
12
x y
x y
+ =⎧⎪⎨
− =⎪⎩
Write the augmented matrix:
( )
( )
( )
( )
312 1 121 1
2 2
32
2 1 25 52 2
3 22 2 25
3 32 1 2 12
3 62 31 1 1 1 1
31 0
31 0 1 1
01 0 1 1
R r
R r r
R r
R r r
⎡ ⎤⎡ ⎤⎢ ⎥→ =⎢ ⎥⎢ ⎥⎢ ⎥⎣ − ⎦ ⎣ − ⎦⎡ ⎤⎢ ⎥→ = − +⎢ ⎥− −⎣ ⎦⎡ ⎤
→ = −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
→ = − +⎢ ⎥⎢ ⎥⎣ ⎦
The solution is 3 , 12
x y= = or 3 , 12
⎛ ⎞⎜ ⎟⎝ ⎠
.
44. 1 22
2 8
x y
x y
⎧ + = −⎪⎨⎪ − =⎩
Write the augmented matrix:
( )
( )
( )
( )
12 1 1
2 1 2
12 24
1 2 1
1 2 41 2 281 281 2
1 2 4 0 1241 2 4 0 31
01 2 20 31
R r
R r r
R r
R r r
⎡ ⎤ ⎡ ⎤−− → =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦−⎣ ⎦−
⎡ ⎤−→ = − +⎢ ⎥⎣ ⎦−⎡ ⎤−→ = −⎢ ⎥
−⎣ ⎦⎡ ⎤
→ = − +⎢ ⎥−⎣ ⎦
The solution is 2, 3x y= = − or (2, 3)− .
45. 3 5 3
15 5 21x yx y− =⎧
⎨ + =⎩
Write the augmented matrix:
( )
( )
( )
( )
513 1 13
53
2 1 2
53 1
2 23015
453
1 2 1315
3 5 3 1 1 15 5 21 15 5 21
1 1 150 30 6
1 1 0 1
01 0 1
R r
R r r
R r
R r r
⎡ ⎤⎡ ⎤− −→ =⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤−
→ = − +⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥→ =⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥→ = +⎢ ⎥⎣ ⎦
The solution is 4 1,3 5
x y= = or 4 1,3 5
⎛ ⎞⎜ ⎟⎝ ⎠
.
46. 2 1
1 32 2
x y
x y
− = −⎧⎪⎨
+ =⎪⎩
( )
( )
( )
1 112 2
1 1231 312 2 2 2
1 12 2 2 1 2
1 12 1 2 12
2 1 1 1 1 1
1 10 1 2
01 0 1 2
R r
R r r
R r r
⎡ ⎤⎡ ⎤− − − −⎢ ⎥→ =⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤− −→ = − +⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
→ = +⎢ ⎥⎢ ⎥⎣ ⎦
The solution is 1 , 22
x y= = or 1 , 22
⎛ ⎞⎜ ⎟⎝ ⎠
.
47. 6
2 3 162 4
x yx z
y z
− =⎧⎪ − =⎨⎪ + =⎩
Write the augmented matrix:
( )
( )
2 1 2
3 12 22 2
0 61 10 3 162
0 2 1 4
0 61 10 3 22 40 2 1 4
0 61 1
0 1 20 2 1 4
R r r
R r
⎡ ⎤−⎢ ⎥
−⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤−⎢ ⎥
→ − = − +⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥
−⎢ ⎥→ =⎢ ⎥⎢ ⎥⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
794 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
( )
32
1 2 132
3 2 3
323 1
3 32 4
31 3 12
32 3 22
0 81
0 1 2 2
0 0 04
0 81
0 1 20 0 01
0 0 810 0 1 20 0 01
R r rR r r
R r
R r r
R r r
⎡ ⎤−⎢ ⎥ = +⎛ ⎞⎢ ⎥−→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎣ ⎦⎡ ⎤−⎢ ⎥⎢ ⎥−→ =⎢ ⎥⎣ ⎦⎡ ⎤
⎛ ⎞= +⎢ ⎥⎜ ⎟→ ⎢ ⎥ ⎜ ⎟= +⎢ ⎥ ⎝ ⎠
⎣ ⎦
The solution is 8, 2, 0x y z= = = or (8, 2, 0).
48. 2 4
2 4 03 2 11
x yy z
x z
+ = −⎧⎪ − + =⎨⎪ − = −⎩
Write the augmented matrix:
( )12
11 12
02 1 40 0423 0 112
2010 0 423 0 112
R r
⎡ ⎤−⎢ ⎥
−⎢ ⎥⎢ ⎥−−⎣ ⎦
⎡ ⎤−⎢ ⎥⎢ ⎥→ =−⎢ ⎥−−⎢ ⎥⎣ ⎦
( )
( )
( )
12
3 1 332
12
12 22
32
11 2 12
33 2 32
13 35
2010 0 342
0 52
2010 0 1 2
0 52
201 10 0 1 20 0 5 5
201 10 0 1 20 0 1 1
R r r
R r
R r r
R r r
R r
⎡ ⎤−⎢ ⎥⎢ ⎥→ = − +−⎢ ⎥
− −⎢ ⎥−⎣ ⎦⎡ ⎤−⎢ ⎥⎢ ⎥→ = −−⎢ ⎥
− −⎢ ⎥−⎣ ⎦⎡ ⎤− ⎛ ⎞= − +⎢ ⎥
⎜ ⎟→ −⎢ ⎥ ⎜ ⎟= +⎢ ⎥ ⎝ ⎠− −⎣ ⎦⎡ ⎤−⎢ ⎥
→ = −−⎢ ⎥⎢ ⎥⎣ ⎦
1 3 1
2 3 2
0 0 310 0 1 2
20 0 1 1
R r rR r r
⎡ ⎤−= − +⎢ ⎥ ⎛ ⎞
→ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥⎣ ⎦
The solution is 3, 2, 1x y z= − = = or ( 3, 2, 1)− .
49. 2 3 7
2 43 2 2 10
x y zx y zx y z
− + =⎧⎪ + + =⎨⎪− + − = −⎩
Write the augmented matrix:
( )
2 1 2
3 1 3
12 25
1 2 1
3 2 3
3 71 22 1 1 43 102 2
3 71 2 20 5 5 10
30 7 114
3 71 20 1 1 20 7 114
0 31 12
0 1 1 24
0 0 3 3
R r rR r r
R r
R r rR r r
⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥− −−⎣ ⎦
⎡ ⎤−= − +⎢ ⎥ ⎛ ⎞
→ − − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥−⎣ ⎦⎡ ⎤−⎢ ⎥
→ =− −⎢ ⎥⎢ ⎥−⎣ ⎦⎡ ⎤
= +⎢ ⎥ ⎛ ⎞→ − − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥
⎣ ⎦
( )13 33
0 31 10 1 1 20 0 1 1
R r
⎡ ⎤⎢ ⎥
→ =− −⎢ ⎥⎢ ⎥⎣ ⎦
1 3 1
2 3 2
0 01 20 0 1 10 0 1 1
R r rR r r
⎡ ⎤= − +⎢ ⎥ ⎛ ⎞
→ − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥⎣ ⎦
The solution is 2, 1, 1x y z= = − = or (2, 1, 1)− .
50. 2 3 02 2 73 4 3 7
x y zx y zx y z
+ − =⎧⎪− + + = −⎨⎪ − − =⎩
Write the augmented matrix:
( )31
2 21
1 12
312 2
2 1 2
3 1 33112 2
2 1 3 02 2 1 73 4 3 7
1 02 2 1 7 3 4 3 7
1 02
0 3 2 7 3
0 7
R r
R r rR r r
−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥− −⎣ ⎦
⎡ ⎤−⎢ ⎥→ − − =⎢ ⎥⎢ ⎥− −⎣ ⎦⎡ ⎤−
= +⎛ ⎞⎢ ⎥→ − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.2: Systems of Linear Equations: Matrices
795 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
( )31
2 272 1
2 23 3 3311
2 2
1 00 1
0 7
R r
⎡ ⎤−⎢ ⎥
→ − − =⎢ ⎥⎢ ⎥−⎣ ⎦
( )
7 76 6 1
1 2 12723 3 11
3 2 3213 356 6
7 76 6
7 623 33 3 13
3513
5613 7
1 3 16713 2
2 3 235 313
1 0
0 1
0 0
1 0
0 1
0 0 1
1 0 0
0 1 0
0 0 1
R r r
R r r
R r
R r r
R r r
⎡ ⎤−⎛ ⎞= − +⎢ ⎥⎜ ⎟→ − −⎢ ⎥ ⎜ ⎟= +⎢ ⎥ ⎝ ⎠− −⎢ ⎥⎣ ⎦
⎡ ⎤−⎢ ⎥
→ − − = −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
⎛ ⎞= +⎢ ⎥⎜ ⎟→ −⎢ ⎥ ⎜ ⎟= +⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦
The solution is 56 7 35, ,13 13 13
x y z= = − = or
56 7 35, ,13 13 13⎛ ⎞−⎜ ⎟⎝ ⎠
.
51. 2 2 2 22 3 23 2 0
x y zx y zx y
− − =⎧⎪ + + =⎨⎪ + =⎩
Write the augmented matrix:
( )
( )
11 12
2 1 2
3 1 3
3 2 3
2 22 232 1 2
3 0 02
1 1 1 132 1 2
3 0 02
1 1 1 12
0 5 3 0 3
0 5 3 3
1 1 1 10 5 3 0 0 0 0 3
R r
R r rR r r
R r r
⎡ ⎤− −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤− −⎢ ⎥
→ =⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤− −
= − +⎢ ⎥ ⎛ ⎞→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦
⎡ ⎤− −⎢ ⎥
→ = − +⎢ ⎥⎢ ⎥−⎣ ⎦
There is no solution. The system is inconsistent.
52. 2 3 0
2 53 4 1
x y zx y zx y z
− − =⎧⎪− + + =⎨⎪ − − =⎩
Write the augmented matrix:
( )
1 2
2 1 2
3 1 3
3 2 3
3 02 151 2 1
3 1 14
51 12 Interchange3 02 1
and 3 1 14
51 12 20 10 1 1
30 162 2
51 120 10 21 10 0 0 4
r r
R r rR r r
R r r
⎡ ⎤− −⎢ ⎥−⎢ ⎥⎢ ⎥−−⎣ ⎦
⎡ ⎤−−−⎛ ⎞⎢ ⎥−→ − ⎜ ⎟⎢ ⎥ −⎝ ⎠⎢ ⎥−−⎣ ⎦
⎡ ⎤−−− = − +⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤−−−⎢ ⎥→ = − +⎢ ⎥⎢ ⎥−⎣ ⎦
There is no solution. The system is inconsistent.
53. 1
2 3 43 2 7 0
x y zx y zx y z
− + + = −⎧⎪ − + − = −⎨⎪ − − =⎩
Write the augmented matrix:
( )1 1
2 1 2
3 1 3
1 2 1
3 2 3
1 1 1 1431 2
23 7 0
1 1 1 1431 2
23 7 0
1 1 1 140 3 1
340 31
20 5140 3 1
0 0 0 0
R r
R r rR r r
R r rR r r
⎡ ⎤− −⎢ ⎥−−−⎢ ⎥⎢ ⎥− −⎣ ⎦
⎡ ⎤− −⎢ ⎥−−→ − = −⎢ ⎥⎢ ⎥− −⎣ ⎦⎡ ⎤− −
= +⎛ ⎞⎢ ⎥−→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −⎣ ⎦⎡ ⎤−−
= +⎛ ⎞⎢ ⎥−→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥⎣ ⎦
The matrix in the last step represents the system 5 24 30 0
x zy z− = −⎧
⎪ − = −⎨⎪ =⎩
or, equivalently, 5 24 3
0 0
x zy z= −⎧
⎪ = −⎨⎪ =⎩
The solution is 5 2x z= − , 4 3y z= − , z is any real number or {( , , ) | 5 2, 4 3,x y z x z y z= − = − z is any real number}.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
796 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
54. 2 3 03 2 2 2
5 3 2
x y zx y zx y z
− − =⎧⎪ + + =⎨⎪ + + =⎩
Write the augmented matrix:
1 3
2 1 2
3 1 3
3 2 37 413 13 1
2 213
3 02 13 2 2 2
5 31 2
5 31 2Interchange
3 2 2 2 and
3 02 1
5 31 23
0 13 7 42
0 13 7 4
5 31 2
0 10 0 0 0
r r
R r rR r r
R r r
R r
⎡ ⎤− −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥ ⎛ ⎞
→ ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥− −⎣ ⎦
⎡ ⎤= − +⎢ ⎥ ⎛ ⎞
→ − − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− − −⎣ ⎦⎡ ⎤⎢ ⎥ = − +⎛ ⎞⎢ ⎥→ ⎜ ⎟⎜ ⎟= −⎢ ⎥ ⎝ ⎠⎣ ⎦
( )
6413 137 4
1 2 113 13
01
0 510 0 0 0
R r r
⎡ ⎤⎢ ⎥⎢ ⎥→ = − +⎢ ⎥⎣ ⎦
The matrix in the last step represents the system 4 6
13 137 4
13 130 0
x z
y z
⎧ + =⎪⎪⎨ + =⎪⎪
=⎩
or, equivalently, 4 6
13 137 4
13 130 0
x z
y z
⎧ = − +⎪⎪⎨ = − +⎪⎪
=⎩
The solution is 4 613 13
x z= − + , 7 413 13
y z= − + ,
z is any real number or 4 6( , , ) ,13 13
x y z x z⎧= − +⎨
⎩
7 4 , is any real number13 13
y z z ⎫= − + ⎬⎭
.
55. 2 2 3 64 3 2 02 3 7 1
x y zx y zx y z
− + =⎧⎪ − + =⎨⎪− + − =⎩
Write the augmented matrix: 2 3 623 04 2
2 3 7 1
⎡ ⎤−⎢ ⎥
−⎢ ⎥⎢ ⎥− −⎣ ⎦
( )32
11 12
32
2 1 2
3 1 3
52
1 2 1
3 2 3
31 13 04 2
2 3 7 1
31 14
40 1 122
40 71
0 9140 1 12
0 0 0 19
R r
R r rR r r
R r rR r r
⎡ ⎤−⎢ ⎥⎢ ⎥−→ =⎢ ⎥− −⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥ = − +⎛ ⎞⎢ ⎥−→ − ⎜ ⎟= +⎢ ⎥ ⎝ ⎠−⎢ ⎥⎣ ⎦⎡ ⎤− −⎢ ⎥ = +⎛ ⎞⎢ ⎥−→ − ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦
There is no solution. The system is inconsistent.
56. 3 2 2 67 3 2 12 3 4 0
x y zx y zx y z
− + =⎧⎪ − + = −⎨⎪ − + =⎩
Write the augmented matrix: 3 6227 3 2 1
3 042
⎡ ⎤−⎢ ⎥
− −⎢ ⎥⎢ ⎥−⎣ ⎦
( )2 23 3
11 13
1 27 3 2 1
3 042R r
⎡ ⎤−⎢ ⎥⎢ ⎥→ − =−⎢ ⎥−⎣ ⎦
( )
2 23 3
2 1 25 83 3
3 1 35 83 3
2 23 35 8
3 2 33 3
1 27
0 15 2
0 4
1 2
0 15 0 0 0 19
R r rR r r
R r r
⎡ ⎤−⎢ ⎥ = − +⎛ ⎞⎢ ⎥−→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥⎢ ⎥−→ − = +⎢ ⎥
−⎣ ⎦
There is no solution. The system is inconsistent.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.2: Systems of Linear Equations: Matrices
797 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
57. 6
3 2 53 2 14
x y zx y zx y z
+ − =⎧⎪ − + = −⎨⎪ + − =⎩
Write the augmented matrix: 61 1 1
23 51231 14
⎡ ⎤−⎢ ⎥
− −⎢ ⎥⎢ ⎥−⎣ ⎦
( )
2 1 2
3 1 3
234 15 5 2 25
715 5
234 1 2 15 5
3 2 36355
61 1 13
230 5 40 82 1
61 1 1
0 10 82 1
01
0 12
0 0
R r rR r r
R r
R r rR r r
⎡ ⎤−= − +⎢ ⎥ ⎛ ⎞
−→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥⎣ − ⎦⎡ ⎤−⎢ ⎥
−⎢ ⎥→ = −⎢ ⎥⎣ − ⎦⎡ ⎤−⎢ ⎥ = − +⎛ ⎞⎢ ⎥−→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠−⎢ ⎥⎣ ⎦
( )
715 5
234 55 5 3 33
11 3 15
42 3 25
01
0 120 0 1
0 01 10 0 3 1
20 0 1
R r
R r r
R r r
⎡ ⎤−⎢ ⎥⎢ ⎥−→ =⎢ ⎥
−⎣ ⎦⎡ ⎤
⎛ ⎞= +⎢ ⎥⎜ ⎟→ ⎢ ⎥ ⎜ ⎟= +⎢ ⎥ ⎝ ⎠−⎣ ⎦
The solution is 1, 3, 2x y z= = = − , or (1, 3, 2)− .
58. 4
2 3 4 155 2 12
x y zx y zx y z
− + = −⎧⎪ − + = −⎨⎪ + − =⎩
Write the augmented matrix:
( )
2 1 2
3 1 3
2 2
1 1 1 43 152 4
25 1 12
1 1 1 42
0 7 1 25
0 6 7 32
1 1 1 40 7 1 20 6 7 32
R r rR r r
R r
⎡ ⎤− −⎢ ⎥
− −⎢ ⎥⎢ ⎥−⎣ ⎦
⎡ ⎤− −= − +⎢ ⎥ ⎛ ⎞
→ −− ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦⎡ ⎤− −⎢ ⎥
→ = −−⎢ ⎥⎢ ⎥−⎣ ⎦
( )
1 2 1
3 2 3
13 35
1 3 1
2 3 2
0 31 10 7 1 2
60 0 5 10
0 31 10 7 1 20 0 1 2
0 01 10 0 3 1
20 0 1 2
R r rR r r
R r
R r rR r r
⎡ ⎤−= +⎢ ⎥ ⎛ ⎞
→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦⎡ ⎤−⎢ ⎥
→ =−⎢ ⎥⎢ ⎥−⎣ ⎦⎡ ⎤
= +⎢ ⎥ ⎛ ⎞→ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥−⎣ ⎦
The solution is 1, 3, 2x y z= = = − or (1, 3, 2)− .
59. 2 3
2 4 72 2 3 4
x y zx y zx y z
+ − = −⎧⎪ − + = −⎨⎪− + − =⎩
Write the augmented matrix: 31 2 1
4 72 12 32 4
⎡ ⎤−−⎢ ⎥
− −⎢ ⎥⎢ ⎥− −⎣ ⎦
( )
2 1 2
3 1 3
3 1 18 8 2 28
31 2 12
80 3 12
20 6 5
31 2 1
0 120 6 5
R r rR r r
R r
⎡ ⎤−−= − +⎢ ⎥ ⎛ ⎞
−→ − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥−−⎣ ⎦⎡ ⎤−−⎢ ⎥
−⎢ ⎥→ = −⎢ ⎥−−⎢ ⎥⎣ ⎦
( )
1314 43 1 1 2 18 8
3 2 311 114 4
1314 43 1 48 8 3 311
11 3 11 4
2 32 3 28
012
0 16
0 0
01
0 10 0 1 1
0 0 31
0 0 10 0 1 1
R r rR r r
R r
R r r
R r r
⎡ ⎤− −⎢ ⎥ = − +⎛ ⎞⎢ ⎥−→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠− −⎢ ⎥⎣ ⎦⎡ ⎤− −⎢ ⎥⎢ ⎥−→ = −⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥ ⎛ ⎞= +
⎜ ⎟⎢ ⎥→⎜ ⎟= +⎢ ⎥ ⎝ ⎠⎣ ⎦
The solution is 13, , 12
x y z= − = = or 13, ,12
⎛ ⎞−⎜ ⎟⎝ ⎠
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
798 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
60. 4 3 8
3 3 126 1
x y zx y zx y z
+ − = −⎧⎪ − + =⎨⎪ + + =⎩
Write the augmented matrix: 31 4 8
3 31 1261 1 1
⎡ ⎤− −⎢ ⎥
−⎢ ⎥⎢ ⎥⎣ ⎦
( )
2 1 2
3 1 3
3612 12 213 13 13
9 4013 13
1 2 1361213 13
3 2 381 913 13
31 4 8 30 13 36 120 3 9 9
31 4 8
0 10 3 9 9
014
0 13
0 0
R r rR r r
R r
R r rR r r
⎡ ⎤− −= − +⎢ ⎥ ⎛ ⎞
→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦⎡ ⎤− −⎢ ⎥⎢ ⎥− −→ = −⎢ ⎥
−⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥
= − +⎛ ⎞⎢ ⎥− −→ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦
( )9 40
13 133612 13
3 313 13 8119
91 3 1138
3 123 3 2131
9
01
0 1
0 0 1
0 0 31
0 0 1
0 0 1
R r
R r r
R r r
⎡ ⎤−⎢ ⎥⎢ ⎥− −→ =⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥ ⎛ ⎞= − +
⎜ ⎟⎢ ⎥−→⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥⎣ ⎦
The solution is 8 13, ,3 9
x y z= = − = or 8 13, ,3 9
⎛ ⎞−⎜ ⎟⎝ ⎠
.
61.
233
2 184 23
x y z
x y z
x y
⎧ + − =⎪⎪
− + =⎨⎪⎪ + =⎩
Write the augmented matrix:
( )
23
83
1 1 23 3 9
11 13
83
3 1 12 1 1 1
04 2
12 1 1 1
04 2
R r
⎡ ⎤−⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤−⎢ ⎥⎢ ⎥→ − =⎢ ⎥⎢ ⎥⎣ ⎦
( )
1 1 23 3 95 5 5 2 1 23 3 9
3 1 3162 43 3 9
1 1 23 3 9
1 33 2 25
162 43 3 9
13 1
1 2 11 33 2
3 2 33
12
0 4
0
1
0 1 1
0
0 01
0 1 10 0 2 2
R r rR r r
R r
R r r
R r r
⎡ ⎤−⎢ ⎥ = − +⎛ ⎞⎢ ⎥−→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥⎢ ⎥−→ = −−⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥ ⎛ ⎞= − +⎢ ⎥− ⎜ ⎟→ −⎢ ⎥ ⎜ ⎟= − +⎝ ⎠⎣ ⎦
( )
( )
131 13 3 32
1323 2 3 2
0 01
0 1 10 0 1 1
0 01
0 0 10 0 1 1
R r
R r r
⎡ ⎤⎢ ⎥⎢ ⎥−→ =−⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥→ = +⎢ ⎥⎣ ⎦
The solution is 1 2, , 13 3
x y z= = = or 1 2, , 13 3
⎛ ⎞⎜ ⎟⎝ ⎠
.
62. 83
12 1
2
x yx y z
x y z
⎧ + =⎪
− + =⎨⎪ + + =⎩
Write the augmented matrix:
83
1 1 0 12 1 1 11 2 1
⎡ ⎤⎢ ⎥
−⎢ ⎥⎢ ⎥⎣ ⎦
( )
2 1 2
3 1 353
53
2 3
53 2 33
1 1 0 12
0 3 1 1 0 1 1
1 1 0 1Interchange
0 1 1 and
0 3 1 1
1 1 0 10 1 1 3
0 0 4 4
R r rR r r
r r
R r r
⎡ ⎤= − +⎢ ⎥ ⎛ ⎞
→ − −⎢ ⎥ ⎜ ⎟= − +⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥ ⎛ ⎞
→ ⎢ ⎥ ⎜ ⎟⎝ ⎠⎢ ⎥
− −⎣ ⎦⎡ ⎤⎢ ⎥
→ = +⎢ ⎥⎢ ⎥⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.2: Systems of Linear Equations: Matrices
799 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
( )
( )
( )
153 343
22 2 33
132
1 1 23
1 1 0 10 1 1
0 0 1 1
1 1 0 10 1 0
0 0 1 1
1 0 0
0 1 0
0 0 1 1
R r
R r r
R r r
⎡ ⎤⎢ ⎥
→ =⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥
→ = −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥
→ = −⎢ ⎥⎢ ⎥⎣ ⎦
The solution is 1 2, , 13 3
x y z= = = or 1 2, , 13 3
⎛ ⎞⎜ ⎟⎝ ⎠
.
63.
42 03 2 6
2 2 2 1
x y z wx y z
x y z wx y z w
+ + + =⎧⎪ − + =⎪⎨ + + − =⎪⎪ − − + = −⎩
Write the augmented matrix: 1 1 1 1 4
0 02 1 13 62 1 1
2 21 2 1
⎡ ⎤⎢ ⎥
−⎢ ⎥⎢ ⎥−⎢ ⎥
− −⎣ − ⎦
2 1 2
3 1 3
4 1 4
1 1 1 1 4 22 80 3 1 3
2 4 60 10 3 3 51
R r rR r rR r r
⎡ ⎤= − +⎛ ⎞⎢ ⎥
⎜ ⎟− −− −⎢ ⎥→ = − +⎜ ⎟⎢ ⎥− − −− ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠− − −⎣ ⎦
( )
2 3
2 2
1 2 1
3 2 3
4 2 4
1 1 1 1 4Interchange2 4 60 1
and 2 80 3 10 3 3 51
1 1 1 1 40 61 2 4
2 80 3 10 3 3 51
20 31 10 61 2 4 30 0 5 10 10 30 0 3 13 13
r r
R r
R r rR r rR r r
⎡ ⎤⎢ ⎥ ⎛ ⎞− − −−⎢ ⎥→ ⎜ ⎟⎢ ⎥− −− − ⎝ ⎠⎢ ⎥
− − −⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥→ = −⎢ ⎥− −− −⎢ ⎥
− − −⎣ ⎦⎡ ⎤−−− = − +⎛⎢ ⎥
⎜⎢ ⎥→ = +⎜⎢ ⎥ ⎜ = +⎢ ⎥ ⎝⎣ ⎦
⎞⎟⎟⎟⎠
( )13 35
1 3 1
2 3 2
4 3 4
20 31 10 61 2 4 0 0 1 2 20 0 3 13 13
0 0 01 10 0 01 2 20 0 1 2 2 30 0 0 7 7
R r
R r rR r rR r r
⎡ ⎤−−−⎢ ⎥⎢ ⎥→ =⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤− = +⎛ ⎞⎢ ⎥
⎜ ⎟⎢ ⎥→ = − +⎜ ⎟⎢ ⎥ ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠⎣ ⎦
( )14 47
1 4 1
3 4 3
0 0 01 10 0 01 2 0 0 1 2 20 0 0 1 1
0 0 01 10 0 01 2
20 0 0 010 0 0 1 1
R r
R r rR r r
⎡ ⎤−⎢ ⎥⎢ ⎥→ =⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥ = +⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥⎣ ⎦
The solution is 1, 2, 0, 1x y z w= = = = or (1, 2, 0, 1).
64.
42 0
2 3 62 2 2 1
x y z wx y zx y z wx y z w
+ + + =⎧⎪ − + + =⎪⎨ + + − =⎪⎪− + − + = −⎩
Write the augmented matrix: 1 1 1 1 4
0 01 2 13 62 1 11 2 12 2
⎡ ⎤⎢ ⎥−⎢ ⎥
⎢ ⎥−⎢ ⎥⎣ − ⎦− −
2 1 2
3 1 3
4 1 4
2 3
1 2 1
3
1 1 1 1 40 3 2 1 4 20 31 1 2 20 3 0 74
1 1 1 1 4Interchange0 31 1 2
and 0 3 2 1 40 3 0 74
0 61 2 40 31 1 2 0 0 5 10 100 0 3 13 13
R r rR r rR r r
r r
R r rR
⎡ ⎤= +⎛ ⎞⎢ ⎥
⎜ ⎟⎢ ⎥→ = − +⎜ ⎟⎢ ⎥−− − ⎜ ⎟= +⎢ ⎥ ⎝ ⎠⎣ ⎦⎡ ⎤⎢ ⎥ ⎛ ⎞−− −⎢ ⎥→ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤
= − +⎢ ⎥−− −⎢ ⎥→ = −
⎢ ⎥⎢ ⎥⎣ ⎦
2 3
4 2 4
33
r rR r r
⎛ ⎞⎜ ⎟+⎜ ⎟⎜ ⎟= − +⎝ ⎠
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
800 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
( )4 3 4
14 35
14 435
0 61 2 40 31 1 2 3 50 0 5 10 100 0 0 35 35
0 61 2 40 31 1 2 0 0 1 2 20 0 0 1 1
R r r
R r
R r
⎡ ⎤⎢ ⎥
−− −⎢ ⎥→ = −⎢ ⎥⎢ ⎥
− −⎣ ⎦⎡ ⎤⎢ ⎥ ⎛ ⎞=−− −⎢ ⎥ ⎜ ⎟→
⎜ ⎟⎢ ⎥ = −⎝ ⎠⎢ ⎥⎣ ⎦
Write the matrix as the corresponding system: 2 4 6
3 22 2
1
x z wy z w
z ww
+ + =⎧⎪ − − = −⎪⎨ + =⎪⎪ =⎩
Substitute and solve: 2(1) 2
2 20
zz
z
+ =+ =
=
0 3(1) 2
3 21
yy
y
− − = −− = −
=
2(0) 4(1) 6
0 4 62
xx
x
+ + =+ + =
=
The solution is 2x = , 1y = , 0z = , 1w = or (2, 1, 0, 1).
65. 2 1
2 2 23 3 3
x y zx y zx y z
+ + =⎧⎪ − + =⎨⎪ + + =⎩
Write the augmented matrix: 1 2 1 12 1 2 23 3 31
⎡ ⎤⎢ ⎥
−⎢ ⎥⎢ ⎥⎣ ⎦
( )
2 1 2
3 1 3
3 2 3
1 2 1 12
0 5 0 0 3
0 5 0 0
1 2 1 10 5 0 0 0 0 0 0
R r rR r r
R r r
⎡ ⎤= − +⎢ ⎥ ⎛ ⎞
→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦⎡ ⎤⎢ ⎥
→ − = − +⎢ ⎥⎢ ⎥⎣ ⎦
The matrix in the last step represents the system
2 15 0
0 0
x y zy
+ + =⎧⎪ − =⎨⎪ =⎩
Substitute and solve: 5 0
0yy
− ==
2(0) 11
x zz x
+ + == −
The solution is 0, 1 ,y z x= = − x is any real number or {( , , ) | 0, 1 ,x y z y z x= = − x is any real number}.
66. 2 3
2 2 63 3 4
x y zx y zx y z
+ − =⎧⎪ − + =⎨⎪ − + =⎩
Write the augmented matrix: 31 2 162 1 2
3 31 4
⎡ ⎤−⎢ ⎥
−⎢ ⎥⎢ ⎥−⎣ ⎦
( )
2 1 2
3 1 3
3 2 3
31 2 12
0 5 0 40 5 4 1
31 2 10 5 0 40 0 0 1
R r rR r r
R r r
⎡ ⎤−= − +⎛ ⎞⎢ ⎥
→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦⎡ ⎤−⎢ ⎥
→ − = − +⎢ ⎥⎢ ⎥⎣ ⎦
There is no solution. The system is inconsistent.
67. 5
3 2 2 0x y z
x y z− + =⎧
⎨ + − =⎩
Write the augmented matrix: 1 1 1 53 2 2 0⎡ − ⎤⎢ ⎥−⎣ ⎦
( )
( )
( )
2 1 2
12 25
1 2 1
1 1 1 5 3
0 5 5 15
1 1 1 5
0 1 1 3
1 0 0 2
0 1 1 3
R r r
R r
R r r
⎡ − ⎤→ = − +⎢ ⎥− −⎣ ⎦
⎡ − ⎤→ =⎢ ⎥− −⎣ ⎦
⎡ ⎤→ = +⎢ ⎥− −⎣ ⎦
The matrix in the last step represents the system 2
3x
y z=⎧
⎨ − = −⎩or, equivalently,
23
xy z=⎧
⎨ = −⎩
Thus, the solution is 2x = , 3y z= − , z is any real number or {( , , ) | 2, 3,x y z x y z= = − z is any real number}.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.2: Systems of Linear Equations: Matrices
801 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
68. 2 4
3 1x y z
x y z+ − =⎧
⎨− + + =⎩
Write the augmented matrix: 2 1 1 41 1 3 1
⎡ − ⎤⎢ ⎥−⎣ ⎦
( )
( )
( )
1 2
2 1 2
12 25 3
3
43
1 2 153
interchange1 1 3 1
and 2 1 1 4
1 1 3 1 2
0 3 5 6
1 1 3 1 0 1 2
1 0 1
20 1
r r
R r r
R r
R r r
⎡ − − − ⎤ ⎛ ⎞→ ⎢ ⎥ ⎜ ⎟−− ⎝ ⎠⎣ ⎦
⎡ − − − ⎤→ = − +⎢ ⎥
⎣ ⎦⎡ ⎤− − −
→ =⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤−
→ = +⎢ ⎥⎢ ⎥⎣ ⎦
The matrix in the last step represents the system 4 13 5 23
x z
y z
⎧ − =⎪⎪⎨⎪ + =⎪⎩
or, equivalently,
413523
x z
y z
⎧ = +⎪⎪⎨⎪ = −⎪⎩
Thus, the solution is: 413
x z= + , 523
y z= − , z
is any real number or 4( , , ) 1 ,3
x y z x z⎧= +⎨
⎩
52 , is any real number3
y z z ⎫= − ⎬⎭
.
69.
2 3 300
3 5
x y zx y zx y z
x y z
+ − =⎧⎪ − − =⎪⎨ − + + =⎪⎪ + + =⎩
Write the augmented matrix: 2 3 1 31 1 1 01 1 1 0
1 1 3 5
⎡ − ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
1 2
2 1 2
3 1 3
4 1 4
1 1 1 0interchange2 3 1 3
and 1 1 1 0
1 1 3 5
1 1 1 02
0 5 1 3
0 0 0 00 2 4 5
r r
R r rR r rR r r
⎡ − − ⎤⎢ ⎥− ⎛ ⎞⎢ ⎥→ ⎜ ⎟−⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦⎡ − − ⎤
= − +⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥→ = +⎜ ⎟⎢ ⎥ ⎜ ⎟= − +⎝ ⎠⎢ ⎥
⎣ ⎦
3 4
1 1 1 0interchange0 5 1 3
and 0 2 4 5
0 0 0 0r r
⎡ − − ⎤⎢ ⎥ ⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦
( )
( )
2 3 2
1 2 1
3 2 3
13 319 18
18
1 1 1 00 1 7 7
20 2 4 50 0 0 0
1 0 8 70 1 7 7
20 0 18 19
0 0 0 0
71 0 870 1 7
=0 0 10 0 0 0
R r r
R r rR r r
R r
⎡ − − ⎤⎢ ⎥− −⎢ ⎥→ = − +⎢ ⎥⎢ ⎥⎣ ⎦⎡ − − ⎤⎢ ⎥ = +− − ⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥⎣ ⎦
−⎡ ⎤−⎢ ⎥−−⎢ ⎥→⎢ ⎥⎢ ⎥⎣ ⎦
The matrix in the last step represents the system 8 77 71918
x zy z
z
− = −⎧⎪ − = −⎪⎨⎪ =⎪⎩
Substitute and solve: 197 718
718
y
y
⎛ ⎞− =⎜ ⎟⎝ ⎠
=
198 718
139
x
x
⎛ ⎞− = −⎜ ⎟⎝ ⎠
=
Thus, the solution is 139
x = , 718
y = , 1918
z = or
13 7 19, ,9 18 18
⎛ ⎞⎜ ⎟⎝ ⎠
.
70.
3 12 4 0
3 2 12 5
x y zx y z
x y zx y
− + =⎧⎪ − − =⎪⎨ − + =⎪⎪ − =⎩
Write the augmented matrix: 1 3 1 12 1 4 01 3 2 11 2 0 5
⎡ − ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥−⎢ ⎥
−⎢ ⎥⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
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2 1 2
3 1 2
4 1 2
2 4
1 2 1
4 2
1 3 1 12
0 5 6 2
0 0 1 00 1 1 4
1 3 1 1interchange0 1 1 4
and 0 0 1 0
0 5 6 2
1 0 2 1330 1 1 4
0 0 1 0 50 0 1 22
R r rR r rR r r
r r
R r rR r
⎡ − ⎤= +⎛ ⎞⎢ ⎥− − ⎜ ⎟⎢ ⎥→ = − +⎜ ⎟⎢ ⎥ ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠−⎢ ⎥⎣ ⎦
⎡ − ⎤⎢ ⎥− ⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥
− −⎢ ⎥⎣ ⎦⎡ − ⎤⎢ ⎥ = +−⎢ ⎥→ ⎢ ⎥ = −⎢ ⎥
− −⎢ ⎥⎣ ⎦
4r⎛ ⎞⎜ ⎟+⎝ ⎠
1 3 1
2 3 2
4 3 4
1 0 0 132
0 1 0 4
0 0 1 00 0 0 22
R r rR r rR r r
⎡ ⎤= +⎛ ⎞⎢ ⎥
⎜ ⎟⎢ ⎥→ = +⎜ ⎟⎢ ⎥ ⎜ ⎟= +⎢ ⎥ ⎝ ⎠−⎢ ⎥⎣ ⎦
There is no solution. The system is inconsistent.
71. 4 4
2 3 3x y z w
x y z w+ + − =⎧
⎨ − + + =⎩
Write the augmented matrix:
1 2
4 1 1 1 41 1 2 3 3
interchange1 1 2 3 3
and 4 1 1 1 4 r r
⎡ − ⎤⎢ ⎥−⎣ ⎦
⎡ − ⎤ ⎛ ⎞→ ⎢ ⎥ ⎜ ⎟− ⎝ ⎠⎣ ⎦
( )2 1 21 1 2 3 3
40 5 7 13 8
R r r⎡ − ⎤
→ = − +⎢ ⎥− − −⎣ ⎦
The matrix in the last step represents the system 2 3 3
5 7 13 8
x y z wy z w− + + =⎧
⎨ − − = −⎩
The second equation yields 5 7 13 8
5 7 13 87 13 85 5 5
y z wy z w
y z w
− − = −= + −
= + −
The first equation yields 2 3 3
3 2 3x y z w
x y z w− + + =
= + − −
Substituting for y: 8 7 133 2 35 5 5
3 2 75 5 5
x z w z w
x z w
⎛ ⎞= + − + + − −⎜ ⎟⎝ ⎠
= − − +
Thus, the solution is 3 2 75 5 5
x z w= − − + ,
7 13 85 5 5
y z w= + − , z and w are any real numbers or
3 2 7 7 13 8( , , , ) , ,5 5 5 5 5 5
x y z w x z w y z w⎧= − − + = + −⎨
⎩
and are any real numbersz w ⎫⎬⎭
.
72. 4 5
2 54
x yx y z w
z w
− + =⎧⎪ − + − =⎨⎪ + =⎩
Write the augmented matrix: 4 1 0 0 5
2 1 1 1 50 0 1 1 4
⎡− ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥⎣ ⎦
( )
( )
( )
514 4
11 14
5144
1512 1 22 2
514 4
2 2
1 0 02 1 1 1 5 0 0 1 1 4
1 0 0
0 1 1 20 0 1 1 4
1 0 00 1 2 2 15 20 0 1 1 4
R r
R r r
R r
⎡ ⎤− −⎢ ⎥
→ − − = −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤−−⎢ ⎥
→ − − = − +⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤− −⎢ ⎥
→ − − = −⎢ ⎥⎢ ⎥⎣ ⎦
( )1 12 2
11 2 14
1 0 50 1 2 2 15 0 0 1 1 4
R r r
⎡ ⎤− −⎢ ⎥
→ − − = +⎢ ⎥⎢ ⎥⎣ ⎦
11 3 12
2 3 2
1 0 0 1 30 1 0 4 7
20 0 1 1 4
R r rR r r
⎡ − ⎤ ⎛ ⎞= +⎢ ⎥→ − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥⎣ ⎦
The matrix in the last step represents the system 3
4 7 4
x wy w
z w
+ = −⎧⎪ + = −⎨⎪ + =⎩
or, equivalently, 37 4
4
x wy wz w
= − −⎧⎪ = − −⎨⎪ = −⎩
The solution is 3x w= − − , 7 4y w= − − , 4z w= − , w is any real number or {( , , , ) | 3 ,x y z w x w= − −
7 4 , 4 , is any real number}y w z w w= − − = −
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Section 8.2: Systems of Linear Equations: Matrices
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73. Each of the points must satisfy the equation 2y ax bx c= + + .
(1,2) : 2( 2, 7) : 7 4 2
(2, 3) : 3 4 2
a b ca b ca b c
= + +− − − = − +
− − = + +
Set up a matrix and solve:
( )
2 1 2
3 1 3
1512 262 2
1 12 2
1 2 1512 2
3 2 3
1 1 1 22 74 1
34 2 1
1 1 1 2 40 3 15 6 4
20 3 11
1 1 1 2
0 120 3 11
01
0 1 22 60 0
R r rR r r
R r
R r rR r r
⎡ ⎤⎢ ⎥
− −⎢ ⎥⎢ ⎥−⎣ ⎦
⎡ ⎤= − +⎢ ⎥ ⎛ ⎞
→ − −− ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −⎣ − ⎦⎡ ⎤⎢ ⎥
= −⎢ ⎥→⎢ ⎥− −⎣ − ⎦⎡ ⎤−⎢ ⎥ = − +⎛ ⎞⎢ ⎥→ ⎜ ⎟= +⎢ ⎥ ⎝ ⎠− −⎣ ⎦
( )1 12 2
11 53 322 2
11 3 12
12 3 22
1 0
0 1
0 0 1 3
1 0 0 20 1 0 1 0 0 1 3
R r
R r r
R r r
⎡ ⎤−⎢ ⎥
→ → = −⎢ ⎥⎢ ⎥⎣ ⎦
−⎡ ⎤ ⎛ ⎞= − +⎢ ⎥ ⎜ ⎟→ ⎢ ⎥ ⎜ ⎟= − +⎝ ⎠⎢ ⎥⎣ ⎦
The solution is 2, 1, 3a b c= − = = ; so the
equation is 22 3y x x= − + + .
74. Each of the points must satisfy the equation 2y ax bx c= + + .
(1, 1) : 1(3, 1) : 1 9 3
(– 2,14) : 14 4 2
a b ca b ca b c
− − = + +− − = + +
= − +
Set up a matrix and solve:
2 1 2
3 1 3
1 1 1 19 3 1 14 1 142
1 1 1 1 90 8 6 8 40 3 186
R r rR r r
⎡ ⎤−⎢ ⎥
−⎢ ⎥⎢ ⎥−⎣ ⎦
⎡ ⎤− = − +⎢ ⎥ ⎛ ⎞→ − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−−⎣ ⎦
12 264 4
3 33 2 3
1 13 3
1 2 14 43 3 1
3 35
11 3 13
41 3 23
1 1 1 1
0 10 0 5 10
01
0 10 0 1 2
0 01 10 0 1 40 0 1 2
R r
R r r
R r r
R r
R r r
R r r
⎡ ⎤−⎢ ⎥ = −⎛ ⎞
−⎢ ⎥→ ⎜ ⎟⎜ ⎟= − +⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥ = − +⎛ ⎞⎢ ⎥−→ ⎜ ⎟⎜ ⎟⎢ ⎥ =⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
⎛ ⎞= +⎢ ⎥→ ⎜ ⎟−⎢ ⎥ ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠⎣ ⎦
The solution is 1, – 4, 2a b c= = = ; so the
equation is 2 4 2y x x= − + .
75. Each of the points must satisfy the equation 3 2( )f x ax bx cx d= + + + .
( 3) 112 : 27 9 3 112( 1) 2 : 2(1) 4 : 4(2) 13 : 8 4 2 13
f a b c df a b c df a b c df a b c d
− = − − + − + = −− = − − + − + = −= + + + == + + + =
Set up a matrix and solve: 27 9 3 1 112
21 1 1 11 1 1 1 48 134 2 1
⎡ ⎤− − −⎢ ⎥
−− −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
3 1
1 1 1 1 4Interchange21 1 1 1
and 27 9 3 1 1128 134 2 1
r r
⎡ ⎤⎢ ⎥ ⎛ ⎞−− −⎢ ⎥→ ⎜ ⎟⎢ ⎥− − − ⎝ ⎠⎢ ⎥⎣ ⎦
2 1 2
3 1 3
4 1 4
1 1 1 1 40 02 2 2 27
40 36 2824 84 60 7 19
R r rR r rR r r
⎡ ⎤= +⎛ ⎞⎢ ⎥
⎜ ⎟⎢ ⎥→ = +⎜ ⎟⎢ ⎥− ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠− − − −⎣ ⎦
( )12 22
1 2 1
3 2 3
4 2 4
1 1 1 1 40 01 1 1
40 36 28244 60 7 19
0 0 31 10 01 1 1 36
8 400 0 24 460 0 3 15
R r
R r rR r rR r r
⎡ ⎤⎢ ⎥⎢ ⎥→ =⎢ ⎥−⎢ ⎥
− − − −⎣ ⎦⎡ ⎤
= − +⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥→ = − +⎜ ⎟⎢ ⎥− − ⎜ ⎟= +⎢ ⎥ ⎝ ⎠
− − −⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
804 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
( )13 3245 5
3 3
0 0 31 10 01 1 1 0 0 1
60 0 3 15
R r
⎡ ⎤⎢ ⎥⎢ ⎥→ =⎢ ⎥− −⎢ ⎥⎢ ⎥− − −⎣ ⎦
( )
1 143 3
1 3 1
51 4 3 43 3
1 143 3
14 4551
3 3
11 4 13
2 4 21
3 4 33
0 010 01 1 1
60 0 1250 0 0 5
0 010 01 1 1 0 0 10 0 0 51
0 0 0 3140 0 01
0 0 0 010 0 0 51
R r rR r r
R r
R r r
R r r
R r r
⎡ ⎤⎢ ⎥
= − +⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ = +⎝ ⎠− −⎢ ⎥⎢ ⎥−−⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥→ = −⎢ ⎥
− −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤ ⎛ = − +⎢ ⎥ ⎜−⎢ ⎥→ = − +⎜⎢ ⎥ ⎜⎢ ⎥ = +⎝⎣ ⎦
⎞⎟⎟⎟⎜ ⎟⎠
The solution is 3, 4, 0, 5a b c d= = − = = ; so the
equation is 3 2( ) 3 4 5f x x x= − + .
76. Each of the points must satisfy the equation 3 2( )f x ax bx cx d= + + + .
( 2) 10 : 8 4 2 10( 1) 3 : 3(1) 5 : 5(3) 15 : 27 9 3 15
f a b c df a b c df a b c df a b c d
− = − − + − + = −− = − + − + == + + + == + + + =
Set up a matrix and solve:
3 1
2 1 2
3 1 3
4 1 4
104 18 231 1 1 151 1 1 1
27 9 3 151
51 1 1 1Interchange31 1 1 1
and 104 18 227 9 3 151
51 1 1 10 0 82 2 80 6 9 3012 270 18 12024 26
r r
R r rR r rR r r
⎡ ⎤−− −⎢ ⎥− −⎢ ⎥
⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥ ⎛ ⎞− −⎢ ⎥→ ⎜ ⎟⎢ ⎥− ⎝ ⎠− −⎢ ⎥⎣ ⎦⎡ ⎤
= +⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥→ = +⎜ ⎟⎢ ⎥ ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠
− −⎣ ⎦− −
( )12 22
51 1 1 10 01 1 4 0 6 9 30120 18 12024 26
R r
⎡ ⎤⎢ ⎥⎢ ⎥→ =⎢ ⎥⎢ ⎥
− −⎣ ⎦− −
( )
1 2 1
3 2 3
4 2 4
13 361
2
0 01 1 10 01 1 4 120 0 6 3 18 180 0 24 8 48
0 01 1 10 01 1 4
0 0 310 0 24 8 48
R r rR r rR r r
R r
⎡ ⎤= − +⎛ ⎞⎢ ⎥
⎜ ⎟⎢ ⎥→ = − +⎜ ⎟⎢ ⎥− − ⎜ ⎟= +⎢ ⎥ ⎝ ⎠⎣ ⎦− − −⎡ ⎤⎢ ⎥⎢ ⎥→ =⎢ ⎥
− −⎢ ⎥⎢ ⎥⎣ ⎦− − −
( )
12
1 3 1
1 4 3 42
12
14 4201
2
11 4 12
2 4 21
3 4 32
0 01 40 01 1 4
240 0 310 0 0 12020
0 01 40 01 1 4 0 0 310 0 0 61
0 0 01 10 0 01 2 0 0 0 010 0 0 61
R r rR r r
R r
R r r
R r r
R r r
⎡ ⎤⎢ ⎥
= − +⎢ ⎥ ⎛ ⎞→ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥− −
⎢ ⎥−⎣ ⎦−
⎡ ⎤⎢ ⎥⎢ ⎥
→ = −⎢ ⎥⎢ ⎥− −⎢ ⎥⎣ ⎦⎡ ⎤ ⎛ ⎞= − +⎢ ⎥ ⎜−⎢ ⎥ ⎜→ = − +⎢ ⎥ ⎜
⎜⎢ ⎥ = +⎝⎣ ⎦
⎟⎟⎟⎟⎠
The solution is 1, 2, 0, 6a b c d= = − = = ; so the
equation is 3 2( ) 2 6f x x x= − + .
77. Let x = the number of servings of salmon steak. Let y = the number of servings of baked eggs. Let z = the number of servings of acorn squash. Protein equation: 30 15 3 78x y z+ + = Carbohydrate equation: 20 2 25 59x y z+ + = Vitamin A equation: 2 20 32 75x y z+ + = Set up a matrix and solve:
( )
3 1
11 12
30 15 3 7820 25 592
20 32 752
20 32 752Interchange
20 25 59 2 and r
30 15 3 78
10 16 37.5120 25 59 230 15 3 78
r
R r
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥ ⎛ ⎞
→ ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥
→ =⎢ ⎥⎢ ⎥⎣ ⎦
2 1 2
3 1 3
10 16 37.5120
0 198 295 691 30
0 285 477 1047
R r rR r r
⎡ ⎤= − +⎢ ⎥ ⎛ ⎞
→ − − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− − −⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.2: Systems of Linear Equations: Matrices
805 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
( )
( )
953 2 366
3457 345766 66
663 33457
10 16 37.510 198 295 691
0 0
10 16 37.510 198 295 691 0 0 1 1
R r r
R r
⎡ ⎤⎢ ⎥
→ − − − = − +⎢ ⎥⎢ ⎥− −⎣ ⎦⎡ ⎤⎢ ⎥
→ − − − = −⎢ ⎥⎢ ⎥⎣ ⎦
Substitute 1z = and solve: 198 295(1) 691
198 3962
yyy
− − = −− = −
=
10(2) 16(1) 37.5
36 37.51.5
xx
x
+ + =+ =
=
The dietitian should serve 1.5 servings of salmon steak, 2 servings of baked eggs, and 1 serving of acorn squash.
78. Let x = the number of servings of pork chops. Let y = the number of servings of corn on the cob. Let z = the number of servings of 2% milk. Protein equation: 23 3 9 47x y z+ + = Carbohydrate equation: 16 13 58y z+ = Calcium equation: 10 10 300 630x y z+ + = Set up a matrix and solve:
13 110
3 1 313 2916 8 1
2 216
467 47516 813 2916
23 3 9 470 16 13 58
10 10 300 630
30 631 1 Interchange0 16 13 58
and 23 3 9 47
30 631 123
0 10 140220 6811 0
0 1
r r
R r r
R r
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎛ ⎞⎢ ⎥
→ ⎜ ⎟⎢ ⎥ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦⎡ ⎤⎢ ⎥ = − +⎛ ⎞⎢ ⎥ ⎜ ⎟→
⎜ ⎟⎢ ⎥ =⎝ ⎠−⎣ ⎦− −
→
( )
1 2 18
3 2 32659
4467 47516 813 29 4
3 316 8 2659
4671 3 116
132 3 216
20
0 0 1402
1 0
0 1
0 0 1 2
1 0 0 10 1 0 2 0 0 1 2
R r rR r r
R r
R r r
R r r
⎡ ⎤⎢ ⎥ = − +⎛ ⎞⎢ ⎥ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥− −⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥→ = −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤ ⎛ ⎞= − +⎢ ⎥ ⎜→ ⎢ ⎥ ⎜ = − +⎢ ⎥ ⎝ ⎠⎣ ⎦
⎟⎟
The dietitian should provide 1 serving of pork chops, 2 servings of corn on the cob, and 2 servings of 2% milk.
79. Let x = the amount invested in Treasury bills. Let y = the amount invested in Treasury bonds. Let z = the amount invested in corporate bonds. Total investment equation: 10,000x y z+ + = Annual income equation: 0.06 0.07 0.08 680x y z+ + = Condition on investment equation: 0.5
2 0z x
x z=
− =
Set up a matrix and solve: 10,0001 1 1
0.06 0.07 0.08 6800 021
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦
( )
2 1 2
3 1 3
2 2
10,0001 1 10.06
0 0.01 0.02 80 10,0000 31
10,0001 1 10 8000 1001 2
10,0000 31
R r rR r r
R r
⎡ ⎤= − +⎢ ⎥ ⎛ ⎞
→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−−−⎣ ⎦⎡ ⎤⎢ ⎥
→ =⎢ ⎥⎢ ⎥−−−⎣ ⎦
( )
1 2 1
3 2 3
3 3
1 3 1
2 3 2
0 20001 10 8000 1 20 0 20001
0 20001 10 8000 1 20 0 20001
0 0 400010 0 4000 1
20 0 20001
R r rR r r
R r
R r rR r r
⎡ ⎤−= − +⎢ ⎥ ⎛ ⎞
→ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥−−⎣ ⎦⎡ ⎤−⎢ ⎥
→ = −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
= +⎢ ⎥→ ⎢ ⎥ = − +⎢ ⎥
⎣ ⎦
⎛ ⎞⎜ ⎟⎝ ⎠
Carletta should invest $4000 in Treasury bills, $4000 in Treasury bonds, and $2000 in corporate bonds.
80. Let x = the fixed delivery charge; let y = the cost of each tree, and let z = the hourly labor charge. 1st subdivision: 250 166 7520x y z+ + = 2nd subdivision: 200 124 5945x y z+ + = 3rd subdivision: 300 200 8985x y z+ + = Set up a matrix and solve:
250 166 75201200 59451 124300 200 89851
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
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2 1 2
3 3 1
12 250
13 350
250 166 752010 50 1575 420 50 34 1465
250 166 752010 0.84 31.5 10 0.68 29.31
R r rR r r
R r
R r
⎡ ⎤= −⎢ ⎥ ⎛ ⎞
→ ⎜ ⎟⎢ ⎥ = −⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤ ⎛ ⎞=⎢ ⎥ ⎜ ⎟→ ⎢ ⎥ ⎜ ⎟=⎢ ⎥ ⎝ ⎠⎣ ⎦
( )
1 1 2
3 2 3
13 30.16
1 1 3
2 2 3
0 3551 44250
0 0.84 31.5 10 0 0.18 2.2
0 3551 440 0.84 31.5 10 0 13.751
0 0 250144
0 0 19.95 10.84
0 0 13.751
R r rR r r
R r
R r rR r r
⎡ ⎤−−= −⎢ ⎥ ⎛ ⎞
→ ⎜ ⎟⎢ ⎥ = −⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤−−⎢ ⎥
→ =⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
= +⎢ ⎥ ⎛ ⎞→ ⎜ ⎟⎢ ⎥ = −⎝ ⎠⎢ ⎥
⎣ ⎦
The delivery charge is $250 per job, the cost for each tree is $19.95, and the hourly labor charge is $13.75.
81. Let x = the number of Deltas produced. Let y = the number of Betas produced. Let z = the number of Sigmas produced. Painting equation: 10 16 8 240x y z+ + = Drying equation: 3 5 2 69x y z+ + = Polishing equation: 2 3 41x y z+ + = Set up a matrix and solve:
( )
( )
1 2 1
2 1 2
3 1 3
12 22
10 16 8 2403 5 2 692 3 1 41
1 1 2 333 5 2 69 32 3 1 41
1 1 2 333
0 2 4 30 2
0 1 3 25
1 1 2 330 1 2 15 0 1 3 25
1 0 4 480 1 2 150 0 1 10
R r r
R r rR r r
R r
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥→ = − +⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
= − +⎛ ⎞⎢ ⎥→ − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −⎣ ⎦⎡ ⎤⎢ ⎥→ − − =⎢ ⎥⎢ ⎥− −⎣ ⎦⎡⎢→ − −⎢
− −⎣
1 1 2
3 3 2
R r rR r r
⎤= −⎛ ⎞⎥
⎜ ⎟⎥ = −⎝ ⎠⎢ ⎥⎦
( )3 3
1 3 1
2 3 2
1 0 4 480 1 2 15 0 0 1 10
1 0 0 84
0 1 0 5 2
0 0 1 10
R r
R r rR r r
⎡ ⎤⎢ ⎥→ − − = −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
= − +⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥⎣ ⎦
The company should produce 8 Deltas, 5 Betas, and 10 Sigmas.
82. Let x = the number of cases of orange juice produced; let y = the number of cases of grapefruit juice produced; and let z = the number of cases of tomato juice produced. Sterilizing equation: 9 10 12 398x y z+ + = Filling equation: 6 4 4 164x y z+ + = Labeling equation: 2 58x y z+ + = Set up a matrix and solve:
( )
1 3
2 1 2
3 1 3
1 12 24 8
9 10 398126 1644 4
581 2 1
581 2 1Interchange
6 164 4 4 and
9 10 39812
581 2 16
0 184 8 29
0 3 1248
581 2 1
0 23 10 3 1248
r r
R r rR r r
R r
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥ ⎛ ⎞
→ ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥
⎣ ⎦⎡ ⎤
= − +⎢ ⎥ ⎛ ⎞→ −− − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−−⎣ ⎦
⎡ ⎤⎢ ⎥⎢ ⎥→ = −⎢ ⎥⎣ − ⎦−
( )
12
1 2 114
3 2 3
121 1
3 34 5
01 122
0 23 18
0 0 5 60
01 12
0 23 10 0 1 12
R r rR r r
R r
⎡ ⎤⎢ ⎥ = − +⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥→ =⎢ ⎥⎣ ⎦
11 3 12
12 3 24
0 0 610 0 20 10 0 1 12
R r r
R r r
⎡ ⎤⎛ ⎞= − +⎢ ⎥⎜ ⎟→ ⎢ ⎥ ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠
⎣ ⎦
The company should prepare 6 cases of orange juice, 20 cases of grapefruit juice, and 12 cases of tomato juice.
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Section 8.2: Systems of Linear Equations: Matrices
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83. Rewrite the system to set up the matrix and solve:
2 2
4 1 1 4
3 1 1 3
3 4 1 1 3 4
4 8 2 0 2 48 5 5 84 3 3 4
0
I II I I II I I I
I I I I I I
− + − = =⎧ ⎧⎪ ⎪= + + =⎪ ⎪→⎨ ⎨= + + =⎪ ⎪⎪ ⎪+ = − − =⎩ ⎩
2 1
12 22
3 1 3
4 1 4
0 0 02 40 0 5 810 3 01 40 01 1 1
0 0 5 81Interchange0 0 02 4
and 0 3 01 40 01 1 10 0 5 81
0 0 01 2 40 0 3 5
6 80 0 1
r r
R r
R r rR r r
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ − − ⎦
⎡ ⎤⎢ ⎥ ⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ − − ⎦⎡ ⎤ ⎛ ⎞=⎢ ⎥ ⎜ ⎟⎢ ⎥→ = − +⎜ ⎟⎢ ⎥−− ⎜ ⎟⎜ ⎟= − +⎢ ⎥ ⎝ ⎠− −⎣ − ⎦
( )
3 4
3 3
4 3 4
14 423
2823
0 0 5 81Interchange0 0 01 2
and 6 80 0 140 0 3 5
0 0 5 810 0 01 2
30 0 6 8123 280 0 0
0 0 5 810 0 01 2 0 0 6 81
0 0 0 1
r r
R rR r r
R r
⎡ ⎤⎢ ⎥ ⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥− −− ⎝ ⎠⎢ ⎥
−−⎣ ⎦⎡ ⎤⎢ ⎥ = −⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥
− −⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥→ = −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
4423
1 4 116
3 4 323
2823
0 0 0150 0 01 2
60 0 01
0 0 0 1
R r rR r r
⎡ ⎤⎢ ⎥
= − +⎢ ⎥ ⎛ ⎞→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥
⎢ ⎥⎣ ⎦
The solution is 14423
I = , 2 2I = , 31623
I = ,
42823
I = .
84. Rewrite the system to set up the matrix and solve: 1 3 2 1 2 3
1 3 1 3
1 2 1 2
024 6 3 0 6 3 24
12 24 6 6 0 6 6 36
I I I I I II I I II I I I
= + − − =⎧ ⎧⎪ ⎪− − = → − − = −⎨ ⎨⎪ ⎪+ − − = − − = −⎩ ⎩
( )
2 1 2
3 1 3
3 12 22 6
12
1 2 132
3 2 3
12
1 1 1 06 0 3 246 6 0 36
1 1 1 06
0 6 9 24 6
0 12 6 36
1 1 1 00 1 4 0 12 6 36
1 0 40 1 4
120 0 12 12
1 0 40 1
R r rR r r
R r
R r rR r r
− −⎡ ⎤⎢ ⎥− − −⎢ ⎥⎢ ⎥− − −⎣ ⎦
− −⎡ ⎤= +⎛ ⎞⎢ ⎥→ − − − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥− − −⎣ ⎦
− −⎡ ⎤⎢ ⎥→ = −⎢ ⎥⎢ ⎥− − −⎣ ⎦⎡ ⎤
= +⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥⎣ ⎦
→ ( )3 13 32 12
11 3 12
32 3 22
4 0 0 1 1
1 0 0 3.50 1 0 2.5 0 0 1 1
R r
R r r
R r r
⎡ ⎤⎢ ⎥ =⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤ ⎛ ⎞= − +⎢ ⎥ ⎜ ⎟→ ⎢ ⎥ ⎜ ⎟= − +⎝ ⎠⎢ ⎥⎣ ⎦
The solution is 1 2 33.5, 2.5, 1I I I= = = .
85. Let x = the amount invested in Treasury bills. Let y = the amount invested in corporate bonds. Let z = the amount invested in junk bonds.
a. Total investment equation: 20,000x y z+ + = Annual income equation: 0.07 0.09 0.11 2000x y z+ + = Set up a matrix and solve:
( )
( )
( )
2 2
2 2 1
12 22
1 1 1 20,0000.07 0.09 0.11 2000
1 1 1 20,000 100
7 9 11 200,000
1 1 1 20,000 7
0 2 4 60,000
1 1 1 20,000
0 1 2 30,000
R r
R r r
R r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ =⎢ ⎥
⎣ ⎦⎡ ⎤
→ = −⎢ ⎥⎣ ⎦⎡ ⎤
→ =⎢ ⎥⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
808 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
( )1 1 21 0 1 10,000
0 1 2 30,000
R r r⎡ − − ⎤
→ = −⎢ ⎥⎣ ⎦
The matrix in the last step represents the
system 10,000
2 30,000
x zy z
− = −⎧⎨ + =⎩
Therefore the solution is 10,000x z= − + , 30,000 2y z= − , z is any real number.
Possible investment strategies:
Amount Invested At
7% 9% 11%
0 10,000 10,000
1000 8000 11,000
2000 6000 12,000
3000 4000 13,000
4000 2000 14,000
5000 0 15,000 b. Total investment equation:
25,000x y z+ + = Annual income equation: 0.07 0.09 0.11 2000x y z+ + = Set up a matrix and solve:
( )
( )
( )
( )
2 2
2 2 1
12 22
1 1 2
1 1 1 25,0000.07 0.09 0.11 2000
1 1 1 25,000 100
7 9 11 200,000
1 1 1 25,000 7
0 2 4 25,000
1 1 1 25,000
0 1 2 12,500
1 0 112,500
0 1 2 12,500
R r
R r r
R r
R r r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ =⎢ ⎥
⎣ ⎦⎡ ⎤
→ = −⎢ ⎥⎣ ⎦⎡ ⎤
→ =⎢ ⎥⎣ ⎦⎡ − ⎤
→ = −⎢ ⎥⎣ ⎦
The matrix in the last step represents the
system 12,500
2 12,500
x zy z
− =⎧⎨ + =⎩
Thus, the solution is 12,500x z= + , 2 12,500y z= − + , z is any real number.
Possible investment strategies:
Amount Invested At
7% 9% 11%
12,500 12,500 0
14,500 8500 2000
16,500 4500 4000
18,750 0 6250 c. Total investment equation:
30,000x y z+ + = Annual income equation: 0.07 0.09 0.11 2000x y z+ + = Set up a matrix and solve:
( )
( )
( )
( )
2 2
1 2 1
12 22
1 1 2
1 1 1 30,0000.07 0.09 0.11 2000
1 1 1 30,000 100
7 9 11 200,000
1 1 1 30,000 7
0 2 4 10,000
1 1 1 30,000
0 1 2 5000
1 0 1 35,000
0 1 2 5000
R r
R r r
R r
R r r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ =⎢ ⎥
⎣ ⎦⎡ ⎤
→ = −⎢ ⎥−⎣ ⎦⎡ ⎤
→ =⎢ ⎥−⎣ ⎦⎡ − ⎤
→ = −⎢ ⎥−⎣ ⎦
The matrix in the last step represents the
system 35,000
2 5000
x zy z
− =⎧⎨ + = −⎩
Thus, the solution is 35,000x z= + , 2 5000y z= − − , z is any real number.
However, y and z cannot be negative. From 2 5000y z= − − , we must have 0.y z= =
One possible investment strategy
Amount Invested At
7% 9% 11%
30,000 0 0 This will yield ($30,000)(0.07) = $2100,
which is more than the required income.
d. Answers will vary.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.2: Systems of Linear Equations: Matrices
809 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
86. Let x = the amount invested in Treasury bills. Let y = the amount invested in corporate bonds. Let z = the amount invested in junk bonds. Let I = income Total investment equation: 25,000x y z+ + = Annual income equation: 0.07 0.09 0.11x y z I+ + = Set up a matrix and solve:
( )
( )
( )
( )
2 2
1 2 1
12 22
1 1 2
1 1 1 25,0000.07 0.09 0.11
1 1 1 25,000 100
7 9 11 100
1 1 1 25,000 7
0 2 4 100 175,000
1 1 1 25,000
0 1 2 50 87,500
1 0 1112,500 50
0 1 2 50 87,500
I
R rI
R r rI
R rI
IR r r
I
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ =⎢ ⎥
⎣ ⎦⎡ ⎤
→ = −⎢ ⎥−⎣ ⎦⎡ ⎤
→ =⎢ ⎥−⎣ ⎦⎡ − − ⎤
→ = −⎢ ⎥−⎣ ⎦
The matrix in the last step represents the system 112,500 50
2 50 87,500
x z Iy z I
− = −⎧⎨ + = −⎩
Thus, the solution is 112,500 50x I z= − + , 50 87,500 2y I z= − − , z is any real number.
a. 1500I =
( )112,500 50112,500 50 150037,500
x I zz
z
= − +
= − +
= +
( )50 87,500 250 1500 87,500 2
12,500 2
y I zz
z
= − −
= − −
= − −
z is any real number. Since y and z cannot be negative, we must have 0.y z= = Investing all of the money at 7% yields $1750, which is more than the $1500 needed.
b. 2000I =
( )112,500 50112,500 50 200012,500
x T zz
z
= − +
= − +
= +
( )50 87,500 250 2000 87,500 212,500 2
y I zz
z
= − −
= − −
= −
z is any real number.
Possible investment strategies:
Amount Invested At
7% 9% 11%
12,500 12,500 0
15,500 6500 3000
18,750 0 6250 c. 2500I =
( )112,500 50112,500 50 2500
12,500
x T zz
z
= − +
= − +
= − +
( )50 87,500 250 2500 87,500 237,500 2
y I zz
z
= − −
= − −
= −
z is any real number.
Possible investment strategies:
Amount invested at
7% 9% 11%
0 12,500 12,500
1000 10,500 13,500
6250 0 18,750 87. Let x = the amount of supplement 1.
Let y = the amount of supplement 2. Let z = the amount of supplement 3.
0.20 0.40 0.30 40 Vitamin C0.30 0.20 0.50 30 Vitamin D
x y zx y z+ + =⎧
⎨ + + =⎩
Multiplying each equation by 10 yields 2 4 3 4003 2 5 300
x y zx y z+ + =⎧
⎨ + + =⎩
Set up a matrix and solve:
( )
( )
( )
312
1 12
32
2 2 112
312
2 21 48
2 4 3 4003 2 5 300
1 2 200
3 2 5 300
1 2 200 3
0 4 300
1 2 200
0 1 75
R r
R r r
R r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ =⎢ ⎥
⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥→ = −
− −⎢ ⎥⎣ ⎦⎡ ⎤
→ = −⎢ ⎥−⎢ ⎥⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
810 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
( )74
1 1 218
1 0 50 2
0 1 75R r r
⎡ ⎤→ = −⎢ ⎥
−⎢ ⎥⎣ ⎦
The matrix in the last step represents the system
7418
50
75
x z
y z
⎧ + =⎪⎨
− =⎪⎩
Therefore the solution is 7504
x z= − ,
1758
y z= + , z is any real number.
Possible combinations:
Supplement 1 Supplement 2 Supplement 3
50mg 75mg 0mg
36mg 76mg 8mg
22mg 77mg 16mg
8mg 78mg 24mg 88. Let x = the amount of powder 1.
Let y = the amount of powder 2. Let z = the amount of powder 3.
120.20 0.40 0.30 12 Vitamin B0.30 0.20 0.40 12 Vitamin E
x y zx y z+ + =⎧
⎨ + + =⎩
Multiplying each equation by 10 yields 2 4 3 1203 2 4 120
x y zx y z+ + =⎧
⎨ + + =⎩
Set up a matrix and solve:
( )
( )
32 2 12
1 1 2
2 4 3 1203 2 4 120
2 4 3 120
0 4 0.5 60
2 0 2.5 60
0 4 0.5 60
R r r
R r r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ = −⎢ ⎥− − −⎣ ⎦
⎡ ⎤→ = +⎢ ⎥− − −⎣ ⎦
The matrix in the last step represents the system 2 2.5 60
4 0.5 60
x zy z+ =⎧
⎨− − = −⎩
Thus, the solution is 30 1.25x z= − , 15 0.125y z= − , z is any real number.
Possible combinations:
Powder 1 Powder 2 Powder 3
30 units 15 units 0 units
20 units 14 units 8 units
10 units 13 units 16 units
0 units 12 units 24 units 89 – 91. Answers will vary. Section 8.3
1. determinants
2. ad bc−
3. False
4. False
5. 3 1
3(2) 4(1) 6 4 24 2
= − = − =
6. 6 1
6(2) 5(1) 12 5 75 2
= − = − =
7. 6 4 6(3) ( 1)(4) 18 4 2231= − − = + =
−
8. 8 3 8(2) 4( 3) 16 12 284 2
− = − − = + =
9. 3 1 3(2) 4( 1) 6 4 24 2− − = − − − = − + = −
10. 24 4(3) ( 5)(2) 12 10 25 3
− = − − − = − + = −−
11.
( ) [ ][ ]
3 4 25 51 1 1 151 1 3 4 22 22 1 1 221 2
3 1 ( 2) 2(5) 4 1( 2) 1(5)
2 1(2) 1( 1)3( 8) 4( 7) 2(3)
24 28 610
− −− = − +− −−
= − − − − − −⎡ ⎤⎣ ⎦+ − −
= − − − += − + +=
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.3: Systems of Linear Equations: Determinants
811 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12.
[ ] [ ][ ]
31 25 6 5 61 16 5 1 3 ( 2)13 8 3 82 28 32
1 1(3) 2( 5) 3 6(3) 8( 5)
2 6(2) 8(1)1(13) 3(58) 2(4)13 174 8
169
−− −− = − + −
= − − − − −
− −
= − −= − −= −
13.
[ ] [ ][ ]
4 1 20 6 0 61 16 0 4 ( 1) 21
3 34 1 4 131 4
4 1(4) 0( 3) 1 6(4) 1(0)
2 6( 3) 1( 1)4( 4) 1(24) 2( 17)
16 24 3426
−− −= − − +−− −−
= − − − + −
+ − − −
= − + + −= − + −= −
14.
[ ] [ ][ ]
3 9 40 04 1 1 401 4 3 ( 9) 4
3 8 8 31 18 3 1
3 4(1) ( 3)(0) 9 1(1) 8(0)
4 1( 3) 8(4)3(4) 9(1) 4( 35)12 9 140
119
−= − − +
− −−
= − − + −
+ − −
= + + −= + −= −
15. 84
x yx y+ =⎧
⎨ − =⎩
1 1 1 1 21 18 1 8 4 124 1
81 4 8 41 4
x
y
D
D
D
= = − − = −−
= = − − = −−
= = − = −
Find the solutions by Cramer's Rule: 12 46 22 2
yx DDx y
D D− −
= = = = = =− −
The solution is (6, 2).
16. 2 5
3x yx y+ =⎧
⎨ − =⎩
1 2 1 2 31 15 2 5 6 113 1
51 3 5 231
x
y
D
D
D
= = − − = −−
= = − − = −−
= = − = −
Find the solutions by Cramer's Rule: 11 11 2 23 3 3 3
yx DDx y
D D− −
= = = = = =− −
The solution is 11 2,3 3
⎛ ⎞⎜ ⎟⎝ ⎠
.
17. 5 132 3 12
x yx y− =⎧
⎨ + =⎩
5 1 15 2 1732
13 1 39 12 51312
5 13 60 26 342 12
x
y
D
D
D
−= = + =
−= = + =
= = − =
Find the solutions by Cramer's Rule: 51 343 217 17
yx DDx y
D D= = = = = =
The solution is (3, 2).
18. 3 5
2 3 8x yx y+ =⎧
⎨ − = −⎩
31 3 6 932
5 3 15 ( 24) 98 3
51 8 10 1882
x
y
D
D
D
= = − − = −−
= = − − − =− −
= = − − = −−
Find the solutions by Cramer's Rule: 9 181 29 9
yx DDx y
D D−
= = = − = = =− −
The solution is ( 1, 2)− .
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Chapter 8: Systems of Equations and Inequalities
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19. 3 24
2 0x
x y=⎧
⎨ + =⎩
3 0 6 0 61 2
024 48 0 480 2
3 24 0 24 2401
x
y
D
D
D
= = − =
= = − =
= = − = −
Find the solutions by Cramer's Rule: 48 248 46 6
yx DDx y
D D−
= = = = = = −
The solution is (8, 4)− .
20. 4 5 3
2 4x y
y+ = −⎧
⎨ − = −⎩
54 8 0 80 2
3 5 6 ( 20) 264 2
34 16 0 160 4
x
y
D
D
D
= = − − = −−
−= = − − =− −
−= = − − = −−
Find the solutions by Cramer's Rule: 26 13 16 2
8 4 8yx DD
x yD D
−= = = − = = =
− −
The solution is 13 , 24
⎛ ⎞−⎜ ⎟⎝ ⎠
.
21. 3 6 245 4 12
x yx y− =⎧
⎨ + =⎩
3 6 12 ( 30) 425 4
624 96 ( 72) 16812 43 24 36 120 845 12
x
y
D
D
D
−= = − − =
−= = − − =
= = − = −
Find the solutions by Cramer's Rule: 168 844 242 42
yx DDx y
D D−
= = = = = = −
The solution is (4, 2)− .
22. 2 4 163 5 9x yx y+ =⎧
⎨ − = −⎩
2 4 10 12 223 5
16 4 80 36 449 5
x
D
D
= = − − = −−
= = − + = −− −
162 18 48 663 9
yD = = − − = −−
Find the solutions by Cramer's Rule: 44 662 322 22
yx DDx y
D D− −
= = = = = =− −
The solution is (2, 3).
23. 3 2 46 4 0
x yx y− =⎧
⎨ − =⎩
23 12 ( 12) 046
D −= = − − − =−
Since 0D = , Cramer's Rule does not apply.
24. 2 5
4 8 6x yx y
− + =⎧⎨ − =⎩
1 2 8 8 04 8
D −= = − =−
Since 0D = , Cramer's Rule does not apply.
25. 2 4 23 2 3
x yx y− = −⎧
⎨ + =⎩
42 4 12 163 2
2 4 4 12 83 2
22 6 6 123 3
x
y
D
D
D
−= = + =
− −= = − + =
−= = + =
Find the solutions by Cramer's Rule: 8 1 12 3
16 2 16 4yx DD
x yD D
= = = = = =
The solution is 1 3,2 4
⎛ ⎞⎜ ⎟⎝ ⎠
.
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Section 8.3: Systems of Linear Equations: Determinants
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26. 3 3 3
84 23
x y
x y
+ =⎧⎪⎨
+ =⎪⎩
83
83
3 3 6 12 64 23 3
6 8 22
3 38 12 4
4
x
y
D
D
D
= = − = −
= = − = −
= = − = −
Find the solutions by Cramer's Rule: 2 1 4 26 3 6 3
yx DDx y
D D− −
= = = = = =− −
The solution is 1 2,3 3
⎛ ⎞⎜ ⎟⎝ ⎠
.
27. 2 3 1
10 10 5x yx y− = −⎧
⎨ + =⎩
32 20 ( 30) 5010 10
31 10 ( 15) 55 10
2 1 10 ( 10) 2010 5
x
y
D
D
D
−= = − − =
−−= = − − − =
−= = − − =
Find the solutions by Cramer's Rule: 5 1 20 2
50 10 50 5yx DD
x yD D
= = = = = =
The solution is 1 2,10 5⎛ ⎞⎜ ⎟⎝ ⎠
.
28. 3 2 05 10 4
x yx y− =⎧
⎨ + =⎩
3 2 30 ( 10) 405 10
0 2 0 ( 8) 8104
3 0 12 0 125 4
x
y
D
D
D
−= = − − =
−= = − − =
= = − =
Find the solutions by Cramer's Rule: 8 1 12 340 5 40 10
yx DDx y
D D= = = = = =
The solution is 1 3,5 10
⎛ ⎞⎜ ⎟⎝ ⎠
.
29. 2 3 6
12
x y
x y
+ =⎧⎪⎨
− =⎪⎩
12
12
32 2 3 51 16 3 3 156
2 2162 1 6 5
1
x
y
D
D
D
= = − − = −−
= = − − = −−
= = − = −
Find the solutions by Cramer's Rule: 15
3 52 15 2 5
yx DDx y
D D
− −= = = = = =
− −
The solution is 3 ,2
1⎛ ⎞⎜ ⎟⎝ ⎠
.
30. 1 22
2 8
x y
x y
⎧ + = −⎪⎨⎪ − =⎩
12
12
1 1 1 21 22 1 4 8 48 2
2 4 ( 2) 681
x
y
D
D
D
= = − − = −−
−= = − = −−
−= = − − =
Find the solutions by Cramer's Rule: 4 62 32 2
yx DDx y
D D−
= = = = = = −− −
The solution is (2, 3)− .
31. 3 5 3
15 5 21x yx y− =⎧
⎨ + =⎩
3 5 15 ( 75) 9015 5
3 5 15 ( 105) 120521
3 3 63 45 1815 21
x
y
D
D
D
−= = − − =
−= = − − =
= = − =
Find the solutions by Cramer's Rule: 120 4 18 190 3 90 5
yx DDx y
D D= = = = = =
The solution is 4 1,3 5
⎛ ⎞⎜ ⎟⎝ ⎠
.
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Chapter 8: Systems of Equations and Inequalities
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32. 2 1
1 32 2
x y
x y
− = −⎧⎪⎨
+ =⎪⎩
12
3 12 2
32
2 1 1 1 21
1 1 1 3 12 2
2 1 3 1 41
x
y
D
D
D
−= = + =
− −= = − + =
−= = + =
Find the solutions by Cramer's Rule: 1 4 22 2
yx DDx y
D D= = = = =
The solution is 1 ,2
2⎛ ⎞⎜ ⎟⎝ ⎠
.
33. 6
3 2 53 2 14
x y zx y zx y z
+ − =⎧⎪ − + = −⎨⎪ + − =⎩
1 1 123 1
231
2 23 31 11 1 ( 1)2 23 31 1
1(4 3) 1( 6 1) 1(9 2)1 7 11
3
D−
−=−
− −= − + −− −
= − − − − − += + −= −
6 1 125 1
2314
2 25 51 16 1 ( 1)2 23 314 14
6(4 3) 1(10 14) 1( 15 28)6 4 13
3
xD−
−= −−
− −− −= − + −− −
= − − − − − += + −= −
61 13 5 1
21 14
5 3 3 51 11 6 ( 1)2 214 1 1 14
1(10 14) 6( 6 1) 1(42 5)4 42 479
yD−
= −−
− −= − + −− −
= − − − − − += − + −= −
61 123 531 14
2 25 3 5 31 1 63 314 1 14 1
1( 28 15) 1(42 5) 6(9 2)13 47 66
6
zD −= −
− −− −= − +
= − + − + + += − − +=
Find the solutions by Cramer's Rule: 3 91 33 3
6 23
yx
z
DDx y
D DD
zD
− −= = = = = =
− −
= = = −−
The solution is (1, 3, 2)− .
34. 4
2 3 4 155 2 12
x y zx y zx y z
− + = −⎧⎪ − + = −⎨⎪ + − =⎩
1 1 132 4
5 1 2
3 34 2 4 21 ( 1) 12 25 51 1
1(6 4) 1( 4 20) 1(2 15)2 24 17
5
D−−=
−
− −= − − +− −
= − + − − + += − += −
1 1415 3 412 1 2
3 15 15 34 44 ( 1) 12 21 12 12 1
4(6 4) 1(30 48) 1( 15 36)8 18 215
xD−−
= − −−
− − − −= − − − +− −
= − − + − + − += − − += −
1 14152 4
5 12 2
15 154 2 4 21 ( 4) 12 25 512 12
1(30 48) 4( 4 20) 1(24 75)18 96 9915
yD−−=
−
− −= − − +− −
= − + − − + += − − += −
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Section 8.3: Systems of Linear Equations: Determinants
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1 1 43 152
5 1 12
3 15 15 32 21 ( 1) ( 4)5 51 12 12 1
1( 36 15) 1(24 75) 4(2 15)21 99 68
10
zD− −− −=
− − − −= − − + −
= − + + + − += − + −=
Find the solutions by Cramer's Rule: 5 151 35 5
10 25
yx
z
DDx y
D DD
zD
− −= = = = = =
− −
= = = −−
The solution is (1, 3, 2)− .
35. 2 3
2 4 72 2 3 4
x y zx y zx y z
+ − = −⎧⎪ − + = −⎨⎪− + − =⎩
1 2 142 1
2 32
4 41 2 1 21 2 ( 1)2 23 32 2
1(12 2) 2( 6 2) 1(4 8)10 8 422
3 2 147 1
34 2
4 47 71 13 2 ( 1)3 32 4 4 2
3(12 2) 2(21 4) 1( 14 16)30 34 266
x
D
D
−−=
− −
− −= − + −− −− −
= − − − + − −= + +=
− −−= −
−
− −− −= − − + −− −
= − − − − − − += − − −= −
31 172 1
2 34
7 71 2 1 21 ( 3) ( 1)2 23 34 4
1(21 4) 3( 6 2) 1(8 14)17 12 611
yD− −−=
− −
− −= − − + −− −− −
= − + − + − −= − +=
31 24 72
2 2 4
4 47 72 21 2 ( 3)2 22 4 4 2
1( 16 14) 2(8 14) 3(4 8)2 12 12
22
zD−
− −=−
− −− −= − + −− −
= − + − − − −= − + +=
Find the solutions by Cramer's Rule: 66 11 13
22 22 222 122
yx
z
DDx y
D DD
zD
−= = = − = = =
= = =
The solution is 13, , 12
⎛ ⎞−⎜ ⎟⎝ ⎠
.
36. 4 3 8
3 3 126 1
x y zx y z
x y z
+ − = −⎧⎪ − + =⎨⎪ + + =⎩
31 43 31
61 1
3 3 3 31 11 4 ( 3)6 61 1 1 1
1( 6 3) 4(18 3) 3(3 1)9 60 1281
D−
= −
− −= − + −
= − − − − − += − − −= −
348312 161 1
3 31 12 12 18 4 ( 3)6 61 1 1 1
8( 6 3) 4(72 3) 3(12 1)72 276 39
243
xD−−
= −
− −= − − + −
= − − − − − − += − −= −
31 83 312
61 1
3 3 3 312 121 ( 8) ( 3)6 61 1 1 1
1(72 3) 8(18 3) 3(3 12)69 120 27216
yD−−
=
= − − + −
= − + − − −= + +=
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Chapter 8: Systems of Equations and Inequalities
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1 4 83 1 121 1 1
3 31 12 12 11 4 ( 8)1 1 1 1 1 1
1( 1 12) 4(3 12) 8(3 1)13 36 329
zD−
= −
− −= − + −
= − − − − − += − + −= −
Find the solutions by Cramer's Rule: 243 216 83 81 81 39 181 9
yx
z
DDx yD DDzD
−= = = = = = −
− −−
= = =−
The solution is 8 13, ,3 9
⎛ ⎞−⎜ ⎟⎝ ⎠
.
37. 2 3 1
3 2 02 4 6 2
x y zx y zx y z
− + =⎧⎪ + − =⎨⎪ − + =⎩
2 3123 1
4 62
2 23 31 11 ( 2) 36 64 2 2 4
1(6 8) 2(18 4) 3( 12 2)2 44 42
0
D−
−=−
− −= − − +− −
= − + + + − −= − + −=
Since 0D = , Cramer's Rule does not apply.
38. 2 5
3 2 42 2 4 10
x y zx yx y z
− + =⎧⎪ + =⎨⎪ − + − = −⎩
1 1 23 02
22 4
0 3 0 32 21 ( 1) 22 24 2 4 2
1( 8 0) 1( 12 0) 2(6 4)8 12 20
0
D−
=
− −
= − − +− − − −
= − − + − − + += − − +=
Since 0D = , Cramer's Rule does not apply.
39. 2 0
2 4 02 2 3 0
x y zx y zx y z
+ − =⎧⎪ − + =⎨⎪− + − =⎩
1 2 142 1
2 32
4 41 2 1 21 2 ( 1)2 23 32 2
1(12 2) 2( 6 2) 1(4 8)10 8 422
D−
−=− −
− −= − + −− −− −
= − − − + − −= + +=
0 2 140 0 [By Theorem (12)]1
0 32xD
−−= =
−
01 102 1 0 [By Theorem (12)]
2 0 3yD
−= =
− −
01 24 02 0 [By Theorem (12)]
2 02zD −= =
−
Find the solutions by Cramer's Rule: 0 00 022 220 022
yx
z
DDx y
D DD
zD
= = = = = =
= = =
The solution is (0, 0, 0).
40. 4 3 0
3 3 06 0
x y zx y z
x y z
+ − =⎧⎪ − + =⎨⎪ + + =⎩
31 43 31
61 1
3 3 3 31 11 4 ( 3)6 61 1 1 1
1( 6 3) 4(18 3) 3(3 1)9 60 1281
0 340 3 0 [By Theorem (12)]10 61
x
D
D
−= −
− −= − + −
= − − − − − += − − −= −
−= =−
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Section 8.3: Systems of Linear Equations: Determinants
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0 313 0 3 0 [By Theorem (12)]
0 61
01 43 0 0 [By Theorem (12)]1
01 1
y
z
D
D
−= =
= =−
Find the solutions by Cramer's Rule: 0 00 081 810 081
yx
z
DDx y
D DD
zD
= = = = = =− −
= = =−
The solution is (0, 0, 0).
41. 2 3 0
3 2 02 4 6 0
x y zx y zx y z
− + =⎧⎪ + − =⎨⎪ − + =⎩
2 3123 1
4 62
2 23 31 11 ( 2) 34 46 62 2
1(6 8) 2(18 4) 3( 12 2)2 44 42
0
D−
−=−
− −= − − +− −
= − + + + − −= − + −=
Since 0D = , Cramer's Rule does not apply.
42. 2 0
3 2 02 2 4 0
x y zx yx y z
− + =⎧⎪ + =⎨⎪ − + − =⎩
1 1 23 02
22 4
0 3 0 32 21 ( 1) 22 24 2 4 2
1( 8 0) 1( 12 0) 2(6 4)8 12 20
0
D−
=
− −
= − − +− − − −
= − − + − − + += − − +=
Since 0D = , Cramer's Rule does not apply.
43. 41 2 3
x y zu v w =
By Theorem (11), the value of a determinant changes sign if any two rows are interchanged.
Thus, 1 2 3
4u v wx y z
= − .
44. 41 2 3
x y zu v w =
By Theorem (14), if any row of a determinant is multiplied by a nonzero number k, the value of the determinant is also changed by a factor of k.
Thus, 2 2(4) 82 4 6 1 2 3
x y z x y zu v w u v w= = = .
45. Let 41 2 3
x y zu v w = .
3 6 9 3 1 2 3 [Theorem (14)]
3( 1) [Theorem (11)]1 2 3
3(4) 12
x y z x y z
u v w u v w
x y zu v w
− − − = −
= − −
= =
46. Let 41 2 3
x y zu v w =
( )2 2 3
1 2 3 1 2 3[Theorem (15)]
( 1) 1 2 3 [Theorem (11)]
( 1)( 1) [Theorem (11)]1 2 3
1 2 3
x u y v z w x y zR r r
u v w u v w
x y z
u v w
x y zu v w
x y zu v w
− − − == +
= −
= − −
= 4=
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Chapter 8: Systems of Equations and Inequalities
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47. Let 41 2 3
x y zu v w =
1 2 33 6 9
2 2 2
1 2 32 3 6 9 [Theorem (14)]
3 6 92( 1) 1 2 3 [Theorem (11)]
x y zu v w
x y zu v w
x y z
u v w
− − −
= − − −
− − −= −
1 3 1
3 6 92( 1)( 1) [Theorem (11)]
1 2 3
[Theorem (15)]2( 1)( 1)
( 3 )1 2 3
x y zu v w
x y zu v w
R r r
− − −= − −
= − −= − +
2( 1)( 1)(4) 8= − − =
48. Let 41 2 3
x y zu v w =
3 1 3
[Theorem (15)] ( )
1 2 2 1 2 34
x y z x x y zu v w u u v w
C c c
−− =
= +
=
49. Let 41 2 3
x y zu v w =
1 2 32 2 2
1 2 3
1 2 32 [Theorem (14)]
1 2 3
2( 1) 1 2 3 [Theorem (11)]1 2 3
x y zu v w
x y zu v w
x y z
u v w
− − −
=− − −
= −− − −
2 3 2
2( 1)( 1) 1 2 3 [Theorem (11)]1 2 3
[Theorem (15)]2( 1)( 1)
( )1 2 3
x y zu v w
x y zu v w
R r r
= − − − − −
= − −= − +
2( 1)( 1)(4)8
= − −=
50. Let 41 2 3
x y zu v w =
1 3 1
2 3 2
3 6 93 1 3 2 3 3
1 2 3
[Theorem (15)]3 1 3 2 3 3
( =3 )1 2 3
[Theorem (15)]3 3 3
( )1 2 3
x y zu v w
x y zu v w
R r r
x y zu v w
R r r
+ + +− − −
= − − −+
== − +
3 [Theorem (14)]1 2 3
3(4)12
x y zu v w=
==
51. Solve for x:
534
3 4 55
5
x x
x xxx
=
− =− =
= −
52. Solve for x:
2
2
1 23
3 2
1 0( 1)( 1) 0
xx
x
xx x
= −
− = −
− =− + =
1 0 or 1 01 or 1
x xx x
− = + == = −
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Section 8.3: Systems of Linear Equations: Determinants
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53. Solve for x:
( ) ( ) ( )
1 134 2 2
51 2
3 32 4 2 41 1 25 52 1 1 2
15 4 20 2 8 3 211 22 11 2
11 131311
x
x
xx
x
x
=−
− + =− −
− − + + + =
− + ==
=
54. Solve for x:
( ) ( ) ( )
3 2 451 020 1
5 51 13 2 4 02 20 01 1
3 2 5 2 2 4 1 06 15 4 4 0
6 7 06 7
76
x
x x
xx
xx
x
=−
− + =− −
− − − − + =
− − + + =− − =
− =
= −
55. Solve for x:
( ) ( ) ( )
( )
2
2
3201 726 1
0 01 12 3 72 26 61 1
2 2 2 3 1 6 7
2 4 3 18 7
2 18 02 9 0
xx
x xx
x x x
x x
x xx x
=−
− + =− −− − − + − =
− + + − =
− − =− + =
0 or 9x x= = −
56. Solve for x:
( ) ( ) ( )
( )
2
2
1 231 4
0 1 2
3 31 11 2 40 01 2 2 1
2 3 1 2 2 1 4
2 3 2 2 4
2 02 1 0
xx x
x xx x
x x x
x x x
x xx x
= −
− + = −
− − + = −
− − + = −
+ =
+ =
10 or 2
x x= = −
57. Expanding the determinant:
( )
1 1
2 2
1 1 1 1
2 2 2 2
1 2 1 2 1 2 2 1
11 01
1 11 0
1 1
( ) ( ) 0
x yx yx y
y x x yx y
y x x y
x y y y x x x y x y
=
− + =
− − − + − =
( )( )
1 2 2 1 2 1 1 2
2 1 2 1 1 2 2 1
( )
( )
x y y y x x x y x y
y x x x y x y x y y
− + − = −
− = − + −
( )
( )
2 1 1 2 1
2 1 1 2 2 1 1 2 1
2 1 1
2 1 2 1 1 2 1 2 1 1
( ) ( ) ( )
( ) ( )
y x x y x xx y x y x y y y x x
x x y yx y y x y x y y x y x
− − −
= − + − − −
− −
= − + − − +
( )( )
( )
2 1 1 2 1 2 1 1
2 1 1 2 1 1
2 11 1
2 1
( ) ( ) ( )
( ) ( )( )( )
( ) ( )
x x y y y y x y y x
x x y y y y x xy y
y y x xx x
− − = − − −
− − = − −
−− = −
−
This is the 2-point form of the equation for a line.
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Chapter 8: Systems of Equations and Inequalities
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58. Any point ( , )x y on the line containing 2 2( , )x y and 3 3( , )x y satisfies:
2 2
3 3
11 01
x yx yx y
=
If the point 1 1( , )x y is on the line containing
2 2( , )x y and 3 3( , )x y [the points are collinear],
then 1 1
2 2
3 3
11 01
x yx yx y
= .
Conversely, if 1 1
2 2
3 3
11 01
x yx yx y
= , then 1 1( , )x y is
on the line containing 2 2( , )x y and 3 3( , )x y , and the points are collinear.
59. Let A = 1 1( , )x y , B = 2 2( , )x y , and C = 3 3( , )x y represent the vertices of a triangle. For simplicity, we position our triangle in the first quadrant. See figure. Let A be the point closest to the y-axis, C be the point farthest from the y-axis, and B be the point “between” points A and C.
y
x1( ,0)x 2( ,0)x 3( ,0)x
1 1( , )x y
2 2( , )x y
3 3( , )x yA
B
C
D E F
Let D = 1( ,0)x , E = 2( ,0)x , and F = 3( ,0)x . We find the area of triangle ABC by subtracting the areas of trapezoids ADEB and BEFC from the area of trapezoid ADFC. Note: 1AD y= ,
2BE y= , 3CF y= , 3 1DF x x= − , 2 1DE x x= − ,
and 3 2EF x x= − . Thus, the areas of the three trapezoids are as follows:
( )( )3 1 1 312ADFCK x x y y= − +
( )( )2 1 1 212ADEBK x x y y= − + , and
( )( )3 2 2 312BEFCK x x y y= − + .
The area of our triangle ABC is ABC ADFC ADEB BEFCK K K K= − −
( )( ) ( )( )
( )( )
3 1 1 3 2 1 1 2
3 2 2 3
1 12 2
1 2
x x y y x x y y
x x y y
= − + − − +
− − +
( )
3 1 3 3 1 1 1 3 2 1
2 2 1 1 1 2 3 2
3 3 2 2 2 3
3 1 1 3 2 1 1 2 3 2 2 3
3 1 1 3 2 1 1 2 3 2 2 3
1 1 1 1 12 2 2 2 2
1 1 1 1 2 2 2 2
1 1 1 2 2 2
1 1 1 1 1 12 2 2 2 2 212
x y x y x y x y x y
x y x y x y x y
x y x y x y
x y x y x y x y x y x y
x y x y x y x y x y x y
= + − − −
− + + −
− + +
= − − + − +
= − − + − +
Expanding D, we obtain
( )
( )
1 2 3
1 2 3
2 3 1 3 1 21 2 3
1 2 3 2 1 3 3 1 2
1 2 1 3 2 1 2 3 3 1 3 2
12
1 1 1
11 1 1 1 1 12
1 ( ) ( ) ( )212
x x xD y y y
y y y y y yx x x
x y y x y y x y y
x y x y x y x y x y x y
=
⎛ ⎞= − +⎜ ⎟
⎝ ⎠
= − − − + −
= − − + + −
which is the same as the area of triangle ABC.
If the vertices of the triangle are positioned differently than shown in the figure, the signs may be reversed. Thus, the absolute value of D will give the area of the triangle.
If the vertices of a triangle are (2, 3), (5, 2), and (6, 5), then:
[ ]
[ ]
[ ]
2 5 61 3 2 52
1 1 1
2 5 3 5 3 21 2 5 61 1 1 1 1 12
1 2(2 5) 5(3 5) 6(3 2)21 2( 3) 5( 2) 6(1)21 6 10 625
D =
⎛ ⎞= − +⎜ ⎟
⎝ ⎠
= − − − + −
= − − − +
= − + +
=
The area of the triangle is 5 5= square units.
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Section 8.3: Systems of Linear Equations: Determinants
821 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
60. Expanding the determinant:
[ ]
2
2
2
2 22
2 2
2 2 2 2 2
2
2
1
1
1
1 11
1 1
( ) ( ) 1( )
( ) ( )( ) ( )
( )
( ) ( ) ( )( )( )( )
x x
y y
z z
y y y yx x
z z z z
x y z x y z y z z y
x y z x y z y z yz y z
y z x xy xz yz
y z x x y z x yy z x y x z
= − +
= − − − + −
= − − − + + −
⎡ ⎤= − − − +⎣ ⎦= − − − −
= − − −
61. If 0,a = then 0b ≠ and 0c ≠ since
0ad bc− ≠ , and the system is by scx dy t
=⎧⎨ + =⎩
.
The solution of the system is ,syb
=
( )sbt dt dy tb sdx
c c bc
−− −= = = . Using Cramer’s
Rule, we get 0 b
D bcc d
= = − ,
xs b
D sd tbt d
= = − ,
00y
sD sc sc
c t= = − = − , so
xD ds tb td sdxD bc bc
− −= = =
− and
yD sc syD bc b
−= = =
−, which is the solution. Note
that these solutions agree if 0.d =
If 0,b = then 0a ≠ and 0d ≠ since
0ad bc− ≠ , and the system is ax s
cx dy t=⎧
⎨ + =⎩.
The solution of the system is ,sxa
=
t cx at csyd ad− −
= = . Using Cramer’s Rule, we
get 0a
D adc d
= = , 0
xs
D sdt d
= = , and
ya s
D at csc t
= = − , so xD sd sxD ad a
= = =
and yD at csyD ad
−= = , which is the solution.
Note that these solutions agree if 0.c =
If 0,c = then 0a ≠ and 0d ≠ since
0ad bc− ≠ , and the system is ax by s
dy t+ =⎧
⎨ =⎩.
The solution of the system is ,tyd
=
s by sd tbxa ad− −
= = . Using Cramer’s Rule, we
get 0a b
D add
= = , xs b
D sd tbt d
= = − ,
and 0ya s
D att
= = , so xD sd tbxD ad
−= = and
yD at tyD ad d
= = = , which is the solution. Note
that these solutions agree if 0.b =
If 0,d = then 0b ≠ and 0c ≠ since
0ad bc− ≠ , and the system is
ax by scx t
+ =⎧⎨ =⎩
.
The solution of the system is ,txc
=
s ax cs atyb bc− −
= = . Using Cramer’s Rule, we
get 00
a bD bc bc
c= = − = − ,
00x
s bD tb tb
t= = − = − , and
ya s
D at csc t
= = − , so xD tb txD bc c
−= = =
− and
yD at cs cs atyD bc bc
− −= = =
−, which is the
solution. Note that these solutions agree if 0.a =
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
822 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
62. Evaluating the determinant to show the relationship:
13 12 11
23 22 21
33 32 31
22 21 23 21 23 2213 12 11
32 31 33 31 33 32
13 22 31 21 32 12 23 31 21 33
11 23 32 22 33
( ) ( ) ( )
a a aa a aa a a
a a a a a aa a a
a a a a a aa a a a a a a a a a
a a a a a
= − +
= − − −
+ −
13 22 31 13 21 32 12 23 31 12 21 33
11 23 32 11 22 33
11 22 33 11 23 32 12 21 33 12 23 31
13 21 32 13 22 31
11 22 33 23 32 12 21 33 23 31
( ) ( )
a a a a a a a a a a a aa a a a a a
a a a a a a a a a a a aa a a a a a
a a a a a a a a a a
= − − ++ −
= − + + −
− += − − + −
13 21 32 22 31
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
11 12 13
21 22 23
31 32 33
( )a a a a aa a a a a a
a a aa a a a a a
a a a a a aa a a
a a a a a a
a a aa a aa a a
− −
= − + −
⎛ ⎞= − − +⎜ ⎟⎜ ⎟
⎝ ⎠
= −
63. Evaluating the determinant to show the relationship: 11 12 13
21 22 23
31 32 33
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
11 22 33 23 32 12 21 33 23 31
13 21 32 22 31
11 22 33 23 32 12 21 33
( ) ( ) ( )
( ) (
a a aka ka kaa a a
ka ka ka ka ka kaa a a
a a a a a aa ka a ka a a ka a ka a
a ka a ka aka a a a a ka a a a
= − +
= − − −+ −
= − − −
()
23 31
13 21 32 22 31
11 22 33 23 32 12 21 33 23 31
13 21 32 22 31
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
11 12 13
21 22 23
) ( )
( ) ( )
( )
aka a a a a
k a a a a a a a a a a
a a a a a
a a a a a ak a a a
a a a a a a
a a ak a a a
a
+ −
= − − −
+ −
⎛ ⎞= − +⎜ ⎟⎜ ⎟
⎝ ⎠
=
31 32 33a a
64. Set up a 3 by 3 determinant in which the first column and third column are the same and evaluate:
11 12 11
21 22 21
31 32 31
22 21 21 21 21 2211 12 11
32 31 31 31 31 32
11 22 31 32 21 12 21 31 31 21
11 21 32 31 22
11 22 31 11 32 21 12 11 21 32 11 31 22
( ) ( ) ( )
(0)0
a a aa a aa a a
a a a a a aa a a
a a a a a aa a a a a a a a a a
a a a a aa a a a a a a a a a a a a
= − +
= − − −
+ −
= − − + −=
65. Evaluating the determinant to show the relationship:
11 21 12 22 13 23
21 22 23
31 32 33
22 23 21 2311 21 12 22
32 33 31 33
21 2213 23
31 32
11 21 22 33 23 32
12 22 21 33 23 31
13 23 21 32 22 31
11 22 33
( ) ( )
( )
( )( )( )( )( )( )
(
a ka a ka a kaa a aa a a
a a a aa ka a ka
a a a a
a aa ka
a aa ka a a a a
a ka a a a aa ka a a a a
a a a
+ + +
= + − +
+ +
= + −− + −
+ + −
= 23 32 21 22 33 23 32
12 21 33 23 31 22 21 33 23 31
13 21 32 22 31 23 21 32 22 31
11 22 33 23 32 21 22 33
21 23 32 12 21 33 23 31
22 21 33 22 23 31
13 21 32
) ( )( ) ( )( ) ( )
( )( )
(
a a ka a a a aa a a a a ka a a a aa a a a a ka a a a a
a a a a a ka a aka a a a a a a aka a a ka a aa a a
− + −− − − −
+ − + −
= − +− − −
− +
+ − 22 31 23 21 32
23 22 31
11 22 33 23 32 12 21 33 23 31
13 21 32 22 31
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
11 12 13
21 22 23
31 32 33
)
( ) ( )( )
a a ka a aka a a
a a a a a a a a a aa a a a a
a a a a a aa a a
a a a a a a
a a aa a aa a a
+
−= − − −
+ −
= − +
=
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.4: Matrix Algebra
823 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8.4
1. inverse
2. square
3. identity
4. False
5. False
6. False
7. 0 3 5 4 1 01 2 6 2 3 2
0 4 3 1 5 01 ( 2) 2 3 6 ( 2)
4 4 51 5 4
A B−⎡ ⎤ ⎡ ⎤
+ = +⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦+ + − +⎡ ⎤
= ⎢ ⎥+ − + + −⎣ ⎦−⎡ ⎤
= ⎢ ⎥−⎣ ⎦
8. 0 3 5 4 1 01 2 6 2 3 2
0 4 3 1 5 01 ( 2) 2 3 6 ( 2)
4 2 53 1 8
A B−⎡ ⎤ ⎡ ⎤
− = −⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦− − − −⎡ ⎤
= ⎢ ⎥− − − − −⎣ ⎦− −⎡ ⎤
= ⎢ ⎥−⎣ ⎦
9. 0 3 5
4 41 2 6
4 0 4 3 4( 5)4 1 4 2 4 6
0 12 204 8 24
A−⎡ ⎤
= ⎢ ⎥⎣ ⎦⋅ ⋅ −⎡ ⎤
= ⎢ ⎥⋅ ⋅ ⋅⎣ ⎦−⎡ ⎤
= ⎢ ⎥⎣ ⎦
10. 4 1 0
3 32 3 2
3 4 3 1 3 03 2 3 3 3 212 3 0
6 9 6
B ⎡ ⎤− = − ⎢ ⎥− −⎣ ⎦
− ⋅ − ⋅ − ⋅⎡ ⎤= ⎢ ⎥− ⋅− − ⋅ − ⋅−⎣ ⎦
− −⎡ ⎤= ⎢ ⎥−⎣ ⎦
11. 0 3 5 4 1 0
3 2 3 21 2 6 2 3 2
0 9 15 8 2 03 6 18 4 6 4
8 7 157 0 22
A B−⎡ ⎤ ⎡ ⎤
− = −⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦−⎡ ⎤ ⎡ ⎤
= −⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦− −⎡ ⎤
= ⎢ ⎥⎣ ⎦
12. 0 3 5 4 1 0
2 4 2 41 2 6 2 3 2
0 6 10 16 4 02 4 12 8 12 8
16 10 106 16 4
A B−⎡ ⎤ ⎡ ⎤
+ = +⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦−⎡ ⎤ ⎡ ⎤
= +⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦−⎡ ⎤
= ⎢ ⎥−⎣ ⎦
13. 4 1
0 3 56 2
1 2 62 3
0(4) 3(6) ( 5)( 2) 0(1) 3(2) ( 5)(3)1(4) 2(6) 6( 2) 1(1) 2(2) 6(3)
28 94 23
AC⎡ ⎤
−⎡ ⎤ ⎢ ⎥= ⋅⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥−⎣ ⎦+ + − − + + −⎡ ⎤
= ⎢ ⎥+ + − + +⎣ ⎦−⎡ ⎤
= ⎢ ⎥⎣ ⎦
14. 4 1
4 1 06 2
2 3 22 3
4(4) 1(6) 0( 2) 4(1) 1(2) 0(3)2(4) 3(6) ( 2)( 2) 2(1) 3(2) ( 2)(3)
22 614 2
BC⎡ ⎤
⎡ ⎤ ⎢ ⎥= ⋅⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎢ ⎥−⎣ ⎦+ + − + +⎡ ⎤
= ⎢ ⎥− + + − − − + + −⎣ ⎦⎡ ⎤
= ⎢ ⎥−⎣ ⎦
15. 4 1
0 3 56 2
1 2 62 3
4(0) 1(1) 4(3) 1(2) 4( 5) 1(6)6(0) 2(1) 6(3) 2(2) 6( 5) 2(6)2(0) 3(1) 2(3) 3(2) 2( 5) 3(6)
1 14 142 22 183 0 28
CA⎡ ⎤
−⎡ ⎤⎢ ⎥= ⋅ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥−⎣ ⎦+ + − +⎡ ⎤
⎢ ⎥= + + − +⎢ ⎥⎢ ⎥− + − + − − +⎣ ⎦
−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
16. 4 1
4 1 06 2
2 3 22 3
4(4) 1( 2) 4(1) 1(3) 4(0) 1( 2)6(4) 2( 2) 6(1) 2(3) 6(0) 2( 2)2(4) 3( 2) 2(1) 3(3) 2(0) 3( 2)
14 7 220 12 414 7 6
CB⎡ ⎤
⎡ ⎤⎢ ⎥= ⋅ ⎢ ⎥⎢ ⎥ − −⎣ ⎦⎢ ⎥−⎣ ⎦+ − + + −⎡ ⎤
⎢ ⎥= + − + + −⎢ ⎥⎢ ⎥− + − − + − + −⎣ ⎦
−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥− −⎣ ⎦
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Chapter 8: Systems of Equations and Inequalities
824 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17. 4 1
0 3 5 4 1 0( ) 6 2
1 2 6 2 3 22 3
4 14 4 5
6 21 5 4
2 3
15 21 1622 34 2211 7 22
C A B⎡ ⎤
⎛ ⎞−⎡ ⎤ ⎡ ⎤⎢ ⎥+ = +⎜ ⎟⎢ ⎥ ⎢ ⎥⎢ ⎥ − −⎣ ⎦ ⎣ ⎦⎝ ⎠⎢ ⎥−⎣ ⎦⎡ ⎤
−⎡ ⎤⎢ ⎥= ⋅ ⎢ ⎥⎢ ⎥ −⎣ ⎦⎢ ⎥−⎣ ⎦−⎡ ⎤
⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦
18. 4 1
0 3 5 4 1 0( ) 6 2
1 2 6 2 3 22 3
4 14 4 5
6 21 5 4
2 3
50 318 21
A B C⎡ ⎤
⎛ ⎞−⎡ ⎤ ⎡ ⎤ ⎢ ⎥+ = +⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎢ ⎥−⎣ ⎦⎡ ⎤
−⎡ ⎤ ⎢ ⎥= ⋅⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎢ ⎥−⎣ ⎦−⎡ ⎤
= ⎢ ⎥⎣ ⎦
19. 2
4 10 3 5 1 0
3 6 2 31 2 6 0 1
2 3
28 9 3 04 23 0 3
25 94 20
AC I⎡ ⎤
−⎡ ⎤ ⎡ ⎤⎢ ⎥− = ⋅ −⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥−⎣ ⎦−⎡ ⎤ ⎡ ⎤
= −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
−⎡ ⎤= ⎢ ⎥⎣ ⎦
20. 3
4 1 1 0 00 3 5
5 6 2 5 0 1 01 2 6
2 3 0 0 1
1 14 14 5 0 02 22 18 0 5 03 0 28 0 0 5
6 14 142 27 183 0 33
CA I⎡ ⎤ ⎡ ⎤
−⎡ ⎤⎢ ⎥ ⎢ ⎥+ = ⋅ +⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥= − +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
21. CA CB−
4 1 4 1
0 3 5 4 1 06 2 6 2
1 2 6 2 3 22 3 2 3
1 14 14 14 7 22 22 18 20 12 43 0 28 14 7 6
13 7 1218 10 1417 7 34
⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= ⋅ − ⋅⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ − −⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦
− −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − − −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦− −⎡ ⎤⎢ ⎥= − −⎢ ⎥⎢ ⎥−⎣ ⎦
22. 4 1 4 1
0 3 5 4 1 06 2 6 2
1 2 6 2 3 22 3 2 3
28 9 22 64 23 14 2
50 318 21
AC BC+
⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= ⋅ + ⋅⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦
−⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
−⎡ ⎤= ⎢ ⎥⎣ ⎦
23. 11
12
13
14
21
22
23
24
2(2) ( 2)(3) 22(1) ( 2)( 1) 42(4) ( 2)(3) 22(6) ( 2)(2) 81(2) 0(3) 21(1) 0( 1) 11(4) 0(3) 41(6) 0(2) 6
aaaaaaaa
= + − = −= + − − == + − == + − == + == + − == + =
= + =
2 2 2 1 4 6 2 4 2 81 0 3 1 3 2 2 1 4 6
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
24. 11
12
13
14
21
22
23
24
4( 6) 1(2) 224(6) 1(5) 294(1) 1(4) 84(0) 1( 1) 12( 6) 1(2) 102(6) 1(5) 172(1) 1(4) 62(0) 1( 1) 1
aaaaaaaa
= − + = −= + == + =
= + − = −= − + = −= + == + =
= + − = −
4 1 6 6 1 0 22 29 8 12 1 2 5 4 1 10 17 6 1
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.4: Matrix Algebra
825 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
25. 1 2
1 2 31 0
0 1 42 4
1(1) 2( 1) 3(2) 1(2) 2(0) 3(4)0(1) ( 1)( 1) 4(2) 0(2) ( 1)(0) 4(4)
5 149 16
⎡ ⎤⎡ ⎤ ⎢ ⎥−⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎢ ⎥⎣ ⎦
+ − + + +⎡ ⎤= ⎢ ⎥+ − − + + − +⎣ ⎦⎡ ⎤
= ⎢ ⎥⎣ ⎦
26. 1 1
2 8 13 2
3 6 00 5
1(2) ( 1)(3) 1(8) ( 1)(6) 1( 1) ( 1)(0)( 3)(2) 2(3) ( 3)(8) 2(6) ( 3)( 1) 2(0)
0(2) 5(3) 0(8) 5(6) 0( 1) 5(0)
−⎡ ⎤−⎡ ⎤⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦
+ − + − − + −⎡ ⎤⎢ ⎥= − + − + − − +⎢ ⎥⎢ ⎥+ + − +⎣ ⎦
1 2 10 12 3
15 30 0
− −⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
27. 1 0 1 1 32 4 1 6 23 6 1 8 1
1(1) 0(6) 1(8) 1(3) 0(2) 1( 1)2(1) 4(6) 1(8) 2(3) 4(2) 1( 1)3(1) 6(6) 1(8) 3(3) 6(2) 1( 1)
9 234 1347 20
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
+ + + + −⎡ ⎤⎢ ⎥= + + + + −⎢ ⎥⎢ ⎥+ + + + −⎣ ⎦⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
28. 4 2 3 2 60 1 2 1 11 0 1 0 2
4(2) ( 2)(1) 3(0) 4(6) ( 2)( 1) 3(2)0(2) 1(1) 2(0) 0(6) 1( 1) 2(2)1(2) 0(1) 1(0) 1(6) 0( 1) 1(2)
−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
+ − + + − − +⎡ ⎤⎢ ⎥= + + + − +⎢ ⎥⎢ ⎥− + + − + − +⎣ ⎦
6 321 32 4
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥− −⎣ ⎦
29. 2 11 1
A ⎡ ⎤= ⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
1 2
2 1 2
2 1 1 01 1 0 1
Interchange1 1 0 1
and 2 1 1 0
1 1 0 1 2
0 1 1 2
r r
R r r
⎡ ⎤⎢ ⎥⎣ ⎦
⎛ ⎞⎡ ⎤→ ⎜ ⎟⎢ ⎥
⎣ ⎦ ⎝ ⎠⎡ ⎤
→ = − +⎢ ⎥− −⎣ ⎦
( )
( )
2 2
1 2 1
1 1 0 1
0 1 1 2
1 0 1 1
0 1 1 2
R r
R r r
⎡ ⎤→ = −⎢ ⎥−⎣ ⎦
−⎡ ⎤→ = − +⎢ ⎥−⎣ ⎦
Thus, 1 1 11 2
A− −⎡ ⎤= ⎢ ⎥−⎣ ⎦
.
30. 3 12 1
A−⎡ ⎤
= ⎢ ⎥−⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
1 2 1
2 1 2
3 1 1 02 1 0 1
1 0 1 1
2 1 0 1
1 0 1 1 2
0 1 2 3
R r r
R r r
−⎡ ⎤⎢ ⎥−⎣ ⎦
⎡ ⎤→ = +⎢ ⎥−⎣ ⎦
⎡ ⎤→ = +⎢ ⎥
⎣ ⎦
Thus, 1 1 12 3
A− ⎡ ⎤= ⎢ ⎥⎣ ⎦
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
826 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
31. 6 52 2
A ⎡ ⎤= ⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
1 2
2 1 2
111 122
2 2
52
1 2 1
6 5 1 02 2 0 1
Interchange2 2 0 1
and 6 5 1 0
2 2 0 1 3
0 1 1 3
1 1 0
0 1 1 3
1 0 1
0 1 1 3
r r
R r r
R r
R r
R r r
⎡ ⎤⎢ ⎥⎣ ⎦
⎛ ⎞⎡ ⎤→ ⎜ ⎟⎢ ⎥
⎣ ⎦ ⎝ ⎠⎡ ⎤
→ = − +⎢ ⎥− −⎣ ⎦⎛ ⎞=⎡ ⎤
→ ⎜ ⎟⎢ ⎥ ⎜ ⎟− = −⎣ ⎦ ⎝ ⎠⎡ ⎤−
→ = − +⎢ ⎥−⎣ ⎦
Thus, 5
1 211 3
A− ⎡ ⎤−= ⎢ ⎥−⎣ ⎦
.
32. 4 16 2
A−⎡ ⎤
= ⎢ ⎥−⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
( )
( )
1 114 4
1 14
1 14 4
2 1 2312 2
1 14 4
2 2
112
1 24
4 1 1 06 2 0 1
1 0
6 2 0 1
1 0 6
0 1
1 0 2
0 1 3 2
1 0 1
0 1 3 2
R r
R r r
R r
R r r
−⎡ ⎤⎢ ⎥−⎣ ⎦
⎡ ⎤− −→ = −⎢ ⎥−⎣ ⎦
⎡ ⎤− −→ = − +⎢ ⎥−⎢ ⎥⎣ ⎦
⎡ ⎤− −→ = −⎢ ⎥− −⎣ ⎦
⎡ ⎤− −→ = +⎢ ⎥− −⎣ ⎦
Thus, 1
1 213 2
A− ⎡ ⎤− −= ⎢ ⎥− −⎣ ⎦
.
33. 2 1
where 0.A aa a⎡ ⎤
= ≠⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
( )
( )
1 112 2
1 12
1 12 2
2 1 21 12 2
1 122 2
2 22
11
1 2 122
2 1 1 00 1
1 0
0 1
1 0
0 1
1 0
0 1 1
1 0 1
0 1 1
aa
a
a
a a
R ra a
R a r ra a
R r
R r r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ =⎢ ⎥
⎣ ⎦⎡ ⎤
→ = − +⎢ ⎥−⎢ ⎥⎣ ⎦⎡ ⎤
→ =⎢ ⎥−⎢ ⎥⎣ ⎦⎡ ⎤−
→ = − +⎢ ⎥−⎢ ⎥⎣ ⎦
Thus, 1
12
1
1a
a
A−⎡ ⎤−
= ⎢ ⎥−⎢ ⎥⎣ ⎦
.
34. 3
where 0.2
bA b
b⎡ ⎤
= ≠⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
( )
( )
2 1 2
3 11
1 1
3 1
2 2
323
1 2 1
3 1 02 0 1
3 1 0
0 1 1 1
1 0
0 1 1 1
1 0
0 1 1 1
1 0
0 1 1 1
b bb
b b
b bb
bb
bR r r
R r
R r
R r r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ = − +⎢ ⎥− −⎣ ⎦
⎡ ⎤→ =⎢ ⎥
− −⎢ ⎥⎣ ⎦⎡ ⎤
→ = −⎢ ⎥−⎢ ⎥⎣ ⎦
⎡ ⎤−→ = − +⎢ ⎥
−⎢ ⎥⎣ ⎦
Thus, 32
1
1 1b bA− ⎡ ⎤−
= ⎢ ⎥−⎢ ⎥⎣ ⎦
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.4: Matrix Algebra
827 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
35. 1 1 10 2 12 3 0
A−⎡ ⎤
⎢ ⎥= −⎢ ⎥⎢ ⎥− −⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
3 1 3
1 1 12 22 2 2
1 12 2
1 2 11 12 2
3 2 3512 2
1 12 2
1 1 1 1 0 00 2 1 0 1 02 3 0 0 0 1
1 1 1 1 0 00 2 1 0 1 0 20 5 2 2 0 1
1 1 1 1 0 00 1 0 0 0 5 2 2 0 1
1 0 1 00 1 0 0
50 0 2 1
1 0 1
R r r
R r
R r rR r r
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− −⎣ ⎦
−⎡ ⎤⎢ ⎥→ − = +⎢ ⎥⎢ ⎥−⎣ ⎦
−⎡ ⎤⎢ ⎥→ − − = −⎢ ⎥⎢ ⎥−⎣ ⎦⎡ ⎤−
= +⎛ ⎞⎢ ⎥→ − − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥− −⎣ ⎦
−→ ( )1 1
3 32 2
11 3 12
12 3 22
00 1 0 0 20 0 1 4 5 2
1 0 0 3 3 10 1 0 2 2 1 0 0 1 4 5 2
R r
R r r
R r r
⎡ ⎤⎢ ⎥− − = −⎢ ⎥⎢ ⎥− −⎣ ⎦
−⎡ ⎤ ⎛ ⎞= − +⎢ ⎥ ⎜ ⎟→ − −⎢ ⎥ ⎜ ⎟= +⎝ ⎠⎢ ⎥− −⎣ ⎦
Thus, 13 3 12 2 14 5 2
A−−⎡ ⎤
⎢ ⎥= − −⎢ ⎥⎢ ⎥− −⎣ ⎦
.
36. 1 0 21 2 31 1 0
A⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
2 1 2
3 1 3
5 1 1 12 22 2 2 2
1 0 2 1 0 01 2 3 0 1 01 1 0 0 0 1
1 0 2 1 0 00 2 5 1 1 0 0 1 2 1 0 1
1 0 2 1 0 00 1 0 0 1 2 1 0 1
R r rR r r
R r
⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦
⎡ ⎤= +⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− − −⎣ ⎦
⎡ ⎤⎢ ⎥→ =⎢ ⎥⎢ ⎥− − −⎣ ⎦
( )5 1 13 2 32 2 2
1 1 12 2 2
1 0 2 1 0 00 1 0 0 0 1
R r r⎡ ⎤⎢ ⎥→ = +⎢ ⎥⎢ ⎥−⎣ ⎦
( )5 1 13 32 2 2
1 3 15
2 3 22
1 0 2 1 0 00 1 0 20 0 1 1 1 2
1 0 0 3 2 4 20 1 0 3 2 5 0 0 1 1 1 2
R r
R r r
R r r
⎡ ⎤⎢ ⎥→ =⎢ ⎥⎢ ⎥−⎣ ⎦
− −⎡ ⎤ = − +⎛ ⎞⎢ ⎥→ − − ⎜ ⎟⎢ ⎥ ⎜ ⎟= − +⎝ ⎠⎢ ⎥−⎣ ⎦
Thus, 13 2 43 2 51 1 2
A−− −⎡ ⎤
⎢ ⎥= − −⎢ ⎥⎢ ⎥−⎣ ⎦
.
37. 1 1 13 2 13 1 2
A⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
1 1 1 1 0 03 2 1 0 1 03 1 2 0 0 1
⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
2 1 2
3 1 3
1 1 1 1 0 03
0 1 4 3 1 0 3
0 2 1 3 0 1
R r rR r r
⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− − −⎣ ⎦
( )2 2
1 1 1 1 0 00 1 4 3 1 0 0 2 1 3 0 1
R r⎡ ⎤⎢ ⎥→ − = −⎢ ⎥⎢ ⎥− − −⎣ ⎦
( )13 37
3 2 17 7 7
5 317 7 7
1 3 19 1 47 7 7
2 3 23 2 17 7 7
1 0 3 2 1 00 1 4 3 1 0 0 0 1
1 0 03
0 1 0 4
0 0 1
R r
R r rR r r
⎡ ⎤− −⎢ ⎥→ − =⎢ ⎥⎢ ⎥−⎣ ⎦⎡ ⎤−⎢ ⎥ = +⎛ ⎞
→ −⎢ ⎥ ⎜ ⎟= − +⎝ ⎠⎢ ⎥−⎢ ⎥⎣ ⎦
Thus,
5 317 7 7
1 9 1 47 7 73 2 17 7 7
A−
⎡ ⎤−⎢ ⎥
= −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
828 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
38. 3 3 11 2 12 1 1
A⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥−⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
3 3 1 1 0 01 2 1 0 1 02 1 1 0 0 1
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦
( )
1 2
2 1 2
3 1 3
2 1 12 23 3 3
1 23 32 13 37 53 3
1 2 1 0 1 0Interchange
3 3 1 1 0 0 and
2 1 1 0 0 1
1 2 1 0 1 03
0 3 2 1 3 0 2
0 5 1 0 2 1
1 2 1 0 1 0
0 1 1 0
0 5 1 0 2 1
1 0 1 0
0 1 1 0
0 0 3 1
r r
R r rR r r
R r
⎡ ⎤⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥−⎣ ⎦
⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− − −⎣ ⎦
⎡ ⎤⎢ ⎥
→ − = −⎢ ⎥⎢ ⎥− − −⎣ ⎦⎡ − −
→ −
−⎣
( )
1 2 1
3 2 3
1 23 3
32 13 33 3 7
5 9 37 7 7
3 4 17 7 7 1
1 3 131 1 27 7 7 2
2 3 235 9 37 7 7
2
5
1 0 1 0
0 1 1 0
0 0 1
1 0 0
0 1 0
0 0 1
R r rR r r
R r
R r r
R r r
⎤⎢ ⎥ = − +⎛ ⎞⎢ ⎥ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥
⎦⎡ ⎤− −⎢ ⎥⎢ ⎥→ − =⎢ ⎥⎢ ⎥−⎣ ⎦⎡ ⎤−⎢ ⎥ ⎛ ⎞= +
⎜ ⎟⎢ ⎥→ −⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦
Thus,
3 4 17 7 7
1 1 1 27 7 75 9 37 7 7
A−
⎡ ⎤−⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦
.
39. 2 8
5x yx y+ =⎧
⎨ + =⎩
Rewrite the system of equations in matrix form: 2 1 8
, ,1 1 5
xA X B
y⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 29, 1 1 11 2
A− −⎡ ⎤= ⎢ ⎥−⎣ ⎦
, so
1 1 1 8 31 2 5 2
X A B− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
.
The solution is 3, 2x y= = or (3, 2) .
40. 3 82 4
x yx y− =⎧
⎨− + =⎩
Rewrite the system of equations in matrix form: 3 1 8
, ,2 1 4
xA X B
y−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 30, 1 1 12 3
A− ⎡ ⎤= ⎢ ⎥⎣ ⎦
, so
1 1 1 8 122 3 4 28
X A B− ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦.
The solution is 12, 28x y= = or (12, 28) .
41. 2 0
5x yx y+ =⎧
⎨ + =⎩
Rewrite the system of equations in matrix form: 2 1 0
, ,1 1 5
xA X B
y⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 29, 1 1 11 2
A− −⎡ ⎤= ⎢ ⎥−⎣ ⎦
, so
1 1 1 0 51 2 5 10
X A B− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
.
The solution is 5, 10x y= − = or ( 5, 10)− .
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.4: Matrix Algebra
829 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
42. 3 42 5
x yx y− =⎧
⎨− + =⎩
Rewrite the system of equations in matrix form: 3 1 4
, ,2 1 5
xA X B
y−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 30, 1 1 12 3
A− ⎡ ⎤= ⎢ ⎥⎣ ⎦
, so
1 1 1 4 92 3 5 23
X A B− ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦.
The solution is 9, 23x y= = or (9, 23) .
43. 6 5 72 2 2
x yx y+ =⎧
⎨ + =⎩
Rewrite the system of equations in matrix form: 6 5 7
, ,2 2 2
xA X B
y⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 31, 5
1 211 3
A− ⎡ ⎤−= ⎢ ⎥−⎣ ⎦
, so
51 2
7 212 11 3
X A B− ⎡ ⎤− ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−− ⎣ ⎦ ⎣ ⎦⎣ ⎦
.
The solution is 2, 1x y= = − or (2, 1)− .
44. 4 06 2 14x yx y
− + =⎧⎨ − =⎩
Rewrite the system of equations in matrix form: 4 1 0
, ,6 2 14
xA X B
y−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 32, 1
1 213 2
A− ⎡ ⎤− −= ⎢ ⎥− −⎣ ⎦
, so
11 2
0 7114 283 2
X A B− −⎡ ⎤− − ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−− − ⎣ ⎦ ⎣ ⎦⎣ ⎦
.
The solution is 7, 28x y= − = − or ( 7, 28)− − .
45. 6 5 132 2 5
x yx y+ =⎧
⎨ + =⎩
Rewrite the system of equations in matrix form: 6 5 13
, ,2 2 5
xA X B
y⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 31, 5
1 211 3
A− ⎡ ⎤−= ⎢ ⎥−⎣ ⎦
, so
5 11 22
1315 21 3
X A B− ⎡ ⎤ ⎡ ⎤− ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎣ ⎦ ⎣ ⎦⎣ ⎦
.
The solution is 1 , 22
x y= = or 1 , 22
⎛ ⎞⎜ ⎟⎝ ⎠
.
46. 4 56 2 9x yx y
− + =⎧⎨ − = −⎩
Rewrite the system of equations in matrix form: 4 1 5
, ,6 2 9
xA X B
y−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 32, 1
1 213 2
A− ⎡ ⎤− −= ⎢ ⎥− −⎣ ⎦
, so
1 11 2 2
5193 2 3
X A B− ⎡ ⎤ ⎡ ⎤− − −⎡ ⎤= = =⎢ ⎥ ⎢ ⎥⎢ ⎥−− − ⎣ ⎦⎣ ⎦ ⎣ ⎦
.
The solution is 1 , 32
x y= − = or 1 , 32
⎛ ⎞−⎜ ⎟⎝ ⎠
.
47. 2 3
0x y
aax ay a+ = −⎧
≠⎨ + = −⎩
Rewrite the system of equations in matrix form: 2 1 3
, ,x
A X Ba a y a
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 33, 1
12
1
1a
a
A−⎡ ⎤−
= ⎢ ⎥−⎢ ⎥⎣ ⎦
, so
11
2
1 3 211
a
a
X A Ba
−⎡ ⎤− − −⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−− ⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦.
The solution is 2, 1x y= − = or ( 2, 1)− .
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
830 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
48. 3 2 3
02 2 2
bx y bb
bx y b+ = +⎧
≠⎨ + = +⎩
Rewrite the system of equations in matrix form: 3 2 3
, ,2 2 2
b x bA X B
b y b+⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥+⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 34, 32
1
1 1b bA− ⎡ ⎤−
= ⎢ ⎥−⎢ ⎥⎣ ⎦
, so
321 2 3 2
2 2 11 1b b b
X A Bb
− ⎡ ⎤− +⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥+−⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦
.
The solution is 2, 1x y= = or (2, 1).
49. 72
05
x yaa
ax ay
⎧ + =⎪ ≠⎨⎪ + =⎩
Rewrite the system of equations in matrix form: 72 1
, ,5ax
A X Ba a y
⎡ ⎤⎡ ⎤ ⎡ ⎤= = = ⎢ ⎥⎢ ⎥ ⎢ ⎥
⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 33, 1
12
1
1a
a
A−⎡ ⎤−
= ⎢ ⎥−⎢ ⎥⎣ ⎦
, so
1 271
32
1
1 5a aa
a a
X A B−⎡ ⎤ ⎡ ⎤− ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
.
The solution is 2 3,x ya a
= = or 2 3,a a
⎛ ⎞⎜ ⎟⎝ ⎠
.
50. 3 14
02 10
bx yb
bx y+ =⎧
≠⎨ + =⎩
Rewrite the system of equations in matrix form: 3 14
, ,2 10
b xA X B
b y⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 34, 32
1
1 1b bA− ⎡ ⎤−
= ⎢ ⎥−⎢ ⎥⎣ ⎦
, so
3 221 14
10 41 1bb bX A B− ⎡ ⎤ ⎡ ⎤− ⎡ ⎤
= = =⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦
.
The solution is 2 , 4x yb
= = or 2 , 4b
⎛ ⎞⎜ ⎟⎝ ⎠
.
51. 0
2 12 3 5
x y zy z
x y
− + =⎧⎪ − + = −⎨⎪− − = −⎩
Rewrite the system of equations in matrix form: 1 1 1 00 2 1 , , 12 3 0 5
xA X y B
z
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 35, 13 3 12 2 14 5 2
A−−⎡ ⎤
⎢ ⎥= − −⎢ ⎥⎢ ⎥− −⎣ ⎦
, so
13 3 1 0 22 2 1 1 34 5 2 5 5
X A B−− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = − − − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
.
The solution is 2, 3, 5x y z= − = = or ( 2, 3, 5)− .
52. 2 6
2 3 56
x zx y zx y
+ =⎧⎪− + + = −⎨⎪ − =⎩
Rewrite the system of equations in matrix form: 1 0 2 61 2 3 , , 51 1 0 6
xA X y B
z
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 36, 13 2 43 2 51 1 2
A−− −⎡ ⎤
⎢ ⎥= − −⎢ ⎥⎢ ⎥−⎣ ⎦
, so
13 2 4 6 43 2 5 5 21 1 2 6 1
X A B−− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = − − − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
.
The solution is 4, 2, 1x y z= = − = or (4, 2, 1)− .
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Section 8.4: Matrix Algebra
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53. 2
2 212 32
x y zy z
x y
− + =⎧⎪⎪ − + =⎨⎪− − =⎪⎩
Rewrite the system of equations in matrix form:
12
1 1 1 20 2 1 , , 22 3 0
xA X y B
z
⎡ ⎤−⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= − = = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 35, 13 3 12 2 14 5 2
A−−⎡ ⎤
⎢ ⎥= − −⎢ ⎥⎢ ⎥− −⎣ ⎦
, so
12
1 12
12
3 3 1 22 2 1 24 5 2 1
X A B−
⎡ ⎤ ⎡ ⎤−⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= = − − = −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
The solution is 1 1, , 12 2
x y z= = − = or
1 1, , 12 2
⎛ ⎞−⎜ ⎟⎝ ⎠
.
54.
2 232 322
x z
x y z
x y
+ =⎧⎪⎪− + + = −⎨⎪
− =⎪⎩
Rewrite the system of equations in matrix form:
32
1 0 2 21 2 3 , ,1 1 0 2
xA X y B
z
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 36, 13 2 43 2 51 1 2
A−− −⎡ ⎤
⎢ ⎥= − −⎢ ⎥⎢ ⎥−⎣ ⎦
, so
1 32
12
3 2 4 2 13 2 5 11 1 2 2
X A B−
⎡ ⎤− −⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= = − − − = −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
.
The solution is 11, 1,2
x y z= = − = or 11, 1,2
⎛ ⎞−⎜ ⎟⎝ ⎠
.
55. 9
3 2 83 2 1
x y zx y zx y z
+ + =⎧⎪ + − =⎨⎪ + + =⎩
Rewrite the system of equations in matrix form: 1 1 1 93 2 1 , , 83 1 2 1
xA X y B
z
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 37,
5 317 7 7
1 9 1 47 7 73 2 17 7 7
A−
⎡ ⎤−⎢ ⎥
= −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
, so
5 3 3417 7 7 7
1 9 851 47 7 7 73 2 1 127 7 7 7
981
X A B−
⎡ ⎤ ⎡ ⎤− −⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= = − =⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
.
The solution is 34 85 12, ,7 7 7
x y z= − = = or
34 85 12, ,7 7 7
⎛ ⎞−⎜ ⎟⎝ ⎠
.
56. 3 3 8
2 52 4
x y zx y zx y z
+ + =⎧⎪ + + =⎨⎪ − + =⎩
Rewrite the system of equations in matrix form: 3 3 1 81 2 1 , , 52 1 1 4
xA X y B
z
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 38,
3 4 17 7 7
1 1 1 27 7 75 9 37 7 7
A−
⎡ ⎤−⎢ ⎥
= −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
, so
3 84 17 7 7 7
1 51 1 27 7 7 75 9 3 177 7 7 7
854
X A B−
⎡ ⎤ ⎡ ⎤− ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= = − =⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
.
The solution is 8 5 17, ,7 7 7
x y z= = = or
8 5 17, ,7 7 7
⎛ ⎞⎜ ⎟⎝ ⎠
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
832 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
57.
273 23
103 23
x y z
x y z
x y z
+ + =⎧⎪⎪ + − =⎨⎪⎪ + + =⎩
Rewrite the system of equations in matrix form:
73
103
1 1 1 23 2 1 , ,3 1 2
xA X y B
z
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= − = = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 37,
5 317 7 7
1 9 1 47 7 73 2 17 7 7
A−
⎡ ⎤−⎢ ⎥
= −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
, so
5 31 17 7 7 3
1 9 71 47 7 7 3
23 102 137 7 7 3
21X A B−
⎡ ⎤ ⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥ ⎢ ⎥
= = − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦
.
The solution is 1 2, 1,3 3
x y z= = = or 1 2, 1,3 3
⎛ ⎞⎜ ⎟⎝ ⎠
.
58. 3 3 1
2 02 4
x y zx y zx y z
+ + =⎧⎪ + + =⎨⎪ − + =⎩
Rewrite the system of equations in matrix form: 3 3 1 11 2 1 , , 02 1 1 4
xA X y B
z
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Find the inverse of 1 and solve A X A B−= :
From Problem 38,
3 4 17 7 7
1 1 1 27 7 75 9 37 7 7
A−
⎡ ⎤−⎢ ⎥
= −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
, so
3 4 17 7 7
1 1 1 27 7 75 9 37 7 7
1 10 14 1
X A B−
⎡ ⎤− ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦
.
The solution is 1, 1, 1x y z= = − = or (1, 1, 1)− .
59. 4 22 1
A ⎡ ⎤= ⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
12 1 21 2
2
1 112 4
1 11 42
4 2 1 02 1 0 1
4 2 1 0
0 0 1
1 0
0 0 1
R r r
R r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ = − +⎢ ⎥−⎣ ⎦
⎡ ⎤→ =⎢ ⎥−⎢ ⎥⎣ ⎦
There is no way to obtain the identity matrix on the left. Thus, this matrix has no inverse.
60. 123
6 1A
⎡ ⎤−= ⎢ ⎥−⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
12
12
2 1 2
1 116 3
1 13
3 1 06 1 0 1
3 1 0 2
0 0 2 1
1 0
0 0 2 1
R r r
R r
⎡ ⎤−⎢ ⎥−⎣ ⎦
⎡ ⎤−→ = +⎢ ⎥
⎣ ⎦⎡ ⎤− −
→ = −⎢ ⎥⎢ ⎥⎣ ⎦
There is no way to obtain the identity matrix on the left. Thus, this matrix has no inverse.
61. 15 310 2
A ⎡ ⎤= ⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
22 1 22 3
3
1 15 15 1
1 11523
15 3 1 010 2 0 1
15 3 1 0
0 0 1
1 0
0 0 1
R r r
R r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ = − +⎢ ⎥−⎣ ⎦
⎡ ⎤→ =⎢ ⎥
−⎢ ⎥⎣ ⎦
There is no way to obtain the identity matrix on the left; thus, there is no inverse.
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Section 8.4: Matrix Algebra
833 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
62. 3 04 0
A−⎡ ⎤
= ⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
42 1 24 3
3
13 1
1 1343
3 0 1 04 0 0 1
3 0 1 0
0 0 1
1 0 0
0 0 1
R r r
R r
−⎡ ⎤⎢ ⎥⎣ ⎦
−⎡ ⎤→ = +⎢ ⎥
⎣ ⎦⎡ ⎤−
→ = −⎢ ⎥⎢ ⎥⎣ ⎦
There is no way to obtain the identity matrix on the left; thus, there is no inverse.
63. 3 1 11 4 71 2 5
A− −⎡ ⎤⎢ ⎥= − −⎢ ⎥⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
3 1 1 1 0 01 4 7 0 1 01 2 5 0 0 1
− −⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥⎣ ⎦
( )
1 3
2 1 2
3 1 3
1 1 12 26 6 6
1 2 5 0 0 1Interchange
1 4 7 0 1 0 and
3 1 1 1 0 0
1 2 5 0 0 10 6 12 0 1 1
30 7 14 1 0 3
1 2 5 0 0 10 1 2 0
0 7 14 1 0 3
r r
R r rR r r
R r
⎡ ⎤⎛ ⎞⎢ ⎥→ − − ⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥− −⎣ ⎦
⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − − − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥→ − = −⎢ ⎥⎢ ⎥⎣ ⎦
1 23 3
1 2 11 16 6
3 2 37 116 6
1 0 1 02
0 1 2 0 7
0 0 0 1
R r rR r r
⎡ ⎤⎢ ⎥ = − +⎛ ⎞
→ −⎢ ⎥ ⎜ ⎟= − +⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦
There is no way to obtain the identity matrix on the left; thus, there is no inverse.
64. 1 1 32 4 15 7 1
A−⎡ ⎤
⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
1 1 3 1 0 02 4 1 0 1 05 7 1 0 0 1
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦
( )
2 1 2
3 1 3
7 1 1 12 26 3 6 6
11 2 16 3 6
1 2 17 1 16 3 6
3 2 3
1 1 3 1 0 02
0 6 7 2 1 0 5
0 12 14 5 0 1
1 1 3 1 0 0
0 1 0
0 12 14 5 0 1
1 0 0
0 1 0 12
0 0 0 1 2 1
R r rR r r
R r
R r rR r r
−⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥−⎣ ⎦
−⎡ ⎤⎢ ⎥
→ − − = −⎢ ⎥⎢ ⎥−⎣ ⎦⎡ ⎤−⎢ ⎥ = − +⎛ ⎞⎢ ⎥→ − − ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦
There is no way to obtain the identity matrix on the left; thus, there is no inverse.
65. 25 61 1218 2 48 35 21
A−⎡ ⎤
⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
Thus, 10.01 0.05 0.010.01 0.02 0.010.02 0.01 0.03
A−−⎡ ⎤
⎢ ⎥≈ −⎢ ⎥⎢ ⎥−⎣ ⎦
66. 18 3 46 20 14
10 25 15A
−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦
Thus, 10.26 0.29 0.201.21 1.63 1.201.84 2.53 1.80
A−− −⎡ ⎤
⎢ ⎥≈ −⎢ ⎥⎢ ⎥−⎣ ⎦
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Chapter 8: Systems of Equations and Inequalities
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67.
44 21 18 62 10 15 5
21 12 12 48 16 4 9
A
⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥− −⎣ ⎦
Thus, 1
0.02 0.04 0.01 0.010.02 0.05 0.03 0.030.02 0.01 0.04 0.000.02 0.06 0.07 0.06
A−
− −⎡ ⎤⎢ ⎥− −⎢ ⎥≈⎢ ⎥−⎢ ⎥−⎣ ⎦
.
68.
16 22 3 521 17 4 82 8 27 205 15 3 10
A
−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥⎢ ⎥
− −⎣ ⎦
Thus, 1
0.01 0.04 0.00 0.030.02 0.02 0.01 0.010.04 0.02 0.04 0.060.05 0.02 0.00 0.09
A−
⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥
− −⎣ ⎦
.
69. 25 61 12 1018 12 7 ; 93 4 1 12
A B−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
Enter the matrices into a graphing utility and use 1A B− to solve the system. The result is shown
below:
Thus, the solution to the system is 4.57x ≈ ,
6.44y ≈ − , 24.07z ≈ − or (4.57, 6.44, 24.07)− − .
70. 25 61 12 1518 12 7 ; 33 4 1 12
A B−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
Enter the matrices into a graphing utility and use 1A B− to solve the system. The result is shown
below:
Thus, the solution to the system is 4.56x ≈ ,
6.06y ≈ − , 22.55z ≈ − or (4.56, 6.06, 22.55)− − .
71. 25 61 12 2118 12 7 ; 73 4 1 2
A B−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥= − =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦
Thus, the solution to the system is 1.19x ≈ − ,
2.46y ≈ , 8.27z ≈ or ( 1.19, 2.46, 8.27)− .
72. 25 61 12 2518 12 7 ; 103 4 1 4
A B−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥= − =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦
Thus, the solution to the system is 2.05x ≈ − ,
3.88y ≈ , 13.36z ≈ or ( 2.05, 3.88, 13.36)− .
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.4: Matrix Algebra
835 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
73. 2 3 115 7 24
x yx y+ =⎧
⎨ + =⎩
Multiply each side of the first equation by 5, and each side of the second equation by 2− . Then add the equations to eliminate x:
10 15 5510 14 48
x yx y+ =⎧⎪
⎨− − = −⎪⎩
7y = Substitute and solve for x:
( )2 3 7 112 21 11
2 105
xx
xx
+ =
+ == −= −
The solution of the system is 5, 7x y= − = or using ordered pairs ( )5,7− .
74. 2 8 8
7 13x yx y+ = −⎧
⎨ + = −⎩
Multiply each side of the second equation by 2− . Then add the equations to eliminate x: 2 8 8
2 14 26x yx y+ = −⎧⎪
⎨− − =⎪⎩
6 183
yy
− == −
Substitute and solve for x: ( )7 3 13
21 138
xx
x
+ − = −
− = −=
The solution of the system is 8, 3x y= = − or using ordered pairs ( )8, 3− .
75. 2 4 2
3 5 2 174 3 22
x y zx y zx y
− + =⎧⎪− + − =⎨⎪ − = −⎩
Write the augmented matrix: 1 2 4 23 5 172
3 04 22
⎡ ⎤−⎢ ⎥− −⎢ ⎥⎢ ⎥− −⎣ ⎦
2 1 23 1 3
1 2 4 20 10 231 30 5 16 30 4
R r rR r r
⎡ ⎤−⎢ ⎥− = +⎢ ⎥− −⎣ ⎦ = − +
2 2
1 2 4 20 10 2310 5 16 30
R r⎡ ⎤−⎢ ⎥− − = −⎢ ⎥− −⎣ ⎦
3 2 3
1 2 4 20 10 2310 0 34 85 5R r r
⎡ ⎤−⎢ ⎥− −⎢ ⎥⎣ ⎦ = − +
52 3 3
1 2 4 20 10 231
/ 340 0 1 R r
⎡ ⎤−⎢ ⎥− −⎢ ⎥
=⎣ ⎦
1 3 1
2 3 252
0 81 2 40 01 2 100 0 1
R r rR r r
⎡ ⎤−− = − +⎢ ⎥ = +⎢ ⎥⎣ ⎦
1 2 1
52
0 01 4 20 01 20 0 1
R r r⎡ ⎤− = +⎢ ⎥⎢ ⎥⎣ ⎦
The solution is 54, 2,2
x y z= − = = or 54,2,2
⎛ ⎞−⎜ ⎟⎝ ⎠
.
76. 2 3 24 3 6
6 2 2
x y zx z
y z
+ − = −⎧⎪ + =⎨⎪ − =⎩
Write the augmented matrix: 32 1 20 3 64
0 6 2 2
⎡ ⎤− −⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦
3 12 2 1 11 1 / 20 3 64
0 6 2 2
R r⎡ ⎤− − =⎢ ⎥⎢ ⎥⎢ ⎥
−⎢ ⎥⎣ ⎦
3 12 2
2 1 2
1 10 6 5 10 40 6 2 2
R r r
⎡ ⎤− −⎢ ⎥⎢ ⎥− = − +⎢ ⎥
−⎣ ⎦
( )
3 12 2
5 52 26 3
1 1/ 60 1
0 6 2 2
R r
⎡ ⎤− −⎢ ⎥⎢ ⎥− − = −⎢ ⎥⎢ ⎥−⎣ ⎦
3 12 2
5 56 3
3 2 3
1 1
0 160 0 3 12 R r r
⎡ ⎤− −⎢ ⎥⎢ ⎥− −⎢ ⎥
= − +⎢ ⎥⎣ ⎦
3 12 2
5 56 3
3 3
1 1
0 1/30 0 1 4 R r
⎡ ⎤− −⎢ ⎥⎢ ⎥− −⎢ ⎥
=⎢ ⎥⎣ ⎦
3 12 1 3 125 53 2 3 26
01 1
0 010 0 1 4
R r r
R r r
⎡ ⎤ = +⎢ ⎥⎢ ⎥ = +⎢ ⎥⎢ ⎥⎣ ⎦
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Chapter 8: Systems of Equations and Inequalities
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3 32 1 2 1253
0 01
0 010 0 1 4
R r r⎡ ⎤− = − +⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
The solution is 3 5, , 42 3
x y z= − = = or
3 5, , 42 3
⎛ ⎞−⎜ ⎟⎝ ⎠
.
77. 5 4 2
5 4 37 13 4 17
x y zx y z
x y z
− + =⎧⎪ − + − =⎨⎪ + − =⎩
Write the augmented matrix: 5 1 4 2
5 31 47 13 174
⎡ ⎤−⎢ ⎥− −⎢ ⎥⎢ ⎥−⎣ ⎦
1 2
2 1
5 31 45 1 4 27 13 174
R rR r
⎡ ⎤− − = −⎢ ⎥
− =⎢ ⎥⎢ ⎥−⎣ ⎦
2 1 2
3 1 3
5 31 40 16 1724 50 48 32 38 7
R r rR r r
⎡ ⎤− −⎢ ⎥
− = − +⎢ ⎥⎢ ⎥− = − +⎣ ⎦
2 172 23 24
5 31 4/ 240 1
0 48 32 38
R r
⎡ ⎤− −⎢ ⎥
− =⎢ ⎥⎢ ⎥
−⎣ ⎦
2 173 24
3 2 3
5 31 4
0 1480 0 0 4 R r r
⎡ ⎤− −⎢ ⎥
−⎢ ⎥⎢ ⎥ = − +⎣ ⎦
The last row of our matrix is a contradiction. Therefore, the system is inconsistent. The solution set is { } , or ∅ .
78. 3 2 22 6 7
2 2 14 17
x y zx y z
x y z
+ − =⎧⎪ + + = −⎨⎪ + − =⎩
Write the augmented matrix: 3 2 1 2
6 72 1172 2 14
⎡ ⎤−⎢ ⎥
−⎢ ⎥⎢ ⎥−⎣ ⎦
1 1 27 91 16 72 1
172 2 14
R r r⎡ ⎤− = −⎢ ⎥
−⎢ ⎥⎢ ⎥−⎣ ⎦
2 1 2
3 1 3
7 91 10 20 251 20 0 0 1 2
R r rR r r
⎡ ⎤−⎢ ⎥
−− = − +⎢ ⎥⎢ ⎥− = − +⎣ ⎦
The last row of our matrix is a contradiction. Therefore, the system is inconsistent. The solution set is { } , or ∅ .
79. 2 3 43 2 3 5 6
x y zx y z
y z
− + =⎧⎪− + − = −⎨⎪ − + =⎩
Write the augmented matrix: 32 1 4
3 32 10 5 61
⎡ ⎤−⎢ ⎥− −−⎢ ⎥⎢ ⎥−⎣ ⎦
3 12 2 1 11 2 / 2
3 32 10 5 61
R r⎡ ⎤− =⎢ ⎥⎢ ⎥− −−⎢ ⎥
−⎣ ⎦
3 12 25 1
2 1 22 2
1 230 3
0 5 61
R r r
⎡ ⎤−⎢ ⎥⎢ ⎥− = +⎢ ⎥⎢ ⎥−⎣ ⎦
3 12 2
61 25 5 2 25
1 2
0 10 5 61
R r
⎡ ⎤−⎢ ⎥⎢ ⎥− − = −⎢ ⎥⎢ ⎥−⎣ ⎦
3 12 2
615 5
3 2 3
1 2
0 150 0 0 0 R r r
⎡ ⎤−⎢ ⎥⎢ ⎥− −⎢ ⎥
= +⎢ ⎥⎣ ⎦
1 1 35 5 1 2 1261
5 5
01
0 10 0 0 0
R r r⎡ ⎤ = +⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥⎣ ⎦
Since the last row yields an identity, and no contradictions exist in the other rows, there are an infinite number of solutions. The solution is
1 15 5
x z= − + , 1 65 5
y z= − , and z is any real
number. That is,
( )
}
1 1 1 6, , | , ,5 5 5 5
is any real number
x y z x z y z
z
⎧ = − + = −⎨⎩
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Section 8.4: Matrix Algebra
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80. 4 3 2 6 3 2
9 6
x y zx y zx y z
− + + =⎧⎪ + − = −⎨⎪ + + =⎩
Write the augmented matrix: 3 64 2
3 1 1 29 61 1
⎡ ⎤−⎢ ⎥
− −⎢ ⎥⎢ ⎥⎣ ⎦
1 3
3 1
9 61 13 1 1 2
3 64 2
R r
R r
⎡ ⎤ =⎢ ⎥
− −⎢ ⎥⎢ ⎥− =⎣ ⎦
2 1 2
3 1 3
9 61 10 26 204 30 39 6 30 4
R r rR r r
⎡ ⎤⎢ ⎥
− −− = − +⎢ ⎥⎢ ⎥ = +⎣ ⎦
( )1022 213 13
9 61 1/ 260 1
0 39 6 30
R r
⎡ ⎤⎢ ⎥
= −⎢ ⎥⎢ ⎥⎣ ⎦
10213 13
3 2 3
9 61 1
0 1390 0 0 0 R r r
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥ = − +⎣ ⎦
5 1213 13 1 2 1
10213 13
01 9
0 10 0 0 0
R r r⎡ ⎤− − = − +⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
Since the last row yields an identity, and no contradictions exist in the other rows, there are an infinite number of solutions. The solution is
5 1213 13
x z= − , 2 1013 13
y z= − + , and z is any real
number. That is,
( )
}
5 12 2 10, , | , ,13 13 13 13
is any real number
x y z x z y z
z
⎧ = − = − +⎨⎩
81. a. 6 9 71.00
; 3 12 158.60
A B⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
b. 6 9 71.003 12 158.60
6(71.00) 9(158.60) 1853.403(71.00) 12(158.60) 2116.20
AB ⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
+⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥+⎣ ⎦ ⎣ ⎦
Nikki’s total tuition is $1853.40, and Joe’s total tuition is $2116.20.
82. a. 4000 3000 0.011
; 2500 3800 0.006
A B⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
b. 4000 3000 0.011
2500 3800 0.006
4000(0.011) 3000(0.006) 62.002500(0.011) 3800(0.006) 50.30
AB ⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
+⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥+⎣ ⎦ ⎣ ⎦
After one month, Jamal’s loans accrued $62.00 in interest, and Stephanie’s loans accrued $50.30 in interest.
c. 4000 3000 1 0.011
( )2500 3800 1 0.006
4000 3000 1.0112500 3800 1.006
4000(1.011) 3000(1.006)2500(1.011) 3800(1.006)
7062.006350.30
A C B⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎡ ⎤
+ = +⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠⎡ ⎤ ⎡ ⎤
= ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
+⎡ ⎤= ⎢ ⎥+⎣ ⎦⎡ ⎤
= ⎢ ⎥⎣ ⎦
Jamal’s loan balance after one month was $7062.00, and Stephanie’s loan balance was $6350.30.
83. a. The rows of the 2 by 3 matrix represent stainless steel and aluminum. The columns represent 10-gallon, 5-gallon, and 1-gallon.
The 2 by 3 matrix is: 500 350 400700 500 850⎡ ⎤⎢ ⎥⎣ ⎦
.
The 3 by 2 matrix is: 500 700350 500400 850
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
.
b. The 3 by 1 matrix representing the amount of
material is: 1583
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
.
c. The days usage of materials is: 15
500 350 400 11,5008
700 500 850 17,0503
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⋅ =⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦
Thus, 11,500 pounds of stainless steel and 17,050 pounds of aluminum were used that day.
d. The 1 by 2 matrix representing cost is: [ ]0.10 0.05 .
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Chapter 8: Systems of Equations and Inequalities
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e. The total cost of the day’s production was:
[ ] [ ]11,5000.10 0.05 2002.50
17,050⎡ ⎤⋅ =⎢ ⎥⎣ ⎦
.
The total cost of the day’s production was $2002.50.
84. a. The rows of the 2 by 3 matrix represent the location. The columns represent the type of car sold. The 2 by 3 matrix for January is:
400 250 50450 200 140⎡ ⎤⎢ ⎥⎣ ⎦
. The 2 by 3 matrix for
February is: 350 100 30350 300 100⎡ ⎤⎢ ⎥⎣ ⎦
.
b. Adding the matrices: 400 250 50 350 100 30450 200 140 350 300 100
750 350 80800 500 240
⎡ ⎤ ⎡ ⎤+⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦⎡ ⎤
= ⎢ ⎥⎣ ⎦
c. The 3 by 1 matrix representing profit: 100150200
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
.
d. Multiplying to find the profit at each location: 100
750 350 80 143,500150
800 500 240 203,000200
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⋅ =⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦
.
The city location has a two-month profit of $143,500. The suburban location has a two-month profit of $203,000.
85. a. 2 1 11 1 01 1 1
K⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
2 1 1 1 0 01 1 0 0 1 01 1 1 0 0 1
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
1 2
2 1 2
3 1 3
1 1 0 0 1 0Interchange
2 1 1 1 0 0 and
1 1 1 0 0 11 1 0 0 1 0
20 1 1 1 2 0 0 0 1 0 1 1
r r
R r rR r r
⎡ ⎤⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦
⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠−⎢ ⎥⎣ ⎦
( )
( )
( )
2 2
2 2 3
1 1 2
1 1 0 0 1 00 1 1 1 2 0 0 0 1 0 1 1
1 1 0 0 1 00 1 0 1 1 1 0 0 1 0 1 1
1 0 0 1 0 10 1 0 1 1 1 0 0 1 0 1 1
R r
R r r
R r r
⎡ ⎤⎢ ⎥→ − − = −⎢ ⎥⎢ ⎥−⎣ ⎦⎡ ⎤⎢ ⎥→ − = +⎢ ⎥⎢ ⎥−⎣ ⎦
−⎡ ⎤⎢ ⎥→ − = −⎢ ⎥⎢ ⎥−⎣ ⎦
Thus, 11 0 11 1 10 1 1
K −−⎡ ⎤
⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦
.
b. 147 34 33 1 0 144 36 27 1 1 147 41 20 0 1 1
M E K −−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥= ⋅ = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
13 1 208 9 196 21 14
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
because 11
12
13
21
22
23
31
32
33
47(1) 34( 1) 33(0) 1347(0) 34(1) 33( 1) 147( 1) 34(1) 33(1) 2044(1) 36( 1) 27(0) 844(0) 36(1) 27( 1) 944( 1) 36(1) 27(1) 1947(1) 41( 1) 20(0) 647(0) 41(1) 20( 1) 21
aaaaaaaaa
= + − + == + + − == − + + =
= + − + == + + − == − + + == + − + =
= + + − == 47( 1) 41(1) 20(1) 14− + + =
c. 13 ;1 ; 20 ; 8 ; 9 ;19 ;6 ; 21 ;14
M A T H IS F U N
→ → → → →→ → → →
The message: Math is fun.
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Section 8.5: Partial Fraction Decomposition
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86. 20.4 0.2 0.1 0.4 0.2 0.10.5 0.6 0.5 0.5 0.6 0.50.1 0.2 0.4 0.1 0.2 0.4
0.27 0.22 0.180.55 0.56 0.550.18 0.22 0.27
P⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
because
11
12
13
21
22
23
0.4(0.4) 0.2(0.5) 0.1(0.1) 0.270.4(0.2) 0.2(0.6) 0.1(0.2) 0.220.4(0.1) 0.2(0.5) 0.1(0.4) 0.180.5(0.4) 0.6(0.5) 0.5(0.1) 0.550.5(0.2) 0.6(0.6) 0.5(0.2) 0.560.5(0.1) 0.6(0.5) 0.5
aaaaaa
= + + == + + == + + =
= + + == + + == + +
31
32
33
(0.4) 0.550.1(0.4) 0.2(0.5) 0.4(0.1) 0.180.1(0.2) 0.2(0.6) 0.4(0.2) 0.220.1(0.1) 0.2(0.5) 0.4(0.4) 0.27
aaa
== + + =
= + + == + + =
Each entry represents the probability that a grandchild has a certain income level given his or her grandparents’ income level.
87. a b
Ac d⎡ ⎤
= ⎢ ⎥⎣ ⎦
If 0D ad bc= − ≠ , then 0a ≠ and 0d ≠ , or 0b ≠ and 0c ≠ . Assuming the former, then
( )
( )
11
1 1
1
2 1 2
1
1 00 1
1 0
0 1
1 0
0 1
1 0
0 1
ba a
a
ba a
bc ca a
ba a
ad bc ca a
a bc d
R rc d
R c r rd
−
⎡ ⎤⎢ ⎥⎣ ⎦⎡ ⎤
→ = ⋅⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
→ = − ⋅ +⎢ ⎥− −⎢ ⎥⎣ ⎦
⎡ ⎤→ ⎢ ⎥
−⎢ ⎥⎣ ⎦
( )
( )
1
2 2
1( )
1 2 1
1 0
0 1
1 0
0 1
1 0
0 1
1 0
0 1
ba a a
ad bcc aad bc ad bc
bc ba a ad bc ad bc b
ac aad bc ad bc
d bad bc ad bc
c aad bc ad bc
d bD D
c aD D
R r
R r r
−−− −
−− −
−− −
−− −−− −
⎡ ⎤⎢ ⎥→ = ⋅⎢ ⎥⎣ ⎦⎡ ⎤+⎢ ⎥→ = − ⋅ +⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥→⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥→⎢ ⎥−⎣ ⎦
Thus, 1 1d bD D
c aD D
d bA
c aD− ⎡ ⎤− −⎡ ⎤= =⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎣ ⎦⎣ ⎦
where D ad bc= − .
88. Answers will vary. Section 8.5
1. True
2. True
3. ( )( )
4 3 2 2 2
22
3 6 3 3 2 1
3 1
x x x x x x
x x
+ + = + +
= +
4. True
5. The rational expression 2 1x
x − is proper, since
the degree of the numerator is less than the degree of the denominator.
6. The rational expression 35 2
1x
x+−
is proper, since
the degree of the numerator is less than the degree of the denominator.
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Chapter 8: Systems of Equations and Inequalities
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7. The rational expression 2
254
xx
+−
is improper, so
perform the division:
2 2
2
14 5
49
x x
x
− +
−
The proper rational expression is: 2
2 25 914 4
xx x
+= +
− −
8. The rational expression 2
23 2
1xx
−−
is improper, so
perform the division:
2 2
2
31 3 2
3 31
x x
x
− −
−
The proper rational expression is: 2
2 23 2 13
1 1xx x
−= +
− −
9. The rational expression 3
25 2 1
4x x
x+ −
− is improper,
so perform the division:
2 3
3
54 5 2 1
5 2022 1
xx x x
x xx
− + −−
−
The proper rational expression is: 3
2 25 2 1 22 1
4 45x x x
x xx+ − −
− −= +
10. The rational expression 4 2
33 2
8x x
x+ −
+ is improper,
so perform the division:
3 4 2
4
2
38 3 2
3 24
24 2
xx x x
x x
x x
+ + −+
− −
The proper rational expression is: 4 2 2
3 33 2 24 2
8 83x x x x
x xx+ − − −
+ += +
11. The rational expression 2
2( 1)
( 4)( 3) 12x x x x
x x x x− −
=+ − + −
is improper, so
perform the division:
2 2
2
112 0
122 12
x x x xx x
x
+ − − ++ −
− +
The proper rational expression is:
2( 1) 2 12 2( 6)
( 4)( 3) ( 4)( 3)121 1x x x x
x x x xx x− − + − −
+ − + −+ −= + = +
12. The rational expression 2 3
2 22 ( 4) 2 8
1 1x x x xx x
+ +=
+ + is
improper, so perform the division:
2 3
3
21 2 8
2 26
xx x x
x xx
+ ++
The proper rational expression is:
2
2 22 ( 4) 6
1 12x x x
x xx+
+ += +
13. Find the partial fraction decomposition: 4
( 1) 14
( 1) 1( 1) ( 1)
4 ( 1)
A Bx x x x
A Bx x x x
x x x x
A x Bx
= +− −
⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟− −⎝ ⎠⎝ ⎠− = −
= − +
Let 1x = , then 4 (0)4A B
B= +=
Let 0x = , then 4 ( 1) (0)4
A BA= − += −
4 4 4( 1) 1x x x x
−= +
− −
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Section 8.5: Partial Fraction Decomposition
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14. Find the partial fraction decomposition 3
( 2)( 1) 2 1x A B
x x x x= +
+ − + −
Multiplying both sides by ( 2)( 1)x x+ − , we obtain: 3 ( 1) ( 2)x A x B x= − + +
Let 1x = , then 3(1) (0) (3)3 3
1
A BBB
= +==
Let 2x = − , then 3(– 2) ( 3) (0)3 6
2
A BAA
= − +− = −
=
3 2 1( 2)( 1) 2 1
xx x x x
= ++ − + −
15. Find the partial fraction decomposition:
2 2
2 22 2
2
1( 1) 1
1( 1) 1
( 1) ( 1)
1 ( 1) ( )
A Bx Cxx x x
A Bx Cxx x x
x x x x
A x Bx C x
+= +
+ +
⎛ ⎞ +⎛ ⎞+⎜ ⎟ ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠+ = +
= + + +
Let 0x = , then 21 (0 1) ( (0) )(0)1A B C
A= + + +=
Let 1x = , then 21 (1 1) ( (1) )(1)1 21 2(1)
1
A B CA B C
B CB C
= + + += + += + +
+ = −
Let 1x = − , then 21 (( 1) 1) ( ( 1) )( 1)
1 (1 1) ( )( 1)1 21 2(1)
1
A B CA B C
A B CB C
B C
= − + + − + −= + + − + −= + −= + −
− = −
Solve the system of equations: 11
2 21
B CB CBB
+ = −− = −
= −= −
1 10
CC
− + = −=
2 21 1
( 1) 1x
xx x x−
= ++ +
16. Find the partial fraction decomposition:
2 21
1( 1)( 4) 4A Bx C
xx x x+
= +++ + +
Multiplying both sides by 2( 1)( 4)x x+ + , we obtain: 21 ( 4) ( )( 1)A x Bx C x= + + + +
Let 1x = − , then 1 (5) ( ( 1) )(0)5 1
15
A B CA
A
= + − +=
=
Let 1x = , then
( )
21 (1 4) ( (1) )(1 1)1 5 ( )(2)1 5 1/ 5 2 21 1 2 20 2 20
A B CA B C
B CB C
B CB C
= + + + += + +
= + +
= + += += +
Let 0x = , then
( )
21 (0 4) ( (0) )(0 1)1 41 4 1/ 5
41515
A B CA C
C
C
C
= + + + += +
= +
= +
=
Since 0B C+ = , we have that 1 05
15
B
B
+ =
= −
1 1 15 5 5
2 21
1( 1)( 4) 4
xxx x x
− += +
++ + +
17. Find the partial fraction decomposition:
( 1)( 2) 1 2x A B
x x x x= +
− − − −
Multiplying both sides by ( 1)( 2)x x− − , we obtain: ( 2) ( 1)x A x B x= − + −
Let 1x = , then 1 (1 2) (1 1)1
1
A BA
A
= − + −= −= −
Let 2x = , then 2 (2 2) (2 1)2
A BB
= − + −=
1 2( 1)( 2) 1 2
xx x x x
−= +
− − − −
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Chapter 8: Systems of Equations and Inequalities
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18. Find the partial fraction decomposition: 3
( 2)( 4) 2 4x A B
x x x x= +
+ − + −
Multiplying both sides by ( 2)( 4)x x+ − , we obtain: 3 ( 4) ( 2)x A x B x= − + +
Let 2x = − , then 3( 2) ( 2 4) ( 2 2)6 6
1
A BA
A
− = − − + − +− = −
=
Let 4x = , then 3(4) (4 4) (4 2)12 6
2
A BB
B
= − + +==
3 1 2( 2)( 4) 2 4
xx x x x
= ++ − + −
19. Find the partial fraction decomposition: 2
2 21 1( 1) ( 1) ( 1)x A B C
x xx x x= + +
− +− + −
Multiplying both sides by 2( 1) ( 1)x x− + , we
obtain: 2 2( 1)( 1) ( 1) ( 1)x A x x B x C x= − + + + + −
Let 1x = , then 2 2
2
1 (1 1)(1 1) (1 1) (1 1)
1 (0)(2) (2) (0)1 2
12
A B C
A B CB
B
= − + + + + −
= + +=
=
Let 1x = − , then 2 2
2
( 1) ( 1 1)( 1 1) ( 1 1) ( 1 1)
1 ( 2)(0) (0) ( 2)1 4
14
A B C
A B CC
C
− = − − − + + − + + − −
= − + + −=
=
Let 0x = , then 2 20 (0 1)(0 1) (0 1) (0 1)0
1 1 32 4 4
A B CA B C
A B C
A
= − + + + + −= − + += +
= + =
3 1 124 2 4
2 21 1( 1) ( 1) ( 1)x
x xx x x= + +
− +− + −
20. Find the partial fraction decomposition:
2 21
2( 2)x A B C
x xx x x+
= + +−−
Multiplying both sides by 2( 2)x x − , we obtain: 21 ( 2) ( 2)x Ax x B x Cx+ = − + − +
Let 0x = , then 20 1 (0)(0 2) (0 2) (0)
1 212
A B CB
B
+ = − + − += −
= −
Let 2x = , then 22 1 (2)(2 2) (2 2) (2)
3 434
A B CC
C
+ = − + − +=
=
Let 1x = , then 22
1 3 322 4 4
A B CA B C
A
= − − += − + −
⎛ ⎞= − − + − = −⎜ ⎟⎝ ⎠
3 314 2 4
2 21
2( 2)x
x xx x x
− −+= + +
−−
21. Find the partial fraction decomposition:
3 2
2 2
1 18 ( 2)( 2 4)
12( 2)( 2 4) 2 4
x x x xA Bx C
xx x x x x
=− − + +
+= +
−− + + + +
Multiplying both sides by 2( 2)( 2 4)x x x− + + ,
we obtain: 21 ( 2 4) ( )( 2)A x x Bx C x= + + + + −
Let 2x = , then
( )21 2 2(2) 4 ( (2) )(2 2)
1 121
12
A B C
A
A
= + + + + −
=
=
Let 0x = , then
( )
( )
2
23
1 0 2(0) 4 ( (0) )(0 2)
1 4 21 4 1/12 2
2
13
A B C
A CC
C
C
= + + + + −
= −
= −
− =
= −
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Section 8.5: Partial Fraction Decomposition
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Let 1x = , then
( )
( )
21 1 2(1) 4 ( (1) )(1 2)
1 711 7 1/123
112
A B C
A B C
B
B
= + + + + −
= − −
= − +
= −
( )
1 1112 312
3 2
1 112 12
2
128 2 4
42 2 4
xxx x x
xx x x
− −= +
−− + +− +
= +− + +
22. Find the partial fraction decomposition:
3 2 22 4 2 4
11 ( 1)( 1) 1x x A Bx C
xx x x x x x+ + +
= = +−− − + + + +
Multiplying both sides by 2( 1)( 1)x x x− + + , we
obtain: 22 4 ( 1) ( )( 1)x A x x Bx C x+ = + + + + −
Let 1x = , then
( ) ( )22(1) 4 1 1 1 (1) (1 1)
6 32
A B C
AA
+ = + + + + −
==
Let 0x = , then
( )22(0) 4 0 0 1 ( (0) )(0 1)
44 2
2
A B C
A CC
C
+ = + + + + −
= −= −= −
Let 1x = − , then
( )22( 1) 4 ( 1) ( 1) 1 ( ( 1) )( 1 1)
2 2 22 2 2 2( 2)
2 42
A B C
A B CB
BB
− + = − + − + + − + − −
= + −= + − −= −= −
3 22 4 2 2 2
11 1x x
xx x x+ − −
= +−− + +
23. Find the partial fraction decomposition: 2
2 2 2 21 1( 1) ( 1) ( 1) ( 1)x A B C D
x xx x x x= + + +
− +− + − +
Multiplying both sides by 2 2( 1) ( 1)x x− + , we obtain:
2 2 2
2 2
( 1)( 1) ( 1)
( 1) ( 1) ( 1)
x A x x B x
C x x D x
= − + + +
+ − + + −
Let 1x = , then 2 2 2
2 2
1 (1 1)(1 1) (1 1)
(1 1) (1 1) (1 1)1 4
14
A B
C DB
B
= − + + +
+ − + + −=
=
Let 1x = − , then 2 2 2
2 2
( 1) ( 1 1)( 1 1) ( 1 1)
( 1 1) ( 1 1) ( 1 1)1 4
14
A B
C DD
D
− = − − − + + − +
+ − − − + + − −=
=
Let 0x = , then 2 2 2
2 2
0 (0 1)(0 1) (0 1)
(0 1) (0 1) (0 1)0
1 1 14 4 2
A B
C DA B C D
A C B D
A C
= − + + +
+ − + + −= − + + +
− = +
− = + =
Let 2x = , then 2 2 2
2 2
2 (2 1)(2 1) (2 1)
(2 1) (2 1) (2 1)4 9 9 3
9 3 4 91 1 39 3 4 94 4 2
132
A B
C DA B C D
A C B D
A C
A C
= − + + +
+ − + + −= + + +
+ = − −
⎛ ⎞+ = − − =⎜ ⎟⎝ ⎠
+ =
Solve the system of equations: 1213 2
A C
A C
− =
+ =
4 114
3 14 2
14
A
A
C
C
=
=
+ =
= −
1 1 1 124 4 4 4
2 2 2 21 1( 1) ( 1) ( 1) ( 1)x
x xx x x x
−= + + +
− +− + − +
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Chapter 8: Systems of Equations and Inequalities
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24. Find the partial fraction decomposition:
2 2 2 21
2( 2) ( 2)x A B C D
x xx x x x+
= + + +−− −
Multiplying both sides by 2 2( 2)x x − , we obtain: 2 2 2 21 ( 2) ( 2) ( 2)x Ax x B x Cx x Dx+ = − + − + − +
Let 0x = , then 2 2
2 2
0+1 (0)(0 2) (0 2)
(0) (0 2) (0)1 4
14
A B
C DB
B
= − + −
+ − +=
=
Let 2x = , then 2 2
2 2
2 1 (2)(2 2) (2 2)
(2) (2 2) (2)3 4
34
A B
C DD
D
+ = − + −
+ − +=
=
Let 1x = , then 2 2
2 2
1 1 (1)(1 2) (1 2) (1) (1 2) (1)
22
1 32 14 4
A BC D
A B C DA C B D
A C
+ = − + −
+ − += + − +
− = − −
− = − − =
Let 3x = , then 2 2
2 2
3 1 (3)(3 2) (3 2) (3) (3 2) (3)
4 3 9 93 9 4 9
1 33 9 4 9 34 4
3 1
A BC D
A B C DA C B D
A C
A C
+ = − + −
+ − += + + +
+ = − −
⎛ ⎞+ = − − = −⎜ ⎟⎝ ⎠
+ = −
Solve the system of equations: 1
3 1A C
A C− =⎧
⎨ + = −⎩
( )
11
3 11 3 1
A CA C
A CC C
− == +
+ = −
+ + = −
4 2
12
C
C
= −
= −
1 11 12 2
A C= + = − + =
31 1 12 4 2 4
2 2 2 21
2( 2) ( 2)x
x xx x x x
−+= + + +
−− −
25. Find the partial fraction decomposition:
2 23
2 1( 2)( 1) ( 1)x A B C
x xx x x−
= + ++ ++ + +
Multiplying both sides by 2( 2)( 1)x x+ + , we
obtain: 23 ( 1) ( 2)( 1) ( 2)x A x B x x C x− = + + + + + +
Let 2x = − , then 22 3 ( 2 1) ( 2 2)( 2 1) ( 2 2)
55
A B CA
A
− − = − + + − + − + + − +− =
= −
Let 1x = − , then 21 3 ( 1 1) ( 1 2)( 1 1) ( 1 2)
44
A B CC
C
− − = − + + − + − + + − +− =
= −
Let 0x = , then 20 3 (0 1) (0 2)(0 1) (0 2)
3 2 23 5 2 2( 4)
2 105
A B CA B C
BBB
− = + + + + + +− = + +− = − + + −
==
2 23 5 5 4
2 1( 2)( 1) ( 1)x
x xx x x− − −
= + ++ ++ + +
26. Find the partial fraction decomposition: 2
2 22 1( 2)( 1) ( 1)x x A B C
x xx x x+
= + ++ −+ − −
Multiplying both sides by 2( 2)( 1)x x+ − , we obtain:
2 2( 1) ( 2)( 1) ( 2)x x A x B x x C x+ = − + + − + +
Let 2x = − , then 2 2( 2) ( 2) ( 2 1) ( 2 2)( 2 1)
( 2 2)2 9
29
A BC
A
A
− + − = − − + − + − −+ − +
=
=
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Section 8.5: Partial Fraction Decomposition
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Let 1x = , then 2 21 1 (1 1) (1 2)(1 1) (1 2)
2 323
A B CC
C
+ = − + + − + +=
=
Let 0x = , then 2 20 0 (0 1) (0 2)(0 1) (0 2)
0 2 22 2
2 2 142 29 3 979
A B CA B C
B A C
B
B
+ = − + + − + += − += +
⎛ ⎞= + =⎜ ⎟⎝ ⎠
=
72 229 9 3
2 22 1( 2)( 1) ( 1)x x
x xx x x+
= + ++ −+ − −
27. Find the partial fraction decomposition:
2 2 2 24
( 4) 4x A B Cx D
xx x x x+ +
= + ++ +
Multiplying both sides by 2 2( 4)x x + , we obtain: 2 2 24 ( 4) ( 4) ( )x Ax x B x Cx D x+ = + + + + +
Let 0x = , then ( )2 2 20 4 (0)(0 4) (0 4) (0) (0)
4 41
A B C DB
B
+ = + + + + +==
Let 1x = , then 2 2 21 4 (1)(1 4) (1 4) ( (1) )(1)
5 5 55 5 5
A B C DA B C DA C D
+ = + + + + += + + += + + +
5 0A C D+ + = Let 1x = − , then
2 2
21 4 ( 1)(( 1) 4) (( 1) 4)
( ( 1) )( 1)3 5 53 5 5
A BC D
A B C DA C D
− + = − − + + − +
+ − + −= − + − += − + − +
5 2A C D− − + = −
Let 2x = , then 2 2 22 4 (2)(2 4) (2 4) ( (2) )(2)
6 16 8 8 46 16 8 8 4
A B C DA B C DA C D
+ = + + + + += + + += + + +
16 8 4 2A C D+ + = −
Solve the system of equations: 5 05 2
2 21
A C DA C D
DD
+ + =− − + = −
= −= −
5 1 01 5
A CC A
+ − == −
16 8(1 5 ) 4( 1) 216 8 40 4 2
24 614
A AA A
A
A
+ − + − = −+ − − = −
− = −
=
1 5 11 5 14 4 4
C ⎛ ⎞= − = − = −⎜ ⎟⎝ ⎠
( )
1 14 4
2 2 2 2
1 14 4
2 2
14 1( 4) 4
414
xxxx x x x
xx x x
− −+= + +
+ +
− += + +
+
28. Find the partial fraction decomposition: 2
2 2 2 210 2
1( 1) ( 2) ( 1) 2x x A B Cx D
xx x x x+ +
= + +−− + − +
Multiply both sides by 2 2( 1) ( 2)x x− + : 2 2 2
2
10 2 ( 1)( 2) ( 2)
( )( 1)
x x A x x B x
Cx D x
+ = − + + +
+ + −
Let 1x = , then
( )
2 2 2
2
10(1) 2(1) (1 1)(1 2) (1 2)
(1) (1 1)12 3
4
A B
C DB
B
+ = − + + +
+ + −
==
Let 0x = , then
( )
2 2 2
2
10(0) 2(0) (0 1)(0 2) (0 2)
(0) (0 1)0 2 20 2 8
2 82 8
A B
C DA B DA D
A DD A
+ = − + + +
+ + −
= − + += − + +
− == −
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Chapter 8: Systems of Equations and Inequalities
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Let 1x = − , then 2 2
2
2
10( 1) 2( 1) ( 1 1)(( 1) 2)
(( 1) 2)
( ( 1) )( 1 1)8 6 3 4 48 6 12 4 4
6 4 4 4
A
B
C DA B C DA C D
A C D
− + − = − − − +
+ − +
+ − + − −= − + − += − + − +
− − + = −
Let 2x = , then 2 2 2
2
10(2) 2(2) (2 1)(2 2) (2 2)
( (2) )(2 1)44 6 6 244 6 24 2
6 2 20
A B
C DA B C DA C D
A C D
+ = − + + +
+ + −= + + += + + +
+ + =
Solve the system of equations (Substitute for D): 2 8
6 4 4 46 4 4(2 8) 4
2 4 282 14
D AA C D
A C AA CA C
= −− − + = −
− − + − = −− =− =
( )6 2 20
6 2 2 8 208 2 28
A C DA C A
A C
+ + =
+ + − =
+ =
Add the equations and solve:
2 14
8 2 28A CA C− =+ =
9 42143
A
A
=
=
14 282 14 143 3
143
C A
C
= − = − = −
= −
14 42 8 2 83 3
D A ⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
14 14 423 3 3
2 2 2 210 2 4
1( 1) ( 2) ( 1) 2
xx xxx x x x
− ++= + +
−− + − +
29. Find the partial fraction decomposition: 2
2 22 3
1( 1)( 2 4) 2 4x x A Bx C
xx x x x x+ + +
= +++ + + + +
Multiplying both sides by 2( 1)( 2 4)x x x+ + + , we obtain:
2 22 3 ( 2 4) ( )( 1)x x A x x Bx C x+ + = + + + + +
Let 1x = − , then 2 2( 1) 2( 1) 3 (( 1) 2( 1) 4)
( ( 1) )( 1 1)2 3
23
AB C
A
A
− + − + = − + − ++ − + − +
=
=
Let 0x = , then
( )
2 20 2(0) 3 (0 2(0) 4) ( (0) )(0 1)3 43 4 2/3
13
A B CA C
C
C
+ + = + + + + += +
= +
=
Let 1x = , then
( ) ( )
2 2
14 2 23 3 3
13
1 2(1) 3 (1 2(1) 4) ( (1) )(1 1)6 7 2 26 7 2 / 3 2 2 1/ 3
2 6
A B CA B C
B
B
B
+ + = + + + + += + +
= + +
= − − =
=
2 1 123 3 3
2 2
2 13 3
2
2 31( 1)( 2 4) 2 4
( 1)1 2 4
xx xxx x x x x
xx x x
++ += +
++ + + + +
+= +
+ + +
30. Find the partial fraction decomposition: 2
2 211 18
( 3 3) 3 3x x A Bx C
xx x x x x− − +
= ++ + + +
Multiplying both sides by 2( 3 3)x x x+ + 2( 1)( 2 4)x x x+ + + , we obtain:
2 211 18 ( 3 3) ( )x x A x x Bx C x− − = + + + +
Let 0x = , then
( ) ( )2 20 11(0) 18 0 3(0) 3 (0) (0)
18 36
A B C
AA
− − = + + + +
− == −
Let 1x = , then
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Section 8.5: Partial Fraction Decomposition
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( ) ( )2 21 11(1) 18 1 3(1) 3 (1) (1)
28 728 7( 6)
14
A B C
A B CB C
B C
− − = + + + +
− = + +− = − + ++ =
Let 1x = − , then
( )( )
2 2( 1) 11( 1) 18 ( 1) 3( 1) 3
( 1) ( 1)66 6
0
A
B CA B C
B CB C
− − − − = − + − +
+ − + −
− = + −− = − + −− =
Add the last two equations and solve: 140
B CB C+ =− =
2 147
BB==
147 14
7
B CCC
+ =+ =
=
2
2 211 18 6 7 7
( 3 3) 3 3x x x
xx x x x x− − − +
= ++ + + +
31. Find the partial fraction decomposition:
(3 2)(2 1) 3 2 2 1x A B
x x x x= +
− + − +
Multiplying both sides by (3 2)(2 1)x x− + , we obtain: (2 1) (3 2)x A x B x= + + −
Let 12x = − , then
( )( ) ( )( )1 2 1/ 2 1 3 1/ 2 221 72 2
17
A B
B
B
− = − + + − −
− = −
=
Let 23
x = , then
( )( ) ( )( )2 2 2 / 3 1 3 2 / 3 232 73 3
27
A B
A
A
= + + −
=
=
2 17 7
(3 2)(2 1) 3 2 2 1x
x x x x= +
− + − +
32. Find the partial fraction decomposition: 1
(2 3)(4 1) 2 3 4 1A B
x x x x= +
+ − + −
Multiplying both sides by (2 3)(4 1)x x+ − , we obtain: 1 (4 1) (2 3)A x B x= − + +
Let 32
x = − , then
3 31 4 1 2 32 2
1 717
A B
A
A
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − − + − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
= −
= −
Let 14
x = , then
1 11 4 1 2 34 4
71227
A B
B
B
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
=
=
1 27 71
(2 3)(4 1) 2 3 4 1x x x x−
= ++ − + −
33. Find the partial fraction decomposition:
2 ( 3)( 1) 3 12 3x x A B
x x x xx x= = +
+ − + −+ −
Multiplying both sides by ( 3)( 1)x x+ − , we obtain: ( 1) ( 3)x A x B x= − + +
Let 1x = , then 1 (1 1) (1 3)1 4
14
A BB
B
= − + +=
=
Let 3x = − , then 3 ( 3 1) ( 3 3)3 4
34
A BA
A
− = − − + − +− = −
=
3 14 4
2 3 12 3x
x xx x= +
+ −+ −
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Chapter 8: Systems of Equations and Inequalities
848 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
34. Find the partial fraction decomposition: 2 2
28 8
( 1)( 2)( 3)( 1)( 5 6)
1 2 3
x x x xx x xx x xA B C
x x x
− − − −=
+ + ++ + +
= + ++ + +
Multiplying both sides by ( 1)( 2)( 3)x x x+ + + , we obtain:
2 8 ( 2)( 3) ( 1)( 3) ( 1)( 2)x x A x x B x x
C x x− − = + + + + +
+ + +
Let 1x = − , then 2( 1) ( 1) 8 ( 1 2)( 1 3)
( 1 1)( 1 3) ( 1 1)( 1 2)
6 23
AB
CA
A
− − − − = − + − ++ − + − ++ − + − +
− == −
Let 2x = − , then 2( 2) ( 2) 8 ( 2 2)( 2 3)
( 2 1)( 2 3) ( 2 1)( 2 2)
22
AB
CB
B
− − − − = − + − ++ − + − ++ − + − +
− = −=
Let 3x = − , then 2( 3) ( 3) 8 ( 3 2)( 3 3)
( 3 1)( 3 3) ( 3 1)( 3 2)
4 22
AB
CC
C
− − − − = − + − ++ − + − ++ − + − +
==
2
28 3 2 2
1 2 3( 1)( 5 6)x x
x x xx x x− − −
= + ++ + ++ + +
35. Find the partial fraction decomposition: 2
2 2 2 2 22 3
( 4) 4 ( 4)x x Ax B Cx D
x x x+ + + +
= ++ + +
Multiplying both sides by 2 2( 4)x + , we obtain: 2 2
2 3 2
2 3 2
2 3 ( )( 4)
2 3 4 4
2 3 (4 ) 4
x x Ax B x Cx D
x x Ax Bx Ax B Cx D
x x Ax Bx A C x B D
+ + = + + + +
+ + = + + + + +
+ + = + + + + +
0A = ; 1B = ;
4 24(0) 2
2
A CCC
+ =+ =
=
4 34(1) 3
1
B DDD
+ =+ =
= −
2
2 2 2 2 22 3 1 2 1
( 4) 4 ( 4)x x x
x x x+ + −
= ++ + +
36. Find the partial fraction decomposition: 3
2 2 2 2 21
( 16) 16 ( 16)x Ax B Cx D
x x x+ + +
= ++ + +
Multiplying both sides by 2 2( 16)x + , we obtain: 3 2
3 3 2
3 3 2
1 ( )( 16)
1 16 16
1 (16 ) 16
x Ax B x Cx D
x Ax Bx Ax B Cx D
x Ax Bx A C x B D
+ = + + + +
+ = + + + + +
+ = + + + + +
1A = ; 0B = ;
16 0
16(1) 016
A CCC
+ =+ =
= −
16 1
16(0) 11
B DDD
+ =+ =
=
3
2 2 2 2 21 16 1
( 16) 16 ( 16)x x x
x x x+ − +
= ++ + +
37. Find the partial fraction decomposition:
3 27 3 7 3
( 3)( 1)2 3
3 1
x xx x xx x xA B Cx x x
+ +=
− +− −
= + +− +
Multiplying both sides by ( 3)( 1)x x x− + , we obtain: 7 3 ( 3)( 1) ( 1) ( 3)x A x x Bx x Cx x+ = − + + + + −
Let 0x = , then 7(0) 3 (0 3)(0 1) (0)(0 1) (0)(0 3)
3 31
A B CA
A
+ = − + + + + −= −= −
Let 3x = , then 7(3) 3 (3 3)(3 1) (3)(3 1) (3)(3 3)
24 122
A B CB
B
+ = − + + + + −==
Let 1x = − , then 7( 1) 3 ( 1 3)( 1 1) ( 1)( 1 1)
( 1)( 1 3)4 4
1
A BC
CC
− + = − − − + + − − ++ − − −
− == −
3 27 3 1 2 1
3 12 3x
x x xx x x+ − −
= + +− +− −
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Section 8.5: Partial Fraction Decomposition
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38. Find the partial fraction decomposition: 5 4 3 2
6 4 4
4 3 2
4
2 3 4
1 ( 1)( 1)( 1)( 1)
1( 1)
1
x x x x x xx x x x x
x x x xx x
A B C D Ex xx x x
+ + − + − +=
− − +
− + − +=
−
= + + + +−
Multiplying both sides by 4 ( 1)x x − , we obtain: 4 3 2 3
2
4
1 ( 1)
( 1) ( 1)
( 1)
x x x x Ax x
Bx x Cx x
D x Ex
− + − + = −
+ − + −
+ − +
Let 0x = , then 4 3 2 3 2
4
0 0 0 0 1 0 (0 1) 0 (0 1) 0(0 1)
(0 1) 0
A BC
D E
− + − + = ⋅ − + ⋅ −+ ⋅ −
+ − + ⋅
11D
D= −= −
Let 1x = , then 4 3 2 3 2
4
1 1 1 1 1 1 (1 1) 1 (1 1) 1(1 1)
(1 1) 1
A BC
D E
− + − + = ⋅ − + ⋅ −+ ⋅ −
+ − + ⋅
1 E= Let 1x = − , then
4 3 2
3 2
4
( 1) ( 1) ( 1) ( 1) 1
( 1) ( 1 1) ( 1) ( 1 1)
( 1)( 1 1) ( 1 1) ( 1)
A B
C D E
− − − + − − − +
= − − − + − − −
+ − − − + − − + −
5 2 2 2 25 2 2 2 2( 1) 1
2 2 2 21
A B C D EA B C
A B CA B C
= − + − += − + − − +
− + =− + =
Let 2x = , then 4 3 2 3 2
4
2 2 2 2 1 2 (2 1) 2 (2 1) 2(2 1)
(2 1) 2
A BC
D E
− + − + = ⋅ − + ⋅ −+ ⋅ −
+ − + ⋅
11 8 4 2 1611 8 4 2 ( 1) 16(1)
8 4 2 44 2 2
A B C D EA B C
A B CA B C
= + + + += + + + − +
+ + = −+ + = −
Let 3x = , then 4 3 2 3 2
4
3 3 3 3 1 3 (3 1) 3 (3 1) 3(3 1)
(3 1) 3
A BC
D E
− + − + = ⋅ − + ⋅ −+ ⋅ −
+ − + ⋅
61 54 18 6 2 8161 54 18 6 2( 1) 81(1)18 54 18 6
9 3 3
A B C D EA B CA B C
A B C
= + + + += + + + − +
− = + ++ + = −
Solve the system of equations by using a matrix equation:
4 2 2 1
9 3 3
A B CA B CA B C
+ + = −⎧⎪ − + =⎨⎪ + + = −⎩
1
4 2 1 21 1 1 19 3 1 3
4 2 1 2 01 1 1 1 19 3 1 3 0
ABC
ABC
−
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
So, 0A = , 1B = − , and 0C = . Thus, 5 4 3 2
6 4 4
2 4
1 1( 1)
1 1 11
x x x x xx x x x
xx x
+ − + − +=
− −− −
= + +−
39. Perform synthetic division to find a factor:
2 1 4 5 22 4 2
1 2 1 0
− −−
−
3 2 2
2
4 5 2 ( 2)( 2 1)
( 2)( 1)
x x x x x x
x x
− + − = − − +
= − −
Find the partial fraction decomposition:
2 2
3 2 2
2
4 5 2 ( 2)( 1)
2 1 ( 1)
x xx x x x x
A B Cx x x
=− + − − −
= + +− − −
Multiplying both sides by 2( 2)( 1)x x− − , we obtain:
2 2( 1) ( 2)( 1) ( 2)x A x B x x C x= − + − − + −
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Chapter 8: Systems of Equations and Inequalities
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Let 2x = , then 2 22 (2 1) (2 2)(2 1) (2 2)4
A B CA
= − + − − + −=
Let 1x = , then 2 21 (1 1) (1 2)(1 1) (1 2)1
1
A B CC
C
= − + − − + −= −= −
Let 0x = , then 2 20 (0 1) (0 2)(0 1) (0 2)0 2 20 4 2 2( 1)
2 63
A B CA B C
BBB
= − + − − + −= + −= + − −
− == −
2
3 2 24 3 1
2 14 5 2 ( 1)x
x xx x x x− −
= + +− −− + − −
40. Perform synthetic division to find a factor: 1 1 1 5 3
1 2 3
1 2 3 0
−−
−
3 2 2
2
5 3 ( 1)( 2 3)
( 3)( 1)
x x x x x x
x x
+ − + = − + −
= + −
Find the partial fraction decomposition:
2 2
3 2 2
2
1 15 3 ( 3)( 1)
3 1 ( 1)
x xx x x x x
A B Cx x x
+ +=
+ − + + −
= + ++ − −
Multiplying both sides by 2( 3)( 1)x x+ − , we obtain:
2 21 ( 1) ( 3)( 1) ( 3)x A x B x x C x+ = − + + − + +
Let 3x = − , then 2 2( 3) 1 ( 3 1) ( 3 3)( 3 1)
( 3 3)10 16
58
A BC
A
A
− + = − − + − + − −+ − +
=
=
Let 1x = , then 2 21 1 (1 1) (1 3)(1 1) (1 3)
2 412
A B CC
C
+ = − + + − + +=
=
Let 0x = , then 2 20 1 (0 1) (0 3)(0 1) (0 3)
1 3 35 11 3 38 293838
A B CA B C
B
B
B
+ = − + + − + += − +
⎛ ⎞= − + ⎜ ⎟⎝ ⎠
=
=
5 3 128 8 2
3 2 21
3 15 3 ( 1)x
x xx x x x+
= + ++ −+ − + −
41. Find the partial fraction decomposition: 3
2 3 2 2 2 2 3( 16) 16 ( 16) ( 16)x Ax B Cx D Ex F
x x x x+ + +
= + ++ + + +
Multiplying both sides by 2 3( 16)x + , we obtain: 3 2 2 2
3 4 2 3 2
3 5 4 3 2
( )( 16) ( )( 16)
( )( 32 256) 16 16
32 32 256
x Ax B x Cx D xEx F
x Ax B x x Cx DxCx D Ex F
x Ax Bx Ax Bx Ax
= + + + + ++ +
= + + + + ++ + + +
= + + + +3 2 256
16 16B Cx DxCx D Ex F
+ + ++ + + +
3 5 4 3 2(32 ) (32 ) (256 16 )
(256 16 )
x Ax Bx A C x B D xA C E x
B D F
= + + + + ++ + +
+ + +
0; 0A B= = ; 32 132(0) 1
1
A CCC
+ =+ =
=
32 0
32(0) 00
B DDD
+ =+ =
=
256 16 0256(0) 16(1) 0
16
A C EEE
+ + =+ + =
= −
256 16 0
256(0) 16(0) 00
B D FFF
+ + =+ + =
=
3
2 3 2 2 2 316
( 16) ( 16) ( 16)x x x
x x x−
= ++ + +
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Section 8.5: Partial Fraction Decomposition
851 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
42. Find the partial fraction decomposition: 2
2 3 2 2 2 2 3( 4) 4 ( 4) ( 4)x Ax B Cx D Ex F
x x x x+ + +
= + ++ + + +
Multiplying both sides by 2 3( 4)x + , we obtain: 2 2 2 2
2 4 2 3 2
2 5 4 3 2
3 2
( )( 4) ( )( 4)
( )( 8 16) 4 4
8 8 16 16
4 4
x Ax B x Cx D xEx F
x Ax B x x Cx DxCx D Ex F
x Ax Bx Ax Bx Ax B
Cx Dx Cx D Ex
= + + + + ++ +
= + + + + ++ + + +
= + + + + +
+ + + + +2 5 4 3 2(8 ) (8 )
(16 4 ) (16 4 )
F
x Ax Bx A C x B D xA C E x B D F
+
= + + + + ++ + + + + +
0; 0A B= = ; 8 08(0) 0
0
A CCC
+ =+ =
=
8 18(0) 1
1
B DDD
+ =+ =
=
16 4 016(0) 4(0) 0
0
A C EEE
+ + =+ + =
=
16 4 016(0) 4(1) 0
4
B D FFF
+ + =+ + =
= −
2
2 3 2 2 2 31 4
( 4) ( 4) ( 4)x
x x x−
= ++ + +
43. Find the partial fraction decomposition:
24 4
( 3)(2 1) 3 2 12 5 3A B
x x x xx x= = +
− + − +− −
Multiplying both sides by ( 3)(2 1)x x− + , we obtain: 4 (2 1) ( 3)A x B x= + + −
Let 12
x = − , then
1 14 2 1 32 2
74 287
A B
B
B
⎛ ⎞⎛ ⎞ ⎛ ⎞= − + + − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
= −
= −
Let 3x = , then 4 (2(3) 1) (3 3)4 7
47
A BA
A
= + + −=
=
847 7
24
3 2 12 5 3 x xx x
−= +
− +− −
44. Find the partial fraction decomposition:
24 4
( 2)(2 1) 2 2 12 3 2x x A B
x x x xx x= = +
+ − + −+ −
Multiplying both sides by ( 2)(2 1)x x+ − , we obtain: 4 (2 1) ( 2)x A x B x= − + +
Let 12
x = , then
1 1 14 2 1 22 2 2
52345
A B
B
B
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
=
=
Let 2x = − , then 4( 2) (2( 2) 1) ( 2 2)
8 585
A BA
A
− = − − + − +− = −
=
8 45 5
24
2 2 12 3 2x
x xx x= +
+ −+ −
45. Find the partial fraction decomposition:
4 2 2
2
2 3 2 39 ( 3)( 3)
3 3
x xx x x x x
A B C Dx x xx
+ +=
− − +
= + + +− +
Multiplying both sides by 2 ( 3)( 3)x x x− + , we obtain:
2 2
2 3 ( 3)( 3) ( 3)( 3)
( 3) ( 3)
x Ax x x B x x
Cx x Dx x
+ = − + + − +
+ + + −
Let 0x = , then
2 2
2 0 3 0(0 3)(0 3) (0 3)(0 3)
0 (0 3) 0 (0 3)3 9
13
A B
C DB
B
⋅ + = ⋅ − + + − +
+ ⋅ + + ⋅ −= −
= −
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Chapter 8: Systems of Equations and Inequalities
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Let 3x = , then
2 2
2 3 3 3(3 3)(3 3) (3 3)(3 3)
3 (3 3) 3 (3 3)9 54
16
A B
C DC
C
⋅ + = ⋅ − + + − +
+ ⋅ + + ⋅ −=
=
Let 3x = − , then
2
2
2( 3) 3 ( 3)( 3 3)( 3 3) ( 3 3)( 3 3)
( 3) ( 3 3)
( 3) ( 3 3)3 54
118
AB
C
DD
D
− + = − − − − ++ − − − +
+ − − +
+ − − −− = −
=
Let 1x = , then
( ) ( ) ( )
2 2
2 1 3 1(1 3)(1 3) (1 3)(1 3)
1 (1 3) 1 (1 3)5 8 8 4 25 8 8 1/ 3 4 1/ 6 2 1/18
8 2 15 83 3 9
1689
29
A B
C DA B C DA
A
A
A
⋅ + = ⋅ − + + − +
+ ⋅ + + ⋅ −= − − + −
= − − − + −
= − + + −
− =
= −
2 1 1 19 3 6 18
4 2 22 3
3 39x
x x xx x x
− −+= + + +
− +−
46. Find the partial fraction decomposition: 2 2
4 2 2
2
9 92 8 ( 2)( 2)( 2)
2 2 2
x xx x x x x
A B Cx Dx x x
+ +=
− − + − ++
= + +− + +
Multiplying both sides by 2( 2)( 2)( 2)x x x+ − + , we obtain:
2 2 29 ( 2)( 2) ( 2)( 2) ( )( 2)( 2)x A x x B x x
Cx D x x+ = + + + − +
+ + − +
Let 2x = , then 2 2 22 9 (2 2)(2 2) (2 2)(2 2)
( (2) )(2 2)(2 2)13 24
1324
A BC D
A
A
+ = + + + − ++ + − +
=
=
Let 2x = − , then 2 2
2
( 2) 9 (( 2) 2)( 2 2)
( 2 2)(( 2) 2) ( ( 2) )( 2 2)( 2 2)
13 241324
A
BC D
B
B
− + = − + − +
+ − − − ++ − + − − − +
= −
= −
Let 0x = , then 2 2 20 9 (0 2)(0 2) (0 2)(0 2)
( (0) )(0 2)(0 2)9 4 4 4
13 139 4 4 424 24
144376
A BC D
A B D
D
D
D
+ = + + + − ++ + − +
= − −
⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
= −
Let 1x = , then 2 2 21 9 (1 2)(1 2) (1 2)(1 2)
( (1) )(1 2)(1 2)10 9 3 3 3
39 13 710 38 8 2
3 00
A BC D
A B C D
C
CC
+ = + + + − ++ + − +
= − − −
= + − +
==
713 132624 24
4 2 29
2 22 8 2x
x xx x x
−−+= + +
− +− − +
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Section 8.6: Systems of Nonlinear Equations
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Section 8.6
1. 3 2y x= + The graph is a line. x-intercept:
0 3 23 2
23
xx
x
= += −
= −
y-intercept: ( )3 0 2 2y = + =
2
−2
y
x
(0, 2)2 ,03
⎛ ⎞−⎜ ⎟⎝ ⎠
2
−2
2. 2
2
4
4
y x
y x
+ =
= −
The graph is a parabola. x-intercepts:
2
2
0 4
42, 2
x
xx x
= −
== − =
y-intercept: 20 4 4y = − = − The vertex has x-coordinate:
( )0 0
2 2 1bxa
= − = − = .
The y-coordinate of the vertex is 20 4 4y = − = − .
3. 2 2
2 2
22
2 2
1
1
11 1
y x
x y
yx
= −
− =
− =
The graph is a hyperbola with center (0, 0), transverse axis along the x-axis, and vertices at ( 1,0)− and (1,0) . The asymptotes are y x= − and y x= .
y
x−5(−1, 0)
5
−5
5
(1, 0)
4. 2 2
2 2
22
22
2 2
4 4
4 44 4
14
12 1
x y
x y
x y
yx
+ =
+=
+ =
+ =
The graph is an ellipse with center (0,0) , major axis along the x-axis, vertices at ( 2,0)− and (2,0) . The graph also has y-intercepts at (0, 1)− and (0,1) .
y
x−5
(−2, 0)
5
−5
5
(2, 0)
(0, −1)
(0, 1)
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Chapter 8: Systems of Equations and Inequalities
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5. 2 1
1y xy x
⎧ = +⎪⎨
= +⎪⎩
(0, 1) and (1, 2) are the intersection points.
Solve by substitution: 2
2
1 1
0( 1) 0
0 or 11 2
x x
x xx x
x xy y
+ = +
− =− =
= == =
Solutions: (0, 1) and (1, 2)
6. 2 1
4 1y xy x
⎧ = +⎪⎨
= +⎪⎩
(0, 1) and (4, 17) are the intersection points.
Solve by substitution: 2
2
1 4 1
4 0( 4) 0
0 or 41 17
x x
x xx x
x xy y
+ = +
− =− =
= == =
Solutions: (0, 1) and (4, 17)
7. 236
8y xy x
⎧⎪ = −⎨
= −⎪⎩
(2.59, 5.41) and (5.41, 2.59) are the intersection points.
Solve by substitution: 2
2 2
2
2
36 8
36 64 16
2 16 28 0
8 14 0
8 64 562
8 2 22
4 2
x x
x x x
x x
x x
x
− = −
− = − +
− + =
− + =
± −=
±=
= ±
( )( )
If 4 2, 8 4 2 4 2
If 4 2, 8 4 2 4 2
x y
x y
= + = − + = −
= − = − − = +
Solutions: ( ) ( )4 2, 4 2 and 4 2, 4 2+ − − +
8. 24
2 4y xy x
⎧⎪ = −⎨
= +⎪⎩
(–2, 0) and (–1.2, 1.6) are the intersection points.
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Section 8.6: Systems of Nonlinear Equations
855 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Solve by substitution: 2
2 2
2
4 2 4
4 4 16 16
5 16 12 0( 2)(5 6) 0
62 or 5
80 or 5
x x
x x x
x xx x
x x
y y
− = +
− = + +
+ + =+ + =
= − = −
= =
Solutions: ( ) 6 82, 0 and ,5 5
⎛ ⎞− −⎜ ⎟⎝ ⎠
9. 2
y xy x
⎧ =⎪⎨
= −⎪⎩
(1, 1) is the intersection point.
Solve by substitution:
2
2
2
4 4
5 4 0( 4)( 1) 0
4 or 12 or =1
x x
x x x
x xx x
x xy y
= −
= − +
− + =− − =
= == −
Eliminate (4, –2); we must have 0y ≥ . Solution: (1, 1)
10. 6
y xy x
⎧ =⎪⎨
= −⎪⎩
(4, 2) is the intersection point.
Solve by substitution:
2
2
6
36 12
13 36 0( 4)( 9) 0
4 or 92 or 3
x x
x x x
x xx x
x xy y
= −
= − +
− + =− − =
= == = −
Eliminate (9, –3); we must have 0y ≥ . Solution: (4, 2)
11. 2
2
2
x y
x y y
=⎧⎪⎨
= −⎪⎩
(0, 0) and (8, 4) are the intersection points.
Solve by substitution: 2
2
2 2
4 0( 4) 0
0 or =40 or =8
y y y
y yy y
y yx x
= −
− =− =
==
Solutions: (0, 0) and (8, 4)
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Chapter 8: Systems of Equations and Inequalities
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12. 2
1
6 9
y x
y x x
= −⎧⎪⎨
= − +⎪⎩
(2, 1) and (5, 4) are the intersection points.
Solve by substitution: 2
2
6 9 1
7 10 0( 2)( 5) 0
2 or =51 or =4
x x x
x xx x
x xy y
− + = −
− + =− − =
==
Solutions: (2, 1) and (5, 4)
13. 2 2
2 2
4
2 0
x y
x x y
⎧ + =⎪⎨
+ + =⎪⎩
(–2, 0) is the intersection point.
Substitute 4 for 2 2x y+ in the second equation.
2
2 4 02 4
2
4 ( 2) 0
xxx
y
+ == −= −
= − − =
Solution: (–2, 0)
14. 2 2
2 2
8
4 0
x y
x y y
⎧ + =⎪⎨
+ + =⎪⎩
(–2, –2) and (2, –2) are the intersection points.
Substitute 8 for 2 2x y+ in the second equation.
2
8 4 04 8
2
8 ( 2) 2
yyy
x
+ == −= −
= ± − − = ±
Solution: (–2, –2) and (2, –2)
15. 2 2
3 5
5
y x
x y
= −⎧⎪⎨
+ =⎪⎩
(1, –2) and (2, 1) are the intersection points.
Solve by substitution: 2 2
2 2
2
2
(3 5) 5
9 30 25 5
10 30 20 0
3 2 0( 1)( 2) 0
x x
x x x
x x
x xx x
+ − =
+ − + =
− + =
− + =− − =
1 or 2
2 1x xy y
= == − =
Solutions: (1, –2) and (2, 1)
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16. 2 2 10
2x y
y x⎧ + =⎪⎨
= +⎪⎩
(1, 3) and (–3, –1) are the intersection points.
Solve by substitution: 2 2
2 2
2
( 2) 10
4 4 10
2 4 6 02( 3)( 1) 0
x x
x x x
x xx x
+ + =
+ + + =
+ − =+ − =
3 or 11 3
x xy y
= − == − =
Solutions: (–3, –1) and (1, 3)
17. 2 2
2
4
4
x y
y x
⎧ + =⎪⎨
− =⎪⎩
(–1, 1.73), (–1, –1.73), (0, 2), and (0, –2) are the intersection points.
2Substitute 4 for in the first equation:x y+ 2
2
2 2
4 4
0( 1) 0
0 or 1 4 3
2 3
x x
x xx x
x xy yy y
+ + =
+ =+ =
= = −
= =
= ± = ±
Solutions: ( ) ( )(0, – 2), (0, 2), 1, 3 , 1, 3− − −
18. 2 2
2
16
2 8
x y
x y
⎧ + =⎪⎨
− =⎪⎩
(–3.46, 2), (0, –4), and (3.46, 2) are the intersection points.
Substitute 2 8y + for 2x in the first equation. 2
2
2 8 16
2 8 0( 4)( 2) 0
y y
y yy y
+ + =
+ − =+ − =
2 2
4 or 2
0 or 12
0 2 3
y y
x x
x x
= − =
= =
= = ±
Solutions: ( ) ( )(0, – 4), 2 3, 2 , 2 3, 2−
19. 2 2
4
8
xy
x y
=⎧⎪⎨
+ =⎪⎩
(–2, –2) and (2, 2) are the intersection points.
Solve by substitution: 2
2
22
4 2
4 2
2 2
2
2
4 8
16 8
16 88 16 0
( 4) 04 0
42 or 22 or 2
xx
xx
x xx x
xx
xx xy y
⎛ ⎞+ =⎜ ⎟⎝ ⎠
+ =
+ =− + =
− =
− === = −= = −
Solutions: (–2, –2) and (2, 2)
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Chapter 8: Systems of Equations and Inequalities
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20. 2
1x yxy
⎧ =⎪⎨
=⎪⎩
(1, 1) is the intersection point.
Solve by substitution: 2
3
2
1
11
(1) 1
xx
xx
y
=
==
= =
Solution: (1, 1)
21. 2 2
2
4
9
x y
y x
⎧ + =⎪⎨
= −⎪⎩
No solution; Inconsistent.
Solve by substitution: 2 2 2
2 4 2
4 2
( 9) 4
18 81 4
17 77 0
x x
x x x
x x
+ − =
+ − + =
− + =
2 17 289 4(77)2
17 192
x± −
=
± −=
There are no real solutions to this expression. Inconsistent.
22. 12 1
xyy x
=⎧⎨ = +⎩
(–1, –1) and (0.5, 2) are the intersection points.
Solve by substitution:
2
(2 1) 1
2 1 0( 1)(2 1) 0
11 or 2
1 2
x x
x xx x
x x
y y
+ =
+ − =+ − =
= − =
= − =
Solutions: (–1, –1) and 1 , 22
⎛ ⎞⎜ ⎟⎝ ⎠
23. 2 4
6 13y xy x
⎧ = −⎪⎨
= −⎪⎩
(3, 5) is the intersection point.
Solve by substitution: 2
2
2
2
4 6 13
6 9 0
( 3) 03 0
3
(3) 4 5
x x
x x
xx
x
y
− = −
− + =
− =− =
=
= − =
Solution: (3,5)
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24. 2 2 103
x yxy
⎧ + =⎪⎨
=⎪⎩
(1, 3), (3, 1), (–3, –1), and (–1, –3) are the intersection points.
Solve by substitution:
( )22
2
4 2
4 2
2 2
2
3
9
10
10
9 10
10 9 0
( 9)( 1) 0( 3)( 3)( 1)( 1) 0
x
x
x
x
x x
x x
x xx x x x
+ =
+ =
+ =
− + =
− − =− + − + =
3 or –3 or 1 or 11 1 3 –3
x x x xy y y y
= = = = −= = − = =
Solutions: (3, 1), (–3, –1), (1, 3), (–1, –3)
25. Solve the second equation for y, substitute into the first equation and solve:
2 22 1844
x y
xy yx
⎧ + =⎪⎨
= ⇒ =⎪⎩
( )( )
22
22
4 2
4 2
4 2
2 2
42 18
162 18
2 16 18
2 18 16 0
9 8 0
8 1 0
xx
xx
x x
x x
x x
x x
⎛ ⎞+ =⎜ ⎟⎝ ⎠
+ =
+ =
− + =
− + =
− − =
2 28 or 1or 18= 2 2
x xxx
= == ±= ± ±
4If 2 2 : 22 2
4If 2 2 : 22 2
4If 1: 414If 1: 41
x y
x y
x y
x y
= = =
= − = = −−
= = =
= − = = −−
Solutions:
( ) ( )2 2, 2 , 2 2, 2 , (1, 4), ( 1, 4)− − − −
26. Solve the second equation for y, substitute into the first equation and solve:
2 2 217 7
x yx y y x
⎧ − =⎪⎨
+ = ⇒ = −⎪⎩
( )( )
22
2 2
7 21
49 14 21
14 7057 5 2
x x
x x x
xxy
− − =
− − + =
=== − =
Solution: (5, 2)
27. Substitute the first equation into the second equation and solve:
2 2
2 1
2 1
y x
x y
= +⎧⎪⎨
+ =⎪⎩
( )
( )
22
2 2
2
2 2 1 1
2 4 4 1 1
6 4 02 3 2 0
x x
x x x
x xx x
+ + =
+ + + =
+ =
+ =
2 0 or 3 2 020 or 3
x x
x x
= + =
= = −
If 0 : 2(0) 1 12 2 4 1If : 2 1 13 3 3 3
x y
x y
= = + =
⎛ ⎞= − = − + = − + = −⎜ ⎟⎝ ⎠
Solutions: 2 1(0, 1), ,3 3
⎛ ⎞− −⎜ ⎟⎝ ⎠
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Chapter 8: Systems of Equations and Inequalities
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28. Solve the second equation for x and substitute into the first equation and solve:
2 24 162 2 2 2
x yy x x y
⎧ − =⎨ − = ⇒ = −⎩
( )
2 2
2 2
32
(2 2) 4 164 8 4 4 16
8 1232
2 2 5
y yy y y
y
y
x
− − =− + − =
− =
= −
= − − = −
Solutions: ( )325,− −
29. Solve the first equation for y, substitute into the second equation and solve:
2 21 0 1
6 5x y y x
x y y x+ + = ⇒ = − −⎧
⎨ + + − = −⎩
2 2
2 2
2
( 1) 6( 1) 52 1 6 6 5
2 5 0(2 5) 0
x x x xx x x x x
x xx x
+ − − + − − − = −
+ + + − − − = −
− =− =
520 or x x= =
If 0 : (0) 1 15 5 7If : 12 2 2
x y
x y
= = − − = −
= = − − = −
Solutions: ( )5 72 2(0, 1), ,− −
30. Solve the second equation for y, substitute into the first equation and solve:
2 22 844
x xy y
xy yx
⎧ − + =⎪⎨
= ⇒ =⎪⎩
( )( )( )( )
22
22
4 2
4 2
2 2
4 42 8
162 4 8
2 16 126 8 0
4 2 0
( 2)( 2) 2 2 0
x xx x
xx
x xx x
x x
x x x x
⎛ ⎞ ⎛ ⎞− + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
− + =
+ =− + =
− − =
− + − + =
2 or 2 or 2 or 22 2 2 2 2 2
x x x xy y y y
= = − = = −= = − = = −
Solutions:
( ) ( )( 2, 2), 2, 2 2 , 2, 2 2 , (2, 2)− − − −
31. Solve the second equation for y, substitute into the first equation and solve:
2 24 3 9 152 52 3 53 3
x xy y
x y y x
⎧ − + =⎪⎨
+ = ⇒ = − +⎪⎩
2
2
2 2 2
2
2 5 2 54 3 9 153 3 3 3
4 2 5 4 20 25 15
10 25 10 0
x x x x
x x x x x
x x
⎛ ⎞ ⎛ ⎞− − + + − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+ − + − + =
− + =
22 5 2 0(2 1)( 2) 0
x xx x
− + =− − =
1 or 22
x x= =
1 2 1 5 4If :2 3 2 3 3
2 5 1If 2 : (2)3 3 3
x y
x y
⎛ ⎞= = − + =⎜ ⎟⎝ ⎠
= = − + =
Solutions: 1 4 1, , 2,2 3 3
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
32. 22 3 6 2 4 0
2 3 4 0y xy y x
x y⎧ − + + + =⎪⎨
− + =⎪⎩
Solve the second equation for x, substitute into the first equation and solve: 2 3 4 0
2 3 43 4
2
x yx y
yx
− + == −
−=
2
2 2
2
2
3 4 3 42 3 6 2 42 2
92 6 6 3 4 42
5 15 025 30 05 ( 6) 0
y yy y y
y y y y y
y y
y yy y
− −⎛ ⎞ ⎛ ⎞− + + = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
− + + + − = −
− + =
− + =− − =
0 or 6y y= = ( )3 0 4
If 0 : 22
y x−
= = −
( )3 6 4If 6 : 7
2y x
−= = =
Solutions: (–2, 0), (7, 6)
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33. Multiply each side of the second equation by 4 and add the equations to eliminate y:
2 2 2 2
42 2 2 2
2
2
4 7 0 4 73 31 12 4 124
13 117 9 3
x y x yx y x y
xxx
⎧ − + = ⎯⎯→ − = −⎪⎨
+ = ⎯⎯→ + =⎪⎩=
== ±
If 3x = : 2 2 23(3) 31 4 2y y y+ = ⇒ = ⇒ = ±
If 3x = − : 2 2 23( 3) 31 4 2y y y− + = ⇒ = ⇒ = ± Solutions: (3, 2), (3, –2), (–3, 2), (–3, –2)
34. 2 2
2 2
3 2 5 0
2 2 0
x y
x y
⎧ − + =⎪⎨
− + =⎪⎩
2 2
2 2
3 2 5
2 2
x y
x y
⎧ − = −⎪⎨
− = −⎪⎩
Multiply each side of the second equation by –2 and add the equations to eliminate y:
2 2
2 2
2
2
3 2 54 2 4
1 1 1
x yx y
xxx
− = −
− + =
− = −
== ±
2 2 2
2 2 2
If 1:
2(1) 2 4 2If 1:
2( 1) 2 4 2
x
y y yx
y y y
=
− = − ⇒ = ⇒ = ±= −
− − = − ⇒ = ⇒ = ±
Solutions: (1, 2), (1, –2), (–1, 2), (–1, –2)
35. 2 2
2 2
7 3 5 0
3 5 12
x y
x y
⎧ − + =⎪⎨
+ =⎪⎩
2 2
2 2
7 3 5
3 5 12
x y
x y
⎧ − = −⎪⎨
+ =⎪⎩
Multiply each side of the first equation by 5 and each side of the second equation by 3 and add the equations to eliminate y:
2 2
2 2
2
2
35 15 25
9 15 36
44 111 4
1 2
x y
x y
x
x
x
− = −
+ =
=
=
= ±
2
2 2
22 2
12
12
If :
1 9 33 5 122 4 2
If :
1 9 33 5 122 4 2
x
y y y
x
y y y
=
⎛ ⎞ + = ⇒ = ⇒ = ±⎜ ⎟⎝ ⎠
= −
⎛ ⎞− + = ⇒ = ⇒ = ±⎜ ⎟⎝ ⎠
Solutions: 1 3 1 3 1 3 1 3, , , , , , ,2 2 2 2 2 2 2 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
36. 2 2
2 2
3 1 0
2 7 5 0
x y
x y
⎧ − + =⎪⎨
− + =⎪⎩
2 2
2 2
3 1
2 7 5
x y
x y
⎧ − = −⎪⎨
− = −⎪⎩
Multiply each side of the first equation by –2 and add the equations to eliminate x:
2 2
2 2
2
2
2 6 2
2 7 5
3
3
3
x y
x y
y
y
y
− + =
− = −
− = −
=
= ±
( )
( )
22 2
22 2
If 3 :
3 3 1 8 2 2
If 3 :
3 3 1 8 2 2
y
x x x
y
x x x
=
− = − ⇒ = ⇒ = ±
= −
− − = − ⇒ = ⇒ = ±
Solutions:
( ) ( ) ( )( )2 2, 3 , 2 2, 3 , 2 2, 3 ,
2 2, 3
− −
− −
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37. Multiply each side of the second equation by 2 and add the equations to eliminate xy:
2 2
22 2
2
2
2 10 2 10
3 2 6 2 4
7 14
2
2
x xy x xy
x xy x xy
x
x
x
⎧ + = ⎯⎯→ + =⎪⎨
− = ⎯⎯→ − =⎪⎩
=
=
= ±
( )
( ) ( )
2
2
If 2 :
3 2 2 2
42 4 2 22
If 2 :
3 2 2 2
42 4 2 22
x
y
y y y
x
y
y y y
=
− ⋅ =
⇒ − ⋅ = − ⇒ = ⇒ =
= −
− − − =
−⇒ ⋅ = − ⇒ = ⇒ = −
Solutions: ( ) ( )2, 2 2 , 2, 2 2− −
38. 2
2
5 13 36 0
7 6
xy y
xy y
⎧ + + =⎪⎨
+ =⎪⎩
2
2
5 13 36
7 6
xy y
xy y
⎧ + = −⎪⎨
+ =⎪⎩
Multiply each side of the second equation by –5 and add the equations to eliminate xy:
2
2
2
2
5 13 36
5 35 30
22 66
3
3
xy y
xy y
y
y
y
+ = −
− − = −
− = −
=
= ±
( ) ( )2
If 3 :
3 7 3 6
153 15 5 33
y
x
x x x
=
+ =
−⇒ ⋅ = − ⇒ = ⇒ = −
( ) ( )2
If 3 :
3 7 3 6
153 15 5 33
y
x
x x x
= −
− + − =
⇒ − ⋅ = − ⇒ = ⇒ =
Solutions: ( ) ( )5 3, 3 , 5 3, 3− −
39. 2 2
2 22 2
2 8 0x y
x y
⎧ + =⎪⎨
− + =⎪⎩
2 2
2 22 2
2 8x y
x y
⎧ + =⎪⎨
− = −⎪⎩
Multiply each side of the first equation by 2 and add the equations to eliminate y:
2 2
2 2
2
2
4 2 42 8
5 44 5
x yx y
x
x
+ =
− = −
= −
= −
No real solution. The system is inconsistent.
40. 2 2
2 24 0
2 3 6y x
x y
⎧ − + =⎪⎨
+ =⎪⎩
2 2
2 24
2 3 6x y
x y
⎧ − =⎪⎨
+ =⎪⎩
Multiply each side of the first equation by 2− and add the equations to eliminate x:
2 2
2 2
2
2
2 2 8 2 3 6
5 22 5
x yx y
y
y
− + = −
+ =
= −
= −
No real solution. The system is inconsistent.
41. 2 2
2 22 16
4 24x y
x y
⎧ + =⎪⎨
− =⎪⎩
Multiply each side of the second equation by 2 and add the equations to eliminate y:
2 2
2 2
2
2
2 168 2 48
9 6464 98 3
x yx y
x
x
x
+ =
− =
=
=
= ±
22 2
2
8If :3
8 802 16 23 9
40 2 109 3
x
y y
y y
=
⎛ ⎞ + = ⇒ =⎜ ⎟⎝ ⎠
⇒ = ⇒ = ±
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Section 8.6: Systems of Nonlinear Equations
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22 2
2
8If :3
8 802 16 23 9
40 2 109 3
x
y y
y y
= −
⎛ ⎞− + = ⇒ =⎜ ⎟⎝ ⎠
⇒ = ⇒ = ±
Solutions: 8 2 10 8 2 10 8 2 10, , , , , ,3 3 3 3 3 3
8 2 10,3 3
⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞
− −⎜ ⎟⎜ ⎟⎝ ⎠
42. 2 2
2 2
4 3 4
2 6 3
x y
x y
⎧ + =⎪⎨
− = −⎪⎩
Multiply each side of the first equation by 2 and add the equations to eliminate y:
2 2
2 2
2
2
8 6 8
2 6 3
10 51 2
2 2
x y
x y
x
x
x
+ =
− = −
=
=
= ±
22 2
2
22 2
2
2If :2
24 3 4 3 22
2 63 32If :
2
24 3 4 3 22
2 63 3
x
y y
y y
x
y y
y y
=
⎛ ⎞+ = ⇒ =⎜ ⎟⎜ ⎟
⎝ ⎠
⇒ = ⇒ = ±
= −
⎛ ⎞− + = ⇒ =⎜ ⎟⎜ ⎟
⎝ ⎠
⇒ = ⇒ = ±
Solutions: 2 6 2 6 2 6, , , , , ,
2 3 2 3 2 3
2 6,2 3
⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞
− −⎜ ⎟⎜ ⎟⎝ ⎠
43. 2 2
2 2
5 2 3 0
3 1 7
x y
x y
⎧ − + =⎪⎪⎨⎪ + =⎪⎩
2 2
2 2
5 2 3
3 1 7
x y
x y
⎧ − = −⎪⎪⎨⎪ + =⎪⎩
Multiply each side of the second equation by 2 and add the equations to eliminate y:
2 2
2 2
2
2
5 2 3
6 2 14
11 11
1 1
x y
x y
xxx
− = −
+ =
=
== ±
22 2 2
22 2 2
If 1:3 1 1 17 4
4(1)12
If 1:3 1 1 17 4
4( 1)12
x
yy y
y
x
yy y
y
=
+ = ⇒ = ⇒ =
⇒ = ±
= −
+ = ⇒ = ⇒ =−
⇒ = ±
Solutions: 1 1 1 11, , 1, , 1, , 1,2 2 2 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
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Chapter 8: Systems of Equations and Inequalities
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44. 2 2
2 2
2 3 1 0
6 7 2 0
x y
x y
⎧ − + =⎪⎪⎨⎪ − + =⎪⎩
2 2
2 2
2 3 1
6 7 2
x y
x y
⎧ − = −⎪⎪⎨⎪ − = −⎪⎩
Multiply each side of the first equation by –3 and add the equations to eliminate x:
2 2
2 2
2
2
6 9 3
6 7 2
2 1
2
2
y y
x y
y
y
y
−+ =
− = −
=
=
= ±
( )
( )
2 2 2
2
2 2 2
2
2 3 2 1If 2 : 122
4 22 3 2 1If 2 : 1
22
4 2
yx x
x x
yx x
x x
= − = − ⇒ =
⇒ = ⇒ = ±
= − − = − ⇒ =−
⇒ = ⇒ = ±
Solutions:
( ) ( ) ( ) ( )2, 2 , 2, 2 , 2, 2 , 2, 2− − − −
45. 4 4
4 4
1 6 6
2 2 19
x y
x y
⎧ + =⎪⎪⎨⎪ − =⎪⎩
Multiply each side of the first equation by –2 and add the equations to eliminate x:
4 4
4 4
4
4
2 12 12
2 2 19
14 7
2
x y
x y
yy
−− = −
− =
− =
= −
There are no real solutions. The system is inconsistent.
46. Add the equations to eliminate y:
4 4
4 4
4
4
4
1 1 1
1 1 4
2 5
2 5
2 5
x y
x y
x
x
x
− =
+ =
=
=
= ±
44 4 4
4
4 4
44 4 4
4
4 4
2 1 1 1 3If : 45 22
5
2 23 3
2 1 1 1 3If : 45 22
5
2 23 3
xy y
y y
xy y
y y
= + = ⇒ =⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
⇒ = ⇒ = ±
= − + = ⇒ =⎛ ⎞
−⎜ ⎟⎜ ⎟⎝ ⎠
⇒ = ⇒ = ±
Solutions:
4 4 4 4 4 4
4 4
2 2 2 2 2 2, , , , , ,5 3 5 3 5 3
2 2,5 3
⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞
− −⎜ ⎟⎜ ⎟⎝ ⎠
47. 2 2
2
3 2 0
6
x xy y
x xy
⎧ − + =⎪⎨
+ =⎪⎩
Subtract the second equation from the first to eliminate the 2x term.
2
2
4 2 6
2 3
xy y
xy y
− + = −
− =
Since 0y ≠ , we can solve for x in this equation to get
2 32
yxy+
= , 0y ≠
Now substitute for x in the second equation and solve for y.
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Section 8.6: Systems of Nonlinear Equations
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2
22 2
4 2 2
2
6
3 3 62 2
6 9 3 624
x xy
y y yy y
y y yy
+ =
⎛ ⎞ ⎛ ⎞+ ++ =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+ + ++ =
( )( )
4 2 4 2 2
4 2
4 2
2 2
6 9 2 6 24
3 12 9 0
4 3 0
3 1 0
y y y y y
y y
y y
y y
+ + + + =
− + =
− + =
− − =
Thus, 3y = ± or 1y = ± . If 1: 2 1 2If 1: 2( 1) 2
y xy x
= = ⋅ == − = − = −
If 3 : 3
If 3 : 3
y x
y x
= =
= − = −
Solutions: ( ) ( )(2, 1), (–2, 1), 3, 3 , 3, 3− − −
48. 2 22 0
6 0x xy y
xy x⎧ − − =⎪⎨
+ + =⎪⎩
Factor the first equation, solve for x, substitute into the second equation and solve:
2 22 0( 2 )( ) 0
2 or
x xy yx y x y
x y x y
− − =− + =
= = −
Substitute 2x y= and solve:
2
2
6 0(2 ) 2 6
2 2 6 0
2( 3) 0
xy xy y y
y y
y y
+ + =+ = −
+ + =
+ + =
21 1 4(1)(3) (No real solution)
2(1)y
− ± −=
Substitute x y= − and solve:
2
6 0( ) 6
6 0( 3)( 2) 0
3 or =2If 3 : 3If 2 : 2
xy xy y y
y yy y
y yy xy x
+ + =− ⋅ + − = −
− − + =− − − =
= −= − == = −
Solutions: (3, –3), (– 2, 2)
49. 2 2 2 021 0
y y x xxy
y
⎧ + + − − =⎪
−⎨+ + =⎪
⎩
Multiply each side of the second equation by –y and add the equations to eliminate y:
( )
2 2
2
2
2 0
2 0
2 0 2 0
y y x x
y y x
x xx x
+ + − − =
− − − + =
− =
− =
0 or 2x x= =
2 2 2
2 2 2
If 0 :
0 0 2 0 2 0( 2)( 1) 0 2 or 1
If 2 :
2 2 2 0 0( 1) 0 0 or 1
Note: 0 because of division by zero.
x
y y y yy y y y
x
y y y yy y y y
y
=
+ + − − = ⇒ + − =⇒ + − = ⇒ = − =
=
+ + − − = ⇒ + =⇒ + = ⇒ = = −
≠
Solutions: (0, –2), (0, 1), (2, –1)
50. 3 2 2
2
2
2 3 4 0
2 0
x x y y
y yxx
⎧ − + + − =⎪⎨ −
− + =⎪⎩
Multiply each side of the second equation by 2x− and add the equations to eliminate x: 3 2 2
3 2 2
2 3 4 0
2 0
4 4 0 4 4 1
x x y y
x x y y
yyy
− + + − =
− + − + =
− ===
3 2 2 3 2
2
If 1:
2 1 3 1 4 0 2 0
( 2) 0 0 or 2Note: 0 because of division by zero.
y
x x x x
x x x xx
=
− + + ⋅ − = ⇒ − =
⇒ − = ⇒ = =≠
Solution: (2, 1)
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Chapter 8: Systems of Equations and Inequalities
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51. Rewrite each equation in exponential form: 3
5
log 3
log (4 ) 5 4x
x
y y x
y y x
⎧ = → =⎪⎨
= → =⎪⎩
Substitute the first equation into the second and solve:
3 5
5 3
3 2
4
4 0
( 4) 0
x x
x x
x x
=
− =
− =
3 20 or 4 0 or 2x x x x= = ⇒ = = ± The base of a logarithm must be positive, thus
0 and 2x x≠ ≠ − .
3If 2 : 2 8x y= = = Solution: (2, 8)
52. Rewrite each equation in exponential form: 3
2
log (2 ) 3 2
log (4 ) 2 4x
x
y y x
y y x
⎧ = ⇒ =⎪⎨
= ⇒ =⎪⎩
Multiply the first equation by 2 then substitute the first equation into the second and solve:
3 2
3 2
2
2
2 0
(2 1) 0
x x
x x
x x
=
− =
− =
2 1 10 or or 02 2
x x x x= = ⇒ = =
The base of a logarithm must be positive, thus 0x ≠ .
21 1 1 1If : 42 2 4 16
x y y⎛ ⎞= = = ⇒ =⎜ ⎟⎝ ⎠
Solution: 1 1,2 16
⎛ ⎞⎜ ⎟⎝ ⎠
53. Rewrite each equation in exponential form: 44ln ln 4ln 4 ln y yx y x e e y= ⇒ = = =
23 3 3
3 3
2 2log 2log log2 2 2
log 2 2log
3 3 3 3 3 9y y y
x y
x y+
= +
= = ⋅ = ⋅ =
So we have the system 4
29
x y
x y
⎧ =⎪⎨
=⎪⎩
Therefore we have : 2 4 2 4 2 29 9 0 (9 ) 0y y y y y y= ⇒ − = ⇒ − =
2 (3 )(3 ) 00 or 3 or 3
y y yy y y
+ − == = − =
Since ln y is undefined when 0y ≤ , the only
solution is 3y = . 4 4If 3 : 3 81y x y x= = ⇒ = =
Solution: ( )81, 3
54. Rewrite each equation in exponential form: 55ln ln 5ln 5ln y yx y x e e y= ⇒ = = =
22 2 2
2 2
3 2log 2log log3 3 2
log 3 2log
2 2 2 2 2 8y y y
x y
x y+
= +
= = ⋅ = ⋅ =
So we have the system 5
28
x y
x y
⎧ =⎪⎨
=⎪⎩
Therefore we have
( )
2 5
2 5
2 3
8
8 0
8 0
y y
y y
y y
=
− =
− =
30 or 8 0 2y y y= − = ⇒ = Since ln y is undefined when 0y ≤ , the only solution is 2y = .
5 5If 2 : 2 32y x y x= = ⇒ = = Solution: ( )32, 2
55.
( ) ( )( ) ( )
2 2
2
22 31 12 2 2
2 21 1 12 2 2
3 2 0
1 0
, 0
x x y y
y yxx
x y
x y x
⎧ + + − + =⎪⎨ −
+ + =⎪⎩⎧ + + − =⎪⎨⎪ + + − = ≠⎩
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Section 8.6: Systems of Nonlinear Equations
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56.
( ) ( )( )
2 2
2 2 51 12 2 2
2 912 4
2 021 0
, 0
y y x xxy
y
x y
x y y
⎧ + + − − =⎪
−⎨+ + =⎪
⎩⎧ − + + =⎪⎨⎪ = − + + ≠⎩
57. Graph: 1 2(2 / 3); ( )y x y e x= ∧ = ∧ − Use INTERSECT to solve:
4.7–4.7
3.1
-3.1 Solution: 0.48, 0.62x y= = or (0.48, 0.62)
58. Graph: 1 2(3 / 2); ( )y x y e x= ∧ = ∧ − Use INTERSECT to solve:
4.7–4.7
3.1
–3.1 Solution: 0.65, 0.52x y= = or (0.65, 0.52)
59. Graph: 3 2 31 22 ; 4 /y x y x= − =
Use INTERSECT to solve:
4.7–4.7
3.1
–3.1 Solution: 1.65, 0.89x y= − = − or (–1.65, –0.89)
60. Graph: 3 31 2
23
2 ; 2 ;
4 /
y x y x
y x
= − = − −
=
Use INTERSECT to solve:
4.7–4.7
3.1
–3.1 Solution: 1.37, 2.14x y= − = or (–1.37, 2.14)
61. Graph: 4 44 41 2
3 4
12 ; 12 ;
2 / ; 2 /
y x y x
y x y x
= − = − −
= = −
Use INTERSECT to solve:
Solutions: 0.58, 1.86;x y= = 1.81, 1.05;x y= =
1.81, 1.05;x y= = − ; 0.58, 1.86x y= = − or (0.58, 1.86), (1.81, 1.05), (1.81, –1.05), (0.58, –1.86)
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Chapter 8: Systems of Equations and Inequalities
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62. Graph: 4 44 41 2
3
6 ; 6 ;1/
y x y xy x= − = − −
=
Use INTERSECT to solve:
Solutions: 0.64, 1.55;x y= = 1.55, 0.64;x y= =
0.64, 1.55;x y= − = − ; 1.55, 0.64x y= − = − or (0.64, 1.55), (1.55, 0.64), (–0.64, –1.55), (–1.55, –0.64)
63. Graph: 1 22 / ; lny x y x= = Use INTERSECT to solve:
4.7–4.7
3.1
–3.1 Solution: 2.35, 0.85x y= = or (2.35, 0.85)
64. Graph: 2 21 2
3
4 ; 4 ;ln
y x y xy x= − = − −
=
Use INTERSECT to solve:
Solution: 1.90, 0.64;x y= = 0.14, 2.00x y= = − or (1.90, 0.64), (0.14, –2.00)
65. Solve the first equation for x, substitute into the second equation and solve:
2 2
2 0 2
( 1) ( 1) 5
x y x y
x y
+ = ⇒ = −⎧⎪⎨
− + − =⎪⎩
2 2
2 2 2
( 2 1) ( 1) 5
4 4 1 2 1 5 5 2 3 0(5 3)( 1) 0
3 =0.6 or 15
6 = 1.2 or 25
y y
y y y y y yy y
y y
x x
− − + − =
+ + + − + = ⇒ + − =− + =
= = −
= − − =
The points of intersection are 6 3, , (2, –1)5 5
⎛ ⎞−⎜ ⎟⎝ ⎠
.
66. Solve the first equation for x, substitute into the
second equation and solve:
2 2
2 6 0 2 6
( 1) ( 1) 5
x y x y
x y
+ + = ⇒ = − −⎧⎪⎨
+ + + =⎪⎩
2 2
2 2
2
( 2 6 1) ( 1) 5
4 20 25 2 1 5
5 22 21 0(5 7)( 3) 0
y y
y y y y
y yy y
− − + + + =
+ + + + + =
+ + =+ + =
7 or 3516 or 05
y y
x x
= − = −
= − =
The points of intersection are 16 7, , (0, 3)5 5
⎛ ⎞− − −⎜ ⎟⎝ ⎠
.
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Section 8.6: Systems of Nonlinear Equations
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67. Complete the square on the second equation.
2
2
4 4 1 4
( 2) 3
y y x
y x
+ + = − +
+ = +
Substitute this result into the first equation. 2
2
2
( 1) 3 4
2 1 3 4
0( 1) 0
0 or 1
x x
x x x
x xx x
x x
− + + =
− + + + =
− =− =
= =2
2
If 0 : ( 2) 0 3
2 3 2 3
If 1: ( 2) 1 32 2 2 2
x y
y y
x yy y
= + = +
+ = ± ⇒ = − ±
= + = ++ = ± ⇒ = − ±
The points of intersection are:
( ) ( ) ( ) ( )0, 2 3 , 0, 2 3 , 1, – 4 , 1, 0− − − + .
68. Complete the square on the second equation, substitute into the first equation and solve:
2 2
2
( 2) ( 1) 4
2 5 0
x y
y y x
⎧ + + − =⎪⎨
− − − =⎪⎩
2
2
2 1 5 1
( 1) 6
y y x
y x
− + = + +
− = +
2
2
2
( 2) 6 4
4 4 6 4
5 6 0( 2)( 3) 0
2 or 3
x x
x x x
x xx x
x x
+ + + =
+ + + + =
+ + =+ + =
= − = −
2
2
If 2 : ( 1) 2 6 1 21 or 3
If 3 : ( 1) 3 6 1 3
1 3
x y yy y
x y y
y
= − − = − + ⇒ − = ±⇒ = − =
= − − = − + ⇒ − = ±
⇒ = ±
The points of intersection are:
( ) ( )3, 1 3 , 3, 1 3 , ( 2, 1), ( 2, 3)− − − + − − − .
69. Solve the first equation for x, substitute into the second equation and solve:
2 2
43
6 1 0
yx
x x y
⎧ =⎪ −⎨⎪ − + + =⎩
43
43
4 3
yx
xy
xy
=−
− =
= +
22
22
22
4 43 6 3 1 0
16 24 249 18 1 0
16 8 0
yy y
yy yy
yy
⎛ ⎞ ⎛ ⎞+ − + + + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+ + − − + + =
+ − =
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Chapter 8: Systems of Equations and Inequalities
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4 2
4 2
2 2
2
2
16 8 0
8 16 0
( 4) 0
4 0
42
y y
y y
y
y
yy
+ − =
− + =
− =
− =
== ±
4If 2 : 3 524If 2 : 3 12
y x
y x
= = + =
= − = + =−
The points of intersection are: (1, –2), (5, 2).
70. Substitute the first equation into the second equation and solve:
2 2
42
4 4 0
yx
x x y
⎧ =⎪ +⎨⎪ + + − =⎩
( )( )( )
( )
22
22
2 2
2 2
4 3 2
4 3 2
2 2
2 2
44 4 02
44 42
( 2) 4 4 16
4 4 4 4 16
8 16 16 16
8 16 0
8 16 0
( 4) 00 or 42 2
x xx
x xx
x x x
x x x x
x x x
x x x
x x x
x xx xy y
⎛ ⎞+ + − =⎜ ⎟+⎝ ⎠
⎛ ⎞+ − = −⎜ ⎟+⎝ ⎠
+ + − = −
+ + + − = −
+ + − = −
+ + =
+ + =
+ == = −= = −
The points of intersection are: (0, 2), (–4, –2).
71. Let and x y be the two numbers. The system of equations is:
2 2
2 2
10
x y x y
x y
− = ⇒ = +⎧⎪⎨
+ =⎪⎩
Solve the first equation for x, substitute into the second equation and solve:
( )
( )( )
2 2
2 2
2
2 10
4 4 10
2 3 03 1 0 3 or 1
y y
y y y
y yy y y y
+ + =
+ + + =
+ − =
+ − = ⇒ = − =
If 3 : 3 2 1If 1: 1 2 3
y xy x
= − = − + = −= = + =
The two numbers are 1 and 3 or –1 and –3.
72. Let and x y be the two numbers. The system of equations is:
2 2
7 7
21
x y x y
x y
+ = ⇒ = −⎧⎪⎨
− =⎪⎩
Solve the first equation for x, substitute into the second equation and solve:
( )2 2
2 2
7 21
49 14 2114 28
2 7 2 5
y y
y y yyy x
− − =
− + − =− = −
= ⇒ = − =
The two numbers are 2 and 5.
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Section 8.6: Systems of Nonlinear Equations
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73. Let and x y be the two numbers. The system of equations is:
2 2
44
8
xy xy
x y
⎧ = ⇒ =⎪⎨⎪ + =⎩
Solve the first equation for x, substitute into the second equation and solve:
( )
22
22
4 2
4 2
22
2
4 8
16 8
16 8
8 16 0
4 0
42
yy
yy
y y
y y
y
yy
⎛ ⎞+ =⎜ ⎟
⎝ ⎠
+ =
+ =
− + =
− =
== ± 4 4If 2 : 2; If 2 : 22 2
y x y x= = = = − = = −−
The two numbers are 2 and 2 or –2 and –2.
74. Let and x y be the two numbers. The system of equations is:
2 2
1010
21
xy xy
x y
⎧ = ⇒ =⎪⎨⎪ − =⎩
Solve the first equation for x, substitute into the second equation and solve:
( )( )
22
22
4 2
4 2
2 2
10 21
100 21
100 21
21 100 0
4 25 0
yy
yy
y y
y y
y y
⎛ ⎞− =⎜ ⎟
⎝ ⎠
− =
− =
+ − =
− + =
2 42
yy
== ±
or 2 25 (no real solution)y = −
10210
2
If 2 : 5
If 2 : 5
y x
y x −
= = =
= − = = −
The two numbers are 2 and 5 or –2 and –5.
75. Let and x y be the two numbers. The system of equations is:
1 1 5
x y xy
x y
− =⎧⎪⎨ + =⎪⎩
Solve the first equation for x, substitute into the second equation and solve:
( )1 1
x xy yyx y y x
y
− =
− = ⇒ =−
1
1 13 3
1 23 3
1 1 5
1 1 5
2 5
2 56 2
1 1 3 21
yy
y
yy y
yy
y yy
y x
−
+ =
−+ =
−=
− ==
= ⇒ = = =−
The two numbers are 12 and 1
3 .
76. Let and x y be the two numbers. The system of equations is:
1 1 3
x y xy
x y
+ =⎧⎪⎨ − =⎪⎩
Solve the first equation for x, substitute into the second equation and solve:
( )1 1
xy x yyx y y x
y
− =
− = ⇒ =−
1
1 1 3
1 1 3
2 3
2 32 2
1 11 1 1 2
yy
y
yy y
yy
y yy
y x
−
− =
−− =
−=
− == −
−= − ⇒ = =
− −
The two numbers are 1− and 12 .
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Chapter 8: Systems of Equations and Inequalities
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77. 2310 10
ab
a b a b
⎧ =⎪⎨⎪ + = ⇒ = −⎩
Solve the second equation for a , substitute into the first equation and solve:
10 23
3(10 ) 230 3 2
30 56 4
bb
b bb b
bb a
−=
− =− =
== ⇒ =
10; 2a b b a+ = − =
The ratio of a b+ to b a− is 102 5= .
78. 4314 14
ab
a b a b
⎧ =⎪⎨⎪ + = ⇒ = −⎩
Solve the second equation for a , substitute into the first equation and solve:
( )
14 43
3 14 442 3 4
42 76 8
bb
b bb b
bb a
−=
− =
− === ⇒ =
2; 14a b a b− = + =
The ratio of a b− to a b+ is 2 1714 = .
79. Let x = the width of the rectangle. Let y = the length of the rectangle.
2 2 1615
x yxy
+ =⎧⎨ =⎩
Solve the first equation for y, substitute into the second equation and solve. 2 2 16
2 16 28
x yy xy x
+ == −= −
( )
( )( )
2
2
8 15
8 15
8 15 05 3 0
x x
x x
x xx x
− =
− =
− + =
− − =
5x = or 3x = The dimensions of the rectangle are 3 inches by 5 inches.
80. Let 2x = the side of the first square. Let 3x = the side of the second square.
( ) ( )2 2
2 2
2
2
2 3 52
4 9 52
13 52
42
x x
x x
x
xx
+ =
+ =
=
== ±
Note that we must have 0x > . The sides of the first square are (2)(2) = 4 feet and the sides of the second square are (3)(2) = 6 feet.
81. Let x = the radius of the first circle. Let y = the radius of the second circle.
2 2
2 2 12
20
x y
x y
π + π = π⎧⎪⎨
π + π = π⎪⎩
Solve the first equation for y, substitute into the second equation and solve: 2 2 12
66
x yx y
y x
π + π = π+ =
= −
2 2
2 2
2 2
2 2 2
2
20
20
(6 ) 20
36 12 20 2 12 16 0
6 8 0 ( 4)( 2) 04 or 22 4
x y
x y
x x
x x x x x
x x x xx xy y
π + π = π
+ =
+ − =
+ − + = ⇒ − + =
− + = ⇒ − − == == =
The radii of the circles are 2 centimeters and 4 centimeters.
82. Let x = the length of each of the two equal sides in the isosceles triangle. Let y = the length of the base. The perimeter of the triangle: 18x x y+ + = Since the altitude to the base y is 3, the Pythagorean theorem produces another equation.
2 22 2 23 9
2 4y yx x⎛ ⎞ + = ⇒ + =⎜ ⎟
⎝ ⎠
Solve the system of equations:
22
2 18 18 2
94
x y y x
y x
+ = ⇒ = −⎧⎪⎨
+ =⎪⎩
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Section 8.6: Systems of Nonlinear Equations
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Solve the first equation for y, substitute into the second equation and solve.
( )
( )
22
22
2 2
18 29
4324 72 4 9
481 18 9
18 905 18 2 5 8
xx
x x x
x x xxx y
−+ =
− ++ =
− + + =− = −
= ⇒ = − =
The base of the triangle is 8 centimeters.
83. The tortoise takes 9 + 3 = 12 minutes or 0.2 hour longer to complete the race than the hare. Let r = the rate of the hare. Let t = the time for the hare to complete the race. Then t + 0.2 = the time for the tortoise and
0.5r − = the rate for the tortoise. Since the length of the race is 21 meters, the distance equations are:
( )( )
2121
0.5 0.2 21
r t rt
r t
⎧ = ⇒ =⎪⎨⎪ − + =⎩
Solve the first equation for r, substitute into the second equation and solve:
( )
( )
( )( )
2
2
21 0.5 0.2 21
4.221 0.5 0.1 21
4.210 21 0.5 0.1 10 21
210 42 5 210
5 42 05 14 3 0
tt
tt
t t tt
t t t t
t tt t
⎛ ⎞− + =⎜ ⎟⎝ ⎠
+ − − =
⎛ ⎞+ − − = ⋅⎜ ⎟⎝ ⎠
+ − − =
+ − =
− + =
5 14 05 14
14 2.85
tt
t
− ==
= =
or 3 03
tt
+ == −
3t = − makes no sense, since time cannot be negative. Solve for r:
21 7.52.8
r = =
The average speed of the hare is 7.5 meters per hour, and the average speed for the tortoise is 7 meters per hour.
84. Let 1 2 3, ,v v v = the speeds of runners 1, 2, 3. Let 1 2 3, ,t t t = the times of runners 1, 2, 3. Then by the conditions of the problem, we have the following system:
1 1
2 1
3 1
2 2
5280527052605280
v tv tv tv t
=⎧⎪ =⎪⎨ =⎪⎪ =⎩
Distance between the second runner and the third runner after 2t seconds is:
2 23 2 3 1
2 15280 5280
52805280 52605270
10.02
v tv t v t
v t⎛ ⎞
− = − ⎜ ⎟⎝ ⎠⎛ ⎞= − ⎜ ⎟⎝ ⎠
≈
The second place runner beats the third place runner by about 10.02 feet.
85. Let x = the width of the cardboard. Let y = the length of the cardboard. The width of the box will be 4x − , the length of the box will be
4y − , and the height is 2. The volume is ( 4)( 4)(2)V x y= − − .
Solve the system of equations: 216216
2( 4)( 4) 224
xy yx
x y
⎧ = ⇒ =⎪⎨⎪ − − =⎩
Solve the first equation for y, substitute into the second equation and solve.
( )
( )( )
2
2
2
2162 8 4 224
1728432 8 32 224
432 8 1728 32 224
8 240 1728 0
30 216 012 18 0
xx
xx
x x x x
x x
x xx x
⎛ ⎞− − =⎜ ⎟⎝ ⎠
− − + =
− − + =
− + =
− + =
− − =
12 012
xx
− ==
or 18 018
xx
− ==
The cardboard should be 12 centimeters by 18 centimeters.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
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86. Let x = the width of the cardboard. Let y = the length of the cardboard. The area of the cardboard is: 216xy =
The volume of the tube is: 2 224V r h= π =
where and 2 or 2xh y r x r= π = =π
.
Solve the system of equations:
2 2
216216
224 2242 4
xy yx
x x yy
⎧ = ⇒ =⎪⎪⎨
⎛ ⎞⎪π = ⇒ =⎜ ⎟⎪ π π⎝ ⎠⎩
Solve the first equation for y, substitute into the second equation and solve.
( )2 216224
4216 896
896 13.03216
xx
x
x
πππ
=
=
= ≈
( )2
896216
216216 216 16.57896
yx π π
= = = ≈
The cardboard should be about 13.03 centimeters by 16.57 centimeters.
87. Find equations relating area and perimeter: 2 2 4500
3 3 ( ) 300x yx y x y
⎧ + =⎪⎨
+ + − =⎪⎩
Solve the second equation for y, substitute into the first equation and solve: 4 2 300
2 300 4150 2
x yy xy x
+ == −= −
2 2
2 2
2
2
2
(150 2 ) 4500
22,500 600 4 4500
5 600 18,000 0
120 3600 0
( 60) 060 0
60150 2(60) 30
x x
x x x
x x
x x
xx
xy
+ − =
+ − + =
− + =
− + =
− =− =
== − =
The sides of the squares are 30 feet and 60 feet.
88. Let x = the length of a side of the square. Let r = the radius of the circle. The area of the square is 2x and the area of the circle is 2rπ . The perimeter of the square is 4x and the circumference of the circle is 2 rπ . Find equations relating area and perimeter:
2 2 1004 2 60x r
x r⎧ + π =⎪⎨
+ π =⎪⎩
Solve the second equation for x, substitute into the first equation and solve: 4 2 60
4 60 21152
x rx r
x r
+ π == − π
= − π
22
2 2 2
2 2
115 10021225 15 1004
1 15 125 04
r r
r r r
r r
⎛ ⎞− π + π =⎜ ⎟⎝ ⎠
− π + π + π =
⎛ ⎞π + π − π + =⎜ ⎟⎝ ⎠
2 2 2
2 2
2
14 ( 15 ) 4 (125)41225 5004
100 500 0
b ac ⎛ ⎞− = − π − π + π⎜ ⎟⎝ ⎠⎛ ⎞= π − π + π⎜ ⎟⎝ ⎠
= π − π <
Since the discriminant is less than zero, it is impossible to cut the wire into two pieces whose total area equals 100 square feet.
89. Solve the system for and l w : 2 2l w P
l w A+ =⎧
⎨ =⎩
Solve the first equation for l , substitute into the second equation and solve.
2
2
2 2
2
2
2
02
l P wPl w
P w w A
P w w A
Pw w A
= −
= −
⎛ ⎞− =⎜ ⎟⎝ ⎠
− =
− + =
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.6: Systems of Nonlinear Equations
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22
2
2
2 42 4
2
164 4
2 2
16162
2 4
P PP P
P P
AA
w
AP P A
± −± −= =
−± ± −
= =
2
2 2
2
2 2
16If then4
16 162 4 4
16If then4
16 162 4 4
P P Aw
P P P A P P Al
P P Aw
P P P A P P Al
+ −=
+ − − −= − =
− −=
− − + −= − =
If it is required that length be greater than width, then the solution is:
2 216 16and4 4
P P A P P Aw l− − + −= =
90. Solve the system for and l b :
22 2
2 2
4
P b l b P l
bh l
= + ⇒ = −⎧⎪⎨
+ =⎪⎩
Solve the first equation for b , substitute into the second equation and solve.
( )
2 2 2
22 2
2 2 2 2
2 2
2 2
4 4
4 2 4
4 4 4 4
4 4
44
h b l
h P l l
h P Pl l l
h P Pl
h PlP
+ =
+ − =
+ − + =
+ =
+=
2 2 2 24 42 2
h P P hb PP P+ −
= − =
91. Solve the equation: 2 4(2 4) 0m m− − =
( )
2
2
8 16 0
4 04
m m
mm
− + =
− =
=
Use the point-slope equation with slope 4 and the point (2, 4) to obtain the equation of the tangent line: 4 4( 2) 4 4 8 4 4y x y x y x− = − ⇒ − = − ⇒ = −
92. Solve the system: 2 2 10x y
y mx b⎧ + =⎪⎨
= +⎪⎩
Solve the system by substitution: ( )
( )
22
2 2 2 2
2 2 2
10
2 10 0
1 2 10 0
x mx b
x m x bmx b
m x bmx b
+ + =
+ + + − =
+ + + − =
Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 (1) 3m b b m= + ⇒ = − There is one solution to the quadratic if the discriminant is zero.
( ) ( ) ( )2 2 2
2 2 2 2 2 2
2 2
2 4 1 10 0
4 4 40 4 40 0
40 4 40 0
bm m b
b m b m m b
m b
− + − =
− + − + =
− + =
Substitute for b and solve: ( )
( )
22
2 2
2
2
2
40 4 3 40 0
40 4 24 36 40 0
36 24 4 0
9 6 1 0
3 1 03 1
13
m m
m m m
m m
m m
mm
m
− − + =
− + − + =
+ + =
+ + =
+ =
= −
= −
1 103 33 3
b m ⎛ ⎞= − = − − =⎜ ⎟⎝ ⎠
The equation of the tangent line is 1 103 3
y x= − + .
93. Solve the system: 2 2y x
y mx b⎧ = +⎪⎨
= +⎪⎩
Solve the system by substitution: 2 22 2 0x mx b x mx b+ = + ⇒ − + − =
Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 (1) 3m b b m= + ⇒ = − Substitute into the quadratic to eliminate b:
2 22 (3 ) 0 ( 1) 0x mx m x mx m− + − − = ⇒ − + − = Find when the discriminant equals 0:
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
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( ) ( )( )
( )
2
2
2
4 1 1 0
4 4 0
2 02 0
2
m m
m m
mm
m
− − − =
− + =
− =
− ==
3 3 2 1b m= − = − = The equation of the tangent line is 2 1y x= + .
94. Solve the system: 2 5x y
y mx b⎧ + =⎪⎨
= +⎪⎩
Solve the system by substitution: 2 25 5 0x mx b x mx b+ + = ⇒ + + − =
Note that the tangent line passes through (–2, 1). Find the relation between m and b:
( )1 2 2 1m b b m= − + ⇒ = + Substitute into the quadratic to eliminate b:
( )
2
2
2 1 5 0
2 4 0
x mx m
x mx m
+ + + − =
+ + − =
Find when the discriminant equals 0: ( ) ( ) ( )
( )
2
2
2
4 1 2 4 0
8 16 0
4 04 0
4
m m
m m
mm
m
− − =
− + =
− =
− ==
( )2 1 2 4 1 9b m= + = + = The equation of the tangent line is 4 9y x= + .
95. Solve the system: 2 22 3 14x y
y mx b⎧ + =⎪⎨
= +⎪⎩
Solve the system by substitution: ( )
( )
22
2 2 2 2
2 2 2
2 3 14
2 3 6 3 14
3 2 6 3 14 0
x mx b
x m x mbx b
m x mbx b
+ + =
+ + + =
+ + + − =
Note that the tangent line passes through (1, 2). Find the relation between m and b: 2 (1) 2m b b m= + ⇒ = − Substitute into the quadratic to eliminate b:
2 2 2
2 2 2 2
(3 2) 6 (2 ) 3(2 ) 14 0
(3 2) (12 6 ) (3 12 2) 0
m x m m x m
m x m m x m m
+ + − + − − =
+ + − + − − = Find when the discriminant equals 0:
2 2 2 2
2
2
2
(12 6 ) 4(3 2)(3 12 2) 0
144 96 16 0
9 6 1 0
(3 1) 03 1 0
13
m m m m m
m m
m m
mm
m
− − + − − =
+ + =
+ + =
+ =+ =
= −
1 72 23 3
b m ⎛ ⎞= − = − − =⎜ ⎟⎝ ⎠
The equation of the tangent line is 1 73 3
y x= − + .
96. Solve the system: 2 23 7x y
y mx b⎧ + =⎪⎨
= +⎪⎩
Solve the system by substitution: ( )
( )
22
2 2 2 2
2 2 2
3 7
3 2 7
3 2 7 0
x mx b
x m x mbx b
m x mbx b
+ + =
+ + + =
+ + + − =
Note that the tangent line passes through (–1, 2). Find the relation between m and b: 2 ( 1) 2m b b m= − + ⇒ = + There is one solution to the quadratic if the discriminant equals 0.
( ) ( )( )2 2 2
2 2 2 2 2 2
2 2
2 2
2 4 3 7 0
4 4 28 12 84 0
28 12 84 0
7 3 21 0
bm m b
b m b m m b
m b
m b
− + − =
− + − + =
− + =
− + =
Substitute for b and solve: ( )
( )
22
2 2
2
2
7 3 2 21 0
7 3 12 12 21 0
4 12 9 0
2 3 02 3
32
m m
m m m
m m
mm
m
− + + =
− − − + =
− + =
− =
=
=
3 72 22 2
b m= + = + =
The equation of the tangent line is 3 72 2
y x= + .
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Section 8.6: Systems of Nonlinear Equations
877 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
97. Solve the system: 2 2 3x y
y mx b⎧ − =⎪⎨
= +⎪⎩
Solve the system by substitution: ( )
( )
22
2 2 2 2
2 2 2
3
2 3
1 2 3 0
x mx b
x m x mbx b
m x mbx b
− + =
− − − =
− − − − =
Note that the tangent line passes through (2, 1). Find the relation between m and b: 1 (2) 1 2m b b m= + ⇒ = − Substitute into the quadratic to eliminate b:
2 2 2
2 2 2 2
2 2 2 2
(1 ) 2 (1 2 ) (1 2 ) 3 0
(1 ) ( 2 4 ) 1 4 4 3 0
(1 ) ( 2 4 ) ( 4 4 4) 0
m x m m x m
m x m m x m m
m x m m x m m
− − − − − − =
− + − + − + − − =
− + − + + − + − =Find when the discriminant equals 0:
( ) ( )( )
( )
22 2 2
2 3 4 4 3
2
2
2
2 4 4 1 4 4 4 0
4 16 16 16 16 16 16 0
4 16 16 0
4 4 0
2 02
m m m m m
m m m m m m
m m
m m
mm
− + − − − + − =
− + − + − + =
− + =
− + =
− =
= The equation of the tangent line is 2 3y x= − .
98. Solve the system: 2 22 14y x
y mx b⎧ − =⎪⎨
= +⎪⎩
Solve the system by substitution: ( )
( )
2 2
2 2 2 2
2 2 2
2 14
2 4 2 14
2 1 4 2 14 0
mx b x
m x mbx b x
m x mbx b
+ − =
+ + − =
− + + − =
Note that the tangent line passes through (2, 3). Find the relation between m and b: 3 (2) 3 2m b b m= + ⇒ = − There is one solution to the quadratic if the discriminant equals 0.
( ) ( ) ( )2 2 2
2 2 2 2 2 2
2 2
2 2
4 4 2 1 2 14 0
16 16 112 8 56 0
112 8 56 0
14 7 0
bm m b
b m b m m b
m b
m b
− − − =
− + + − =
+ − =
+ − =
Substitute for b and solve:
( )
( )
22
2 2
2
2
14 3 2 7 0
14 4 12 9 7 0
18 12 2 0
2 3 1 03 1
13
m m
m m m
m m
mm
m
+ − − =
+ − + − =
− + =
− =
=
=
1 73 2 3 23 3
b m ⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
The equation of the tangent line is 1 73 3
y x= + .
99. Solve for 1 2and r r :
1 2
1 2
br ra
cr ra
⎧ + = −⎪⎪⎨⎪ =⎪⎩
Substitute and solve:
1 2
2 2
22 2 0
br ra
b cr ra a
b cr ra a
= − −
⎛ ⎞− − =⎜ ⎟⎝ ⎠
− − − =
22 2 0ar br c+ + =
2
2
1 2
2
2
2
42
42
4 22 2
42
b b acra
br ra
b b ac ba a
b b ac ba a
b b aca
− ± −=
= − − =
⎛ ⎞− ± −⎜ ⎟= − −⎜ ⎟⎝ ⎠
−= −
− −=
∓
∓
The solutions are: 2 2
1 24 4 and
2 2b b ac b b acr r
a a− + − − − −
= = .
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
878 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
100. Consider the circle with equation ( ) ( )2 2 2x h y k r− + − = and the third degree
polynomial with equation 3 2y ax bx cx d= + + + . Substituting the second equation into the first equation yields
( ) ( )22 3 2 2x h ax bx cx d k r− + + + + − = .
In order to find the roots for this equation we can expand the terms on the left hand side of the equation. Notice that ( )2x h− yields a 2nd degree
polynomial, and ( )23 2ax bx cx d k+ + + − yields a
6th degree polynomial. Therefore, we need to find the roots of a 6th degree equation, and the Fundamental Theorem of Algebra states that there will be at most six real roots. Thus, the circle and the 3rd degree polynomial will intersect at most six times. Now consider the circle with equation ( ) ( )2 2 2x h y k r− + − = and the polynomial of degree n with equation
2 30 1 2 3 ... n
ny a a x a x a x a x= + + + + + . Substituting the first equation into the first equation yields
( ) ( )22 2 3 20 1 2 3 ... n
nx h a a x a x a x a x k r− + + + + + + − =
In order to find the roots for this equation we can expand the terms on the left hand side of the equation. Notice that ( )2x h− yields a 2nd degree polynomial,
and ( )22 30 1 2 3 ... n
na a x a x a x a x k+ + + + + −
yields a polynomial of degree 2n. Therefore, we need to find the roots of an equation of degree 2n, and the Fundamental Theorem of Algebra states that there will be at most 2n real roots. Thus, the circle and the nth degree polynomial will intersect at most 2n times.
101. Since the area of the square piece of sheet metal is 100 square feet, the sheet’s dimensions are 10 feet by 10 feet. Let x = the length of the cut.
10 – 2x 10 – 2x
x
10
10
xx
The dimensions of the box are: length 10 2 ;x= − width 10 2 ; heightx x= − = . Note that each of these expressions must be positive. So we must have 0 and 10 2 0 5,x x x> − > ⇒ < that is, 0 5x< < . So the volume of the box is given by
( ) ( ) ( )( )( )( )( ) ( )2
length width height
10 2 10 2
10 2
V
x x x
x x
= ⋅ ⋅
= − −
= −
a. In order to get a volume equal to 9 cubic feet, we solve ( ) ( )210 2 9.x x− =
( ) ( )( )
2
2
2 3
10 2 9
100 40 4 9
100 40 4 9
x x
x x x
x x x
− =
− + =
− + =
So we need to solve the equation 3 24 40 100 9 0x x x− + − = .
Graphing 3 21 4 40 100 9y x x x= − + − on a
calculator yields the graph
10−2
−40
80
10−2
−40
80
10−2
−40
80
The graph indicates that there are three real zeros on the interval [0, 6]. Using the ZERO feature of a graphing calculator, we find that the three roots shown occur at 0.093x ≈ ,
4.274x ≈ and 5.632x ≈ . But we’ve already noted that we must have 0< <5x , so the only practical values for the sides of the square base are 0.093x ≈ feet and 4.274x ≈ feet.
b. Answers will vary.
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Section 8.7: Systems of Inequalities
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Section 8.7
1. 3 4 84 4
1
x xxx
+ < −<<
{ } ( )1 or ,1x x < −∞
2. 3 2 6x y− = The graph is a line. x-intercept: ( )3 2 0 6
3 62
xxx
− =
==
y-intercept: ( )3 0 2 62 6
3
yyy
− =
− == −
3. 2 2 9x y+ = The graph is a circle. Center: (0, 0) ; Radius: 3
4. 2 4y x= + The graph is a parabola. x-intercepts: 2
2
0 4
4, no intercepts
x
x x
= +
= − −
y-intercept: 20 4 4y = + = The vertex has x-coordinate:
( )0 0
2 2 1bxa
= − = − = .
The y-coordinate of the vertex is 20 4 4y = + = .
5. True
6. 2y x= ; right; 2
7. satisfied
8. half-plane
9. False
10. True
11. 0y ≥ Graph the line 0y = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 2). Since 2 ≥ 0 is true, shade the side of the line containing (0, 2).
12. 0x ≥ Graph the line 0x = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (2, 0). Since 2 ≥ 0 is true, shade the side of the line containing (2, 0).
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Chapter 8: Systems of Equations and Inequalities
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13. 4x ≥ Graph the line 4x = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (5, 0). Since 5 ≥ 0 is true, shade the side of the line containing (5, 0).
14. 2y ≤ Graph the line 2y = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (5, 0). Since 0 ≤ 2 is true, shade the side of the line containing (5, 0).
15. 2 6x y+ ≥ Graph the line 2 6x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 6+ ≥ is false, shade the opposite side of the line from (0, 0).
16. 3 2 6x y+ ≤ Graph the line 3 2 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≤ is true, shade the side of the line containing (0, 0).
17. 2 2 1x y+ >
Graph the circle 2 2 1x y+ = . Use a dashed line since the inequality uses >. Choose a test point not on the circle, such as (0, 0). Since 2 20 0 1+ > is false, shade the opposite side of the circle from (0, 0).
18. 2 2 9x y+ ≤
Graph the circle 2 2 9x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 9+ ≤ is true, shade the same side of the circle as (0, 0).
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Section 8.7: Systems of Inequalities
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19. 2 1y x≤ −
Graph the parabola 2 1y x= − . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since
20 0 1≤ − is false, shade the opposite side of the parabola from (0, 0).
20. 2 2y x> +
Graph the parabola 2 2y x= + . Use a dashed line since the inequality uses >. Choose a test point not on the parabola, such as (0, 0). Since
20 0 2> + is false, shade the opposite side of the parabola from (0, 0).
21. 4xy ≥ Graph the hyperbola 4xy = . Use a solid line since the inequality uses ≥ . Choose a test point not on the hyperbola, such as (0, 0). Since 0 0 4⋅ ≥ is false, shade the opposite side of the hyperbola from (0, 0).
22. 1xy ≤ Graph the hyperbola 1xy = . Use a solid line since the inequality uses ≤ . Choose a test point not on the hyperbola, such as (0, 0). Since 0 0 1⋅ ≤ is true, shade the same side of the hyperbola as (0, 0).
23. 2
2 4x yx y
+ ≤⎧⎨ + ≥⎩
Graph the line 2x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 4x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
24. 3 6
2 2x yx y
− ≥⎧⎨ + ≤⎩
Graph the line 3 6x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 6− ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 2+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
882 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
25. 2 43 2 6
x yx y
− ≤⎧⎨ + ≥ −⎩
Graph the line 2 4x y− = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4− ≤ is true, shade the side of the line containing (0, 0). Graph the line 3 2 6x y+ = − . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≥ − is true, shade the side of the line containing (0, 0). The overlapping region is the solution.
26. 4 5 02 2
x yx y
− ≤⎧⎨ − ≥⎩
Graph the line 4 5 0x y− = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (2, 0). Since 4(2) 5(0) 0− ≤ is false, shade the opposite side of the line from (2, 0). Graph the line 2 2x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2− ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
27. 2 3 03 2 6
x yx y
− ≤⎧⎨ + ≤⎩
Graph the line 2 3 0x y− = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 3). Since 2(0) 3(3) 0− ≤ is true, shade the side of the line containing (0, 3). Graph the line 3 2 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution.
28. 4 2
2 2x yx y
− ≥⎧⎨ + ≥⎩
Graph the line 4 2x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 4(0) 0 2− ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
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Section 8.7: Systems of Inequalities
883 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
29. 2 6
2 4 0x yx y
− ≤⎧⎨ − ≥⎩
Graph the line 2 6x y− = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6− ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 4 0x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 2). Since 2(0) 4(2) 0− ≥ is false, shade the opposite side of the line from (0, 2). The overlapping region is the solution.
30. 4 84 4
x yx y
+ ≤⎧⎨ + ≥⎩
Graph the line 4 8x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 4 4x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 4+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
31. 2 22 2
x yx y
+ ≥ −⎧⎨ + ≥⎩
Graph the line 2 2x y+ = − . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ − is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
32. 4 44 0
x yx y
− ≤⎧⎨ − ≥⎩
Graph the line 4 4x y− = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 4− ≤ is true, shade the side of the line containing (0, 0). Graph the line 4 0x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (1, 0). Since 1 4(0) 0− ≥ is true, shade the side of the line containing (1, 0). The overlapping region is the solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
884 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
33. 2 3 62 3 0
x yx y
+ ≥⎧⎨ + ≤⎩
Graph the line 2 3 6x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 0x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 2). Since 2(0) 3(2) 0+ ≤ is false, shade the opposite side of the line from (0, 2). Since the regions do not overlap, the solution is an empty set.
34. 2 02 2
x yx y
+ ≥⎧⎨ + ≥⎩
Graph the line 2 0x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (1, 0). Since 2(1) 0 0+ ≥ is true, shade the side of the line containing (1, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
35. 2 2 9
3x y
x y⎧ + ≤⎪⎨
+ ≥⎪⎩
Graph the circle 2 2 9x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 9+ ≤ is true, shade the same side of the circle as (0, 0). Graph the line 3x y+ = . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 0 3+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
36. 2 2 9
3x y
x y⎧ + ≥⎪⎨
+ ≤⎪⎩
Graph the circle 2 2 9x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the circle, such as (0, 0). Since
2 20 0 9+ ≥ is false, shade the opposite side of the circle as (0, 0). Graph the line 3x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 3+ ≤ is true, shade the same side of the line as (0, 0). The overlapping region is the solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.7: Systems of Inequalities
885 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
37. 2 4
2y xy x
⎧ ≥ −⎪⎨
≤ −⎪⎩
Graph the parabola 2 4y x= − . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since
20 0 4≥ − is true, shade the same side of the parabola as (0, 0). Graph the line 2y x= − . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 2≤ − is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
38. 2y xy x
⎧ ≤⎪⎨
≥⎪⎩
Graph the parabola 2y x= . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (1, 2). Since 22 1≤ is false, shade the opposite side of the parabola from (1, 2). Graph the line y x= . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (1, 2). Since 2 1≥ is true, shade the same side of the line as (1, 2). The overlapping region is the solution.
39. 2 2
2
16
4
x y
y x
⎧ + ≤⎪⎨
≥ −⎪⎩
Graph the circle 2 2 16x y+ = . Use a sold line since the inequality is not strict. Choose a test point not on the circle, such as (0,0) . Since
2 20 0 16+ ≤ is true, shade the side of the circle containing (0,0) . Graph the parabola
2 4y x= − . Use a solid line since the inequality is not strict. Choose a test point not on the parabola, such as (0,0) . Since 20 0 4≥ − is true, shade the side of the parabola that contains (0,0) . The overlapping region is the solution.
40. 2 2
2
25
5
x y
y x
⎧ + ≤⎪⎨
≤ −⎪⎩
Graph the circle 2 2 25x y+ = . Use a sold line since the inequality is not strict. Choose a test point not on the circle, such as (0,0) . Since
2 20 0 25+ ≤ is true, shade the side of the circle containing (0,0) . Graph the parabola
2 5y x= − . Use a solid line since the inequality is not strict. Choose a test point not on the parabola, such as (0,0) . Since 20 0 5≤ − is false, shade the side of the parabola opposite that which contains the point (0,0) . The overlapping region is the solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
886 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
41. 2
4
1
xy
y x
≥⎧⎪⎨
≥ +⎪⎩
Graph the hyperbola 4xy = . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 0 0 4⋅ ≥ is false, shade the opposite side of the hyperbola from (0, 0). Graph the parabola
2 1y x= + . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 20 0 1≥ + is false, shade the opposite side of the parabola from (0, 0).The overlapping region is the solution.
42. 2
2
1
1
y x
y x
⎧ + ≤⎪⎨
≥ −⎪⎩
Graph the parabola 2 1y x+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since 0 + 02 ≤ 1 is true, shade the same side of the parabola as (0, 0). Graph the parabola
2 1y x= − . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 20 0 1≥ − is true, shade the same side of the parabola as (0, 0). The overlapping region is the solution.
43.
00
2 62 6
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≤⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 6+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices:
The x-axis and y-axis intersect at (0, 0). The intersection of 2 6x y+ = and the y-axis is (0, 3). The intersection of 2 6x y+ = and the x-axis is (3, 0). To find the intersection of 2 6x y+ = and 2 6x y+ = , solve the system:
2 62 6
x yx y
+ =⎧⎨ + =⎩
Solve the first equation for x: 6 2x y= − . Substitute and solve: 2(6 2 ) 6
12 4 612 3 6
3 62
y yy y
yyy
− + =− + =
− =− = −
=
6 2(2) 2x = − = The point of intersection is (2, 2). The four corner points are (0, 0), (0, 3), (3, 0), and (2, 2).
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Section 8.7: Systems of Inequalities
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44.
004
2 3 6
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≥⎪⎪ + ≥⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 4x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 4+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 6x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is unbounded.
Find the vertices: The intersection of 4x y+ = and the y-axis is (0, 4). The intersection of 4x y+ = and the x-axis is (4, 0). The two corner points are (0, 4), and (4, 0).
=
45.
002
2 4
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≥⎪⎪ + ≥⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0). Graph the line 2 4x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is unbounded.
Find the vertices: The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 2 4x y+ = and the y-axis is (0, 4). The two corner points are (2, 0), and (0, 4).
46.
00
3 62 2
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≤⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 3 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 6+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 0 and 0x y= = is (0, 0). The intersection of 2 2x y+ = and the x-axis is (1, 0). The intersection of 2 2x y+ = and the y-axis is (0, 2). The three corner points are (0, 0), (1, 0), and (0, 2).
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
888 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
47.
002
2 3 123 12
xy
x yx yx y
≥⎧⎪ ≥⎪⎪ + ≥⎨⎪ + ≤⎪
+ ≤⎪⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 12x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 12+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 3 12x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 12+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 2 3 12x y+ = and the y-axis is (0, 4). The intersection of 3 12x y+ = and the x-axis is (4, 0).
To find the intersection of 2 3 12x y+ = and 3 12x y+ = , solve the system:
2 3 123 12
x yx y
+ =⎧⎨ + =⎩
Solve the second equation for y: 12 3y x= − . Substitute and solve: 2 3(12 3 ) 12
2 36 9 127 24
247
x xx x
x
x
+ − =+ − =
− = −
=
24 72 1212 3 127 7 7
y ⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
The point of intersection is 24 12,7 7
⎛ ⎞⎜ ⎟⎝ ⎠
.
The five corner points are (0, 2), (0, 4), (2, 0),
(4, 0), and 24 12,7 7
⎛ ⎞⎜ ⎟⎝ ⎠
.
48.
00210
2 3
xy
x yx yx y
≥⎧⎪ ≥⎪⎪ + ≥⎨⎪ + ≤⎪
+ ≤⎪⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 10x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 10+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 3x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 3+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2 3x y+ = and the y-axis is (0, 3). To find the intersection of 2 3 and 2x y x y+ = + = , solve the system:
2 32
x yx y
+ =⎧⎨ + =⎩
Solve the second equation for y: 2y x= − . Substitute and solve: 2 2 3
1x x
x+ − =
=
2 1 1y = − = The point of intersection is (1, 1). The three corner points are (0, 2), (0, 3), and (1, 1).
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 8.7: Systems of Inequalities
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49.
0028
2 10
xy
x yx yx y
≥⎧⎪ ≥⎪⎪ + ≥⎨⎪ + ≤⎪
+ ≤⎪⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 10x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 10+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 8x y+ = and the y-axis is (0, 8). The intersection of 2 10x y+ = and the x-axis is (5, 0). To find the intersection of
8 and 2 10x y x y+ = + = , solve the system: 8
2 10x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 8y x= − . Substitute and solve: 2 8 10
2x x
x+ − =
=
8 2 6y = − = The point of intersection is (2, 6). The five corner points are (0, 2), (0, 8), (2, 0), (5, 0), and (2, 6).
50.
0028
2 1
xy
x yx yx y
≥⎧⎪ ≥⎪⎪ + ≥⎨⎪ + ≤⎪
+ ≥⎪⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 1x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 8x y+ = and the y-axis is (0, 8). The intersection of 8x y+ = and the x-axis is (8, 0). The four corner points are (0, 2), (0, 8), (2, 0), and (8, 0).
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51.
00
2 12 10
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≥⎪⎪ + ≤⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 1x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 10x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 10+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2 1x y+ = and the y-axis is (0, 0.5). The intersection of 2 1x y+ = and the x-axis is (1, 0). The intersection of 2 10x y+ = and the y-axis is (0, 5). The intersection of
2 10x y+ = and the x-axis is (10, 0). The four corner points are (0, 0.5), (0, 5), (1, 0), and (10, 0).
52.
00
2 12 10
28
xy
x yx y
x yx y
≥⎧⎪ ≥⎪⎪ + ≥⎪⎨ + ≤⎪⎪ + ≥⎪
+ ≤⎪⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 1x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 10x y+ = . Use a solid line since the inequality uses ≤. Choose a
test point not on the line, such as (0, 0). Since 0 2(0) 10+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 2 10x y+ = and the y-axis is (0, 5). The intersection of 8x y+ = and the x-axis is (8, 0). To find the intersection of
8x y+ = and 2 10x y+ = , solve the system: 8
2 10x yx y
+ =⎧⎨ + =⎩
Solve the first equation for x: 8x y= − . Substitute and solve: (8 ) 2 10
2y y
y− + =
=
8 2 6x = − = The point of intersection is (6, 2). The five corner points are (0, 2), (0, 5), (2, 0), (8, 0), and (6, 2).
53. The system of linear inequalities is: 4600
xx y
xy
≤⎧⎪ + ≤⎪⎨ ≥⎪⎪ ≥⎩
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54. The system of linear inequalities is: 52600
yx y
xxy
≤⎧⎪ + ≥⎪⎪ ≤⎨⎪ ≥⎪
≥⎪⎩
55. The system of linear inequalities is: 201550
00
xy
x yx y
x
≤⎧⎪ ≥⎪⎪ + ≤⎨⎪ − ≤⎪
≥⎪⎩
56. The system of linear inequalities is: 65
3 4 122 8
0
yx
x yx y
x
≤⎧⎪ ≤⎪⎪ + ≥⎨⎪ − ≤⎪
≥⎪⎩
57. a. Let x = the amount invested in Treasury bills, and let y = the amount invested in corporate bonds. The constraints are:
50,000x y+ ≤ because the total investment cannot exceed $50,000.
35,000x ≥ because the amount invested in Treasury bills must be at least $35,000.
10,000y ≤ because the amount invested in corporate bonds must not exceed $10,000.
0, 0x y≥ ≥ because a non-negative amount must be invested. The system is
50,00035,00010,00000
x yxyxy
+ ≤⎧⎪ ≥⎪⎪ ≤⎨⎪ ≥⎪
≥⎪⎩
b. Graph the system.
The corner points are (35,000, 0),
(35,000, 10,000), (40,000, 10,000), (50,000, 0).
58. a. Let x = the # of standard model trucks, and let y = the # of deluxe model trucks. The constraints are:
0, 0x y≥ ≥ because a non-negative number of trucks must be manufactured. 2 3 80x y+ ≤ because the total painting hours worked cannot exceed 80. 3 4 120x y+ ≤ because the total detailing hours worked cannot exceed 120. The system is
00
2 3 803 4 120
xyx yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≤⎩
b. Graph the system.
The corner points are ( ) ( )800,0 , 0. , 40,03
⎛ ⎞⎜ ⎟⎝ ⎠
.
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Chapter 8: Systems of Equations and Inequalities
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59. a. Let x = the # of packages of the economy blend, and let y = the # of packages of the superior blend. The constraints are:
0, 0x y≥ ≥ because a non-negative # of packages must be produced. 4 8 75 16x y+ ≤ ⋅ because the total amount of “A grade” coffee cannot exceed 75 pounds. (Note: 75 pounds = (75)(16) ounces.) 12 8 120 16x y+ ≤ ⋅ because the total amount of “B grade” coffee cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) Simplifying the inequalities, we obtain:
4 8 75 162 75 42 300
x yx yx y
+ ≤ ⋅+ ≤ ⋅+ ≤
12 8 120 163 2 120 43 2 480
x yx yx y
+ ≤ ⋅+ ≤ ⋅+ ≤
The system is: 002 300
3 2 480
xyx yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≤⎩
b. Graph the system.
The corner points are (0, 0), (0, 150), (90, 105), (160, 0).
60. a. Let x = the # of lower-priced packages, and let y = the # of quality packages. The constraints are:
0, 0x y≥ ≥ because a non-negative # of packages must be produced. 8 6 120 16x y+ ≤ ⋅ because the total amount of peanuts cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) 4 6 90 16x y+ ≤ ⋅ because the total amount of cashews cannot exceed 90 pounds. (Note: 90 pounds = (90)(16) ounces.) Simplifying the inequalities, we obtain:
8 6 120 164 3 120 84 3 960
x yx yx y
+ ≤ ⋅+ ≤ ⋅+ ≤
4 6 90 162 3 90 82 3 720
x yx yx y
+ ≤ ⋅+ ≤ ⋅+ ≤
The system is 4 3 9602 3 720
0; 0
x yx y
x y
+ ≤⎧⎪ + ≤⎨⎪ ≥ ≥⎩
b. Graph the system.
The corner points are (0, 0), (0, 240), (120, 160), (240, 0).
61. a. Let x = the # of microwaves, and let y = the # of printers. The constraints are:
0, 0x y≥ ≥ because a non-negative # of items must be shipped. 30 20 1600x y+ ≤ because a total cargo weight cannot exceed 1600 pounds. 2 3 150x y+ ≤ because the total cargo volume cannot exceed 150 cubic feet. Note that the inequality 30 20 1600x y+ ≤ can be simplified: 3 2 160x y+ ≤ . The system is:
3 2 1602 3 150
0; 0
x yx y
x y
+ ≤⎧⎪ + ≤⎨⎪ ≥ ≥⎩
b. Graph the system.
The corner points are (0, 0), (0, 50), (36, 26),
1603
,0⎛ ⎞⎜ ⎟⎝ ⎠
.
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Section 8.8: Linear Programming
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Section 8.8
1. objective function
2. True
3. z x y= +
Vertex Value of (0, 3) 0 3 3(0, 6) 0 6 6(5, 6) 5 6 11(5, 2) 5 2 7(4, 0) 4 0 4
z x yzzzzz
= += + == + == + == + == + =
The maximum value is 11 at (5, 6), and the minimum value is 3 at (0, 3).
4. 2 3z x y= +
Vertex Value of 2 3(0, 3) 2(0) 3(3) 9(0, 6) 2(0) 3(6) 18(5, 6) 2(5) 3(6) 28(5, 2) 2(5) 3(2) 16(4, 0) 2(4) 3(0) 8
z x yzzzzz
= += + == + == + == + == + =
The maximum value is 28 at (5, 6), and the minimum value is 8 at (4, 0).
5. 10z x y= +
Vertex Value of 10(0, 3) 0 10(3) 30(0, 6) 0 10(6) 60(5, 6) 5 10(6) 65(5, 2) 5 10(2) 25(4, 0) 4 10(0) 4
z x yzzzzz
= += + == + == + == + == + =
The maximum value is 65 at (5, 6), and the minimum value is 4 at (4, 0).
6. 10z x y= +
Vertex Value of 10(0, 3) 10(0) 3 3(0, 6) 10(0) 6 6(5, 6) 10(5) 6 56(5, 2) 10(5) 2 52(4, 0) 10(4) 0 40
z x yzzzzz
= += + == + == + == + == + =
The maximum value is 56 at (5, 6), and the minimum value is 3 at (0, 3).
7. 5 7z x y= +
Vertex Value of 5 7(0, 3) 5(0) 7(3) 21(0, 6) 5(0) 7(6) 42(5, 6) 5(5) 7(6) 67(5, 2) 5(5) 7(2) 39(4, 0) 5(4) 7(0) 20
z x yzzzzz
= += + == + == + == + == + =
The maximum value is 67 at (5, 6), and the minimum value is 20 at (4, 0).
8. 7 5z x y= +
Vertex Value of 7 5(0, 3) 7(0) 5(3) 15(0, 6) 7(0) 5(6) 30(5, 6) 7(5) 5(6) 65(5, 2) 7(5) 5(2) 45(4, 0) 7(4) 5(0) 28
z x yzzzzz
= += + == + == + == + == + =
The maximum value is 65 at (5, 6), and the minimum value is 15 at (0, 3).
9. Maximize 2z x y= + subject to 0,x ≥ 0, 6,y x y≥ + ≤ 1x y+ ≥ . Graph the
constraints.
y
x(6,0)
(0,6)
(0,1)
(1,0)
x + y = 6
x + y = 1 The corner points are (0, 1), (1, 0), (0, 6), (6, 0).
Evaluate the objective function: Vertex Value of 2(0, 1) 2(0) 1 1(0, 6) 2(0) 6 6(1, 0) 2(1) 0 2(6, 0) 2(6) 0 12
z x yzzz
z
= += + == + == + =
= + =
The maximum value is 12 at (6, 0).
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Chapter 8: Systems of Equations and Inequalities
894 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10. Maximize 3z x y= + subject to 0,x ≥ 0y ≥ , 3x y+ ≥ 5, 7x y≤ ≤ . Graph the constraints.
y
x
x = 5
y = 7
x + y = 3
(5,7)(0,7)
(5,0)
(0,3)
(3,0)
The corner points are (0, 3), (3, 0), (0, 7), (5, 0),
(5, 7). Evaluate the objective function: Vertex Value of 3(0, 3) 0 3(3) 9(3, 0) 3 3(0) 3(0, 7) 0 3(7) 21(5, 0) 5 3(0) 5(5, 7) 5 3(7) 26
z x yzz
zz
z
= += + == + =
= + == + =
= + =
The maximum value is 26 at (5, 7).
11. Minimize 2 5z x y= + subject to 0,x ≥ 0, 2, 5,y x y x≥ + ≥ ≤ 3y ≤ . Graph the
constraints.
y
x
(5,3)
(5,0)
(0,3)
(2,0)
(0,2)
x = 5
y = 3
x + y = 2 The corner points are (0, 2), (2, 0), (0, 3), (5, 0),
(5, 3). Evaluate the objective function: Vertex Value of 2 5(0, 2) 2(0) 5(2) 10(0, 3) 2(0) 5(3) 15(2, 0) 2(2) 5(0) 4(5, 0) 2(5) 5(0) 10(5, 3) 2(5) 5(3) 25
z x yzzzzz
= += + == + == + == + == + =
The minimum value is 4 at (2, 0).
12. Minimize 3 4z x y= + subject to 0x ≥ , 0, 2 3 6,y x y≥ + ≥ 8x y+ ≤ . Graph the
constraints.
y
x
x + y = 8
2x + 3y = 6
(0,8)
(0,2)
(3,0) (8,0)
The corner points are (0, 2), (3, 0), (0, 8), (8, 0).
Evaluate the objective function: Vertex Value of 3 4(0, 2) 3(0) 4(2) 8(3, 0) 3(3) 4(0) 9(0, 8) 3(0) 4(8) 32(8, 0) 3(8) 4(0) 24
z x yzz
zz
= += + == + =
= + == + =
The minimum value is 8 at (0, 2).
13. Maximize 3 5z x y= + subject to 0,x ≥ 0,y ≥ 2,x y+ ≥ 2 3 12,x y+ ≤ 3 2 12x y+ ≤ . Graph
the constraints.
y
x
(2.4,2.4)
(0,4)
(0,2)
(2,0) (4,0)
3x + 2y = 12
2x + 3y = 12
x + y = 2 To find the intersection of 2 3 12x y+ = and
3 2 12x y+ = , solve the system:
2 3 123 2 12
x yx y
+ =⎧⎨ + =⎩
Solve the second equation for y: 362
y x= −
Substitute and solve:
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Section 8.8: Linear Programming
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32 3 6 12292 18 1225 62
125
x x
x x
x
x
⎛ ⎞+ − =⎜ ⎟⎝ ⎠
+ − =
− = −
=
3 12 18 126 62 5 5 5
y ⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
The point of intersection is ( )2.4, 2.4 . The corner points are (0, 2), (2, 0), (0, 4), (4, 0), (2.4, 2.4). Evaluate the objective function:
Vertex Value of 3 5(0, 2) 3(0) 5(2) 10(0, 4) 3(0) 5(4) 20(2, 0) 3(2) 5(0) 6(4, 0) 3(4) 5(0) 12
(2.4, 2.4) 3(2.4) 5(2.4) 19.2
z x yzzzz
z
= += + == + == + == + =
= + =
The maximum value is 20 at (0, 4).
14. Maximize 5 3z x y= + subject to 0,x ≥ 0, 2, 8, 2 10y x y x y x y≥ + ≥ + ≤ + ≤ . Graph
the constraints.
y
x
x + y = 8
2x + y = 10
x + y = 2
(2,0)
(0,2)
(0,8)
(5,0)
(2,6)
To find the intersection of 8x y+ = and
2 10x y+ = , solve the system: 8
2 10x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 8y x= − . Substitute and solve: 2 8 10
2x x
x+ − =
=
8 2 6y = − = The point of intersection is (2, 6). The corner points are (0, 2), (2, 0), (0, 8), (5, 0), (2, 6). Evaluate the objective function:
Vertex Value of 5 3(0, 2) 5(0) 3(2) 6(0, 8) 5(0) 3(8) 24(2, 0) 5(2) 3(0) 10(5, 0) 5(5) 3(0) 25(2, 6) 5(2) 3(6) 28
z x yzzzzz
= += + == + == + == + == + =
The maximum value is 28 at (2, 6).
15. Minimize 5 4z x y= + subject to 0,x ≥ 0, 2, 2 3 12, 3 12y x y x y x y≥ + ≥ + ≤ + ≤ . Graph
the constraints.
y
x
(0,4)
(0,2)
(2,0) (4,0)
x + y = 2
2x + 3y = 12
3x + y = 12
247
, 127( )
To find the intersection of 2 3 12x y+ = and
3 12x y+ = , solve the system: 2 3 123 12
x yx y
+ =⎧⎨ + =⎩
Solve the second equation for y: 12 3y x= − Substitute and solve:
247
2 3(12 3 ) 122 36 9 12
7 24
x xx x
xx
+ − =+ − =
− = −=
( ) 7224 127 7 712 3 12y = − = − =
The point of intersection is 24 12,7 7
⎛ ⎞⎜ ⎟⎝ ⎠
.
The corner points are (0, 2), (2, 0), (0, 4), (4, 0), 24 12,7 7
⎛ ⎞⎜ ⎟⎝ ⎠
. Evaluate the objective function:
( ) ( ) ( )24 12 24 127 7 7 7
Vertex Value of 5 4(0, 2) 5(0) 4(2) 8(0, 4) 5(0) 4(4) 16(2, 0) 5(2) 4(0) 10(4, 0) 5(4) 4(0) 20
, 5 4 24
z x yzzzz
z
= += + == + == + == + =
= + =
The minimum value is 8 at (0, 2).
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Chapter 8: Systems of Equations and Inequalities
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16. Minimize 2 3z x y= + subject to 0,x ≥ 0, 3, 9, 3 6y x y x y x y≥ + ≥ + ≤ + ≥ . Graph
the constraints.
y
x
x + y = 9
x + y = 3 x + 3y = 6
(0,9)
(0,3)
(6,0)(9,0)
32 , 3
2( )
To find the intersection of 3x y+ = and 3 6x y+ = , solve the system:
33 6
x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 3y x= − . Substitute and solve:
3(3 ) 69 3 6
2 332
x xx x
x
x
+ − =+ − =
− = −
=
3 332 2
y = − =
The point of intersection is 3 3,2 2
⎛ ⎞⎜ ⎟⎝ ⎠
.
The corner points are (0, 3), (6, 0), (0, 9), (9, 0), 3 3,2 2
⎛ ⎞⎜ ⎟⎝ ⎠
. Evaluate the objective function:
Vertex Value of 2 3(0, 3) 2(0) 3(3) 9(0, 9) 2(0) 3(9) 27(6, 0) 2(6) 3(0) 12(9, 0) 2(9) 3(0) 183 3 3 3 15, 2 42 2 2 2 2
z x yz
zzz
z
= += + =
= + == + == + =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
The minimum value is 152
at 3 3,2 2
⎛ ⎞⎜ ⎟⎝ ⎠
.
17. Maximize 5 2z x y= + subject to 0,x ≥ 0, 10, 2 10, 2 10y x y x y x y≥ + ≤ + ≥ + ≥ .
Graph the constraints.
y
x
(0,10)
(10,0)
x + y = 10
2x + y = 10x + 2y = 10
103 , 10
3( )
To find the intersection of 2 10x y+ = and 2 10x y+ = , solve the system:
2 102 10
x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 10 2y x= − . Substitute and solve:
2(10 2 ) 1020 4 10
3 10103
x xx x
x
x
+ − =+ − =
− = −
=
10 20 1010 2 103 3 3
y ⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
The point of intersection is (10/3 10/3). The corner points are (0, 10), (10, 0), (10/3, 10/3). Evaluate the objective function:
13
Vertex Value of 5 2(0, 10) 5(0) 2(10) 20(10, 0) 5(10) 2(0) 5010 10 10 10 70, 5 2 233 3 3 3 3
z x yzz
z
= += + == + =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
The maximum value is 50 at (10, 0).
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Section 8.8: Linear Programming
897 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18. Maximize 2 4z x y= + subject to 0, 0, 2 4, 9x y x y x y≥ ≥ + ≥ + ≤ .
Graph the constraints.
y
x
x + y = 9
2x + y = 4
(0,9)
(9,0)(2,0)
(0,4)
The corner points are (0, 9), (9, 0), (0, 4), (2, 0).
Evaluate the objective function: Vertex Value of 2 4(0, 9) 2(0) 4(9) 36(9, 0) 2(9) 4(0) 18(0, 4) 2(0) 4(4) 16(2, 0) 2(2) 4(0) 4
z x yzzzz
= += + == + == + == + =
The maximum value is 36 at (0, 9).
19. Let x = the number of downhill skis produced, and let y = the number of cross-country skis produced. The total profit is: 70 50P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of skis must be produced.
2 40x y+ ≤ Manufacturing time available. 32x y+ ≤ Finishing time available. Graph the constraints.
y
x
(8,24)
(20,0)(0,0)
(0,32)
2x + y = 40
x + y = 32
To find the intersection of 32x y+ = and
2 40x y+ = , solve the system: 32
2 40x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 32y x= − . Substitute and solve: 2 (32 ) 40
8x x
x+ − =
=
32 8 24y = − = The point of intersection is (8, 24). The corner points are (0, 0), (0, 32), (20, 0), (8, 24). Evaluate the objective function:
Vertex Value of 70 50(0, 0) 70(0) 50(0) 0(0, 32) 70(0) 50(32) 1600(20, 0) 70(20) 50(0) 1400(8, 24) 70(8) 50(24) 1760
P x yP
PPP
= += + =
= + == + == + =
The maximum profit is $1760, when 8 downhill skis and 24 cross-country skis are produced.
With the increase of the manufacturing time to 48 hours, we do the following: The constraints are:
0, 0x y≥ ≥ A positive number of skis must be produced.
2 48x y+ ≤ Manufacturing time available. 32x y+ ≤ Finishing time available. Graph the constraints.
y
x
(16,16)
(0,32)
(24,0)(0,0)
2x + y = 48
x + y = 32
To find the intersection of 32x y+ = and
2 48x y+ = , solve the system: 32
2 48x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 32y x= − . Substitute and solve: 2 (32 ) 48
16x x
x+ − =
=
32 16 16y = − = The point of intersection is (16, 16). The corner points are (0, 0), (0, 32), (24, 0), (16, 16). Evaluate the objective function:
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Chapter 8: Systems of Equations and Inequalities
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Vertex Value of 70 50(0, 0) 70(0) 50(0) 0(0, 32) 70(0) 50(32) 1600(24, 0) 70(24) 50(0) 1680(16, 16) 70(16) 50(16) 1920
P x yP
PPP
= += + =
= + == + == + =
The maximum profit is $1920, when 16 downhill skis and 16 cross-country skis are produced.
20. Let x = the number of acres of soybeans planted , and let y = the number of acres of wheat planted. The total profit is: 180 100P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of acres must be planted.
70x y+ ≤ Acres available to plant. 60 30 1800x y+ ≤ Money available for
preparation. 3 4 120x y+ ≤ Workdays available. Graph the constraints.
y
x
x + y = 70
60x + 30y = 1800
3x + 4y = 120
(0,30)
(30,0)(0,0)
(24,12)
To find the intersection of 60 30 1800x y+ = and
3 4 120x y+ = , solve the system: 60 30 1800
3 4 120x yx y+ =⎧
⎨ + =⎩
Solve the first equation for y: 60 30 1800
2 6060 2
x yx y
y x
+ =+ =
= −
Substitute and solve: 3 4(60 2 ) 120
3 240 8 1205 120
24
x xx x
xx
+ − =+ − =
− = −=
60 2(24) 12y = − = The point of intersection is (24, 12). The corner points are (0, 0), (0, 30), (30, 0), (24, 12). Evaluate the objective function:
Vertex Value of 180 100(0, 0) 180(0) 100(0) 0(0, 30) 180(0) 100(30) 3000(30, 0) 180(30) 100(0) 5400(24, 12) 180(24) 100(12) 5520
P x yP
PPP
= += + =
= + == + == + =
The maximum profit is $5520, when 24 acres of soybeans and 12 acres of wheat are planted.
With the increase of the preparation costs to $2400, we do the following: The constraints are:
0, 0x y≥ ≥ A non-negative number of acres must be planted.
70x y+ ≤ Acres available to plant. 60 30 2400x y+ ≤ Money available for
preparation. 3 4 120x y+ ≤ Workdays available. Graph the constraints.
y
x
x + y = 70
3x + 4y = 120
60x + 30y = 2400
(0,30)
(40,0)(0,0)
The corner points are (0, 0), (0, 30), (40, 0).
Evaluate the objective function: Vertex Value of 180 100(0, 0) 180(0) 100(0) 0(0, 30) 180(0) 100(30) 3000(40, 0) 180(40) 100(0) 7200
P x yP
PP
= += + =
= + == + =
The maximum profit is $7200, when 40 acres of soybeans and 0 acres of wheat are planted.
21. Let x = the number of rectangular tables rented, and let y = the number of round tables rented. The cost for the tables is: 28 52C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of tables must be used.
35x y+ ≤ Maximum number of tables. 6 10 250x y+ ≥ Number of guests. 15x ≤ Rectangular tables available.
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Graph the constraints.
y
x
6 + 10 = 250x y
x y + = 35
(15, 20)
x = 15
(0, 25)
(0, 35)
(15, 16)
25
10
The corner points are (0, 25), (0, 35), (15, 20),
(15, 16). Evaluate the objective function: Vertex Value of 28 52(0, 25) 28(0) 52(25) 1300(0, 35) 28(0) 52(35) 1820(15, 20) 28(15) 52(20) 1460(15, 16) 28(15) 52(16) 1252
C x yCCCC
= += + == + == + == + =
Kathleen should rent 15 rectangular tables and 16 round tables in order to minimize the cost. The minimum cost is $1252.00.
22. Let x = the number of buses rented, and let y = the number of vans rented. The cost for the vehicles is: 975 350C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of buses and vans must be used.
40 8 320x y+ ≥ Number of regular seats. 3 36x y+ ≥ Number of handicapped seats. Graph the constraints.
y
xx y + 3 = 36
40 + 8 = 320x y
(6, 10)(0, 12)
(0, 40)
(36, 0)
25
25
To find the intersection of 40 8 320x y+ = and
3 36x y+ = , solve the system: 40 8 320
3 36x yx y
+ =⎧⎨ + =⎩
Solve the second equation for x:
3 363 36
x yx y
+ == − +
Substitute and solve: 40( 3 36) 8 320120 1440 8 320
112 112010
y yy y
yy
− + + =− + + =
− = −=
3(10) 36 30 36 6x = − + = − + = The point of intersection is (6, 10). The corner points are (0, 40), (6, 10), and (36, 0). Evaluate the objective function:
Vertex Value of 975 350(0, 40) 975(0) 350(40) 14,000(6, 10) 975(6) 350(10) 9350 (36, 0) 975(36) 350(0) 35,100
C x yCCC
= += + == + == + =
The college should rent 6 buses and 10 vans for a minimum cost of $9350.00.
23. Let x = the amount invested in junk bonds, and let y = the amount invested in Treasury bills. The total income is: 0.09 0.07I x y= + . Income is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative amount must be invested.
20,000x y+ ≤ Total investment cannot exceed $20,000.
12,000x ≤ Amount invested in junk bonds must not exceed $12,000.
8000y ≥ Amount invested in Treasury bills must be at least $8000.
a. y x≥ Amount invested in Treasury bills must be equal to or greater than the amount invested in junk bonds.
Graph the constraints.
x + y = 20000
y = x
x = 12000
y = 8000
(0,20000)
(0,8000) (8000,8000)
(10000,10000)
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Chapter 8: Systems of Equations and Inequalities
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The corner points are (0, 20,000), (0, 8000), (8000, 8000), (10,000, 10,000). Evaluate the objective function:
Vertex Value of 0.09 0.07(0, 20000) 0.09(0) 0.07(20000)
1400(0, 8000) 0.09(0) 0.07(8000)
560(8000, 8000) 0.09(8000) 0.07(8000)
1280(10000, 10000) 0.09(10000) 0.07(10000)
1600
I x yI
I
I
I
= += +== +== +== +=
The maximum income is $1600, when $10,000 is invested in junk bonds and $10,000 is invested in Treasury bills.
b. y x≤ Amount invested in Treasury bills must not exceed the amount invested in junk bonds.
Graph the constraints.
x + y = 20000
y = x
x = 12000
y = 8000(8000,8000)
(10000,10000)
(12000,8000)
The corner points are (12,000, 8000),
(8000, 8000), (10,000, 10,000). Evaluate the objective function:
Vertex Value of 0.09 0.07(12000, 8000) 0.09(12000) 0.07(8000)
1640(8000, 8000) 0.09(8000) 0.07(8000)
1280(10000, 10000) 0.09(10000) 0.07(10000)
1600
I x yI
I
I
= += +== +=
= +=
The maximum income is $1640, when $12,000 is invested in junk bonds and $8000 is invested in Treasury bills.
24. Let x = the number of hours that machine 1 is operated, and let y = the number of hours that machine 2 is operated. The total cost is:
50 30C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of hours must be used.
10x ≤ Time used on machine 1. 10y ≤ Time used on machine 2. 60 40 240x y+ ≥ 8-inch pliers to be produced. 70 20 140x y+ ≥ 6-inch pliers to be produced. Graph the constraints.
(10,10)(0,10)
(10,0)(4,0)
(0,7)
60x + 40y = 24070x + 20y = 140
12 , 21
4( )
To find the intersection of 60 40 240x y+ = and 70 20 140x y+ = , solve the system:
60 40 24070 20 140
x yx y
+ =⎧⎨ + =⎩
Divide the first equation by 2− and add the result to the second equation:
30 20 12070 20 140
x yx y
− − = −+ =
40 2020 140 2
x
x
=
= =
Substitute and solve: 160 40 240230 40 240
40 210210 21 1540 4 4
y
yy
y
⎛ ⎞ + =⎜ ⎟⎝ ⎠
+ ==
= = =
The point of intersection is 1 1, 52 4
⎛ ⎞⎜ ⎟⎝ ⎠
.
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Section 8.8: Linear Programming
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The corner points are (0, 7), (0, 10), (4, 0),
(10, 0), (10, 10), 1 1, 52 4
⎛ ⎞⎜ ⎟⎝ ⎠
. Evaluate the
objective function: Vertex Value of 50 30(0, 7) 50(0) 30(7) 210(0, 10) 50(0) 30(10) 300(4, 0) 50(4) 30(0) 200(10, 0) 50(10) 30(0) 500(10, 10) 50(10) 30(10) 8001 1 1 1, 5 50 30 5 182.502 4 2 4
C x yCCCCC
P
= += + == + == + == + == + =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
The minimum cost is $182.50, when machine 1
is used for 12
hour and machine 2 is used for 154
hours.
25. Let x = the number of pounds of ground beef, and let y = the number of pounds of ground pork. The total cost is: 0.75 0.45C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of pounds must be used.
200x ≤ Only 200 pounds of ground beef are available.
50y ≥ At least 50 pounds of ground pork must be used.
0.75 0.60 0.70( )x y x y+ ≥ + Leanness condition
(Note that the last equation will simplify to 12
y x≤ .) Graph the constraints.
y
x
(200,100)
(200,50)(100,50)
The corner points are (100, 50), (200, 50),
(200, 100). Evaluate the objective function:
Vertex Value of 0.75 0.45(100, 50) 0.75(100) 0.45(50) 97.50 (200, 50) 0.75(200) 0.45(50) 172.50(200, 100) 0.75(200) 0.45(100) 195
C x yCCC
= += + == + == + =
The minimum cost is $97.50, when 100 pounds of ground beef and 50 pounds of ground pork are used.
26. Let x = the number of gallons of regular, and let y = the number of gallons of premium. The
total profit is: 0.75 0.90P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of gallons must be used.
14
y x≥ At least one gallon of premium
for every 4 gallons of regular. 5 6 3000x y+ ≤ Daily shipping weight limit. 24 20 16(725)x y+ ≤ Available flavoring. 12 20 16(425)x y+ ≤ Available milk-fat (Note: the last two inequalities simplify to
6 5 2900x y+ ≤ and 3 5 1700x y+ ≤ .) Graph the constraints.
y
x
5 + 6 = 3000x y
(0, 340) (400, 100)
300
15014
y x=
24 + 20 = 16(725)x y
12 + 20 = 16(425)x y
The corner points are (0, 0), (400, 100), (0, 340). Evaluate the objective function:
Vertex Value of 0.75 0.90(0, 0) 0.75(0) 0.90(0) 0
(400,100) 0.75(400) 0.90(100) 390(0, 340) 0.75(0) 0.90(340) 306
P x yPPP
= += + == + == + =
Mom and Pop should produce 400 gallons of regular and 100 gallons of premium ice cream. The maximum profit is $390.00.
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27. Let x = the number of racing skates manufactured, and let y = the number of figure skates manufactured. The total profit is:
10 12P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of skates must be manufactured.
6 4 120x y+ ≤ Only 120 hours are available for fabrication.
2 40x y+ ≤ Only 40 hours are available for finishing.
Graph the constraints.
y
x
(10,15)
(20,0)(0,0)
(0,20)
To find the intersection of 6 4 120x y+ = and
+2 40x y = , solve the system: 6 4 120
2 40x yx y
+ =⎧⎨ + =⎩
Solve the second equation for x: 40 2x y= − Substitute and solve: 6(40 2 ) 4 120240 12 4 120
8 12015
y yy y
yy
− + =− + =
− = −=
40 2(15) 10x = − = The point of intersection is (10, 15). The corner points are (0, 0), (0, 20), (20, 0), (10, 15). Evaluate the objective function:
Vertex Value of 10 12(0, 0) 10(0) 12(0) 0(0, 20) 10(0) 12(20) 240(20, 0) 10(20) 12(0) 200(10, 15) 10(10) 12(15) 280
P x yP
PPP
= += + =
= + == + == + =
The maximum profit is $280, when 10 racing skates and 15 figure skates are produced.
28. Let x = the amount placed in the AAA bond. Let y = the amount placed in a CD. The total return is: 0.08 0.04R x y= + . Return is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive amount must be invested in each.
50,000x y+ ≤ Total investment cannot exceed $50,000.
20,000x ≤ Investment in the AAA bond cannot exceed $20,000.
15,000y ≥ Investment in the CD must be at least $15,000.
y x≥ Investment in the CD must exceed or equal the investment in the bond.
Graph the constraints.
y
x
y = x
x + y = 50000
x = 20000
y = 15000
(0,50000)
(0,15000)(15000,15000)
(20000,30000)
(20000,20000)
The corner points are (0, 50,000), (0, 15,000),
(15,000, 15,000), (20,000, 20,000), (20,000, 30,000). Evaluate the objective function:
Vertex Value of 0.08 0.04(0, 50000) 0.08(0) 0.04(50000)
2000(0, 15000) 0.08(0) 0.04(15000)
600(15000, 15000) 0.08(15000) 0.04(15000)
1800(20000, 20000) 0.08(20000) 0.04(20000)
2400(20000, 30000
R x yR
R
R
R
= += +== +== +== +=
) 0.08(20000) 0.04(30000)2800
R = +=
The maximum return is $2800, when $20,000 is invested in a AAA bond and $30,000 is invested in a CD.
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Section 8.8: Linear Programming
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29. Let x = the number of metal samples, and let y = the number of plastic samples. The total
cost is: 9 4C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
2, 2x y≥ ≥ At least 2 of each sample must be made.
6x y+ ≥ At least 6 samples are needed. 4 2 24x y+ ≤ Only 24 hours are available. Graph the constraints.
y
x
(2,8)
(5,2)(4,2)
(2,4)
The corner points are (2, 4), (2, 8), (4, 2), (5, 2).
Evaluate the objective function: Vertex Value of 9 4(2, 4) 9(2) 4(4) 34(2, 8) 9(2) 4(8) 50(4, 2) 9(4) 4(2) 44(5, 2) 9(5) 4(2) 53
C x yCCCC
= += + == + == + == + =
The minimum cost is $34, when 2 metal samples and 4 plastic samples are ordered.
30. Let x = the amount of “Gourmet Dog,” and let y = the amount of “Chow Hound.” The total
cost is: 0.40 0.32C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of cans must be purchased.
20 35 1175x y+ ≥ At least 1175 units of vitamins per month.
75 50 2375x y+ ≥ At least 2375 calories per month.
60x y+ ≤ Storage space for 60 cans. Graph the constraints.
y
x
x + y = 60
75x+50y=2375
20x+35y=1175
(0,47.5)
(0,60)
(60,0)
(58.75,0)
(15,25)
The corner points are (0, 47.5), (0, 60), (60, 0),
(58.75, 0), (15, 25). Evaluate the objective function:
Vertex Value of 0.40 0.32(0, 47.5) 0.40(0) 0.32(47.5) 15.20(0, 60) 0.40(0) 0.32(60) 19.20(60, 0) 0.40(60) 0.32(0) 24.00
(58.75, 0) 0.40(58.75) 0.32(0) 23.50(15, 25) 0.40(15) 0.32(25) 14.00
C x yCCC
CC
= += + == + == + =
= + == + =
The minimum cost is $14, when 15 cans of "Gourmet Dog" and 25 cans of “Chow Hound” are purchased.
31. Let x = the number of first class seats, and let y = the number of coach seats. Using the hint,
the revenue from x first class seats and y coach seats is ,Fx Cy+ where 0.F C> > Thus, R Fx Cy= + is the objective function to be maximized. The constraints are: 8 16x≤ ≤ Restriction on first class seats. 80 120y≤ ≤ Restriction on coach seats.
a. 112
xy
≤ Ratio of seats.
The constraints are: 8 16x≤ ≤ 80 120y≤ ≤ 12x y≤ Graph the constraints.
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Chapter 8: Systems of Equations and Inequalities
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The corner points are (8, 96), (8, 120), and (10, 120). Evaluate the objective function:
Vertex Value of (8, 96) 8 96(8, 120) 8 120(10, 120) 10 120
R Fx CyR F CR F CR F C
= += += += +
Since 0,C > 120 96 ,C C> so 8 120 8 96 .F C F C+ > + Since 0,F > 10 8 ,F F> so 10 120 8 120 .F C F C+ > + Thus, the maximum revenue occurs when the aircraft is configured with 10 first class seats and 120 coach seats.
b. 18
xy
≤
The constraints are: 8 16x≤ ≤ 80 120y≤ ≤ 8x y≤ Graph the constraints.
The corner points are (8, 80), (8, 120),
(15, 120), and (10, 80). Evaluate the objective function:
Vertex Value of (8, 80) 8 80(8, 120) 8 120(15, 120) 15 120(10, 80) 10 80
R Fx CyR F CR F CR F CR F C
= += += += += +
Since 0F > and 0,C > 15 10F F> and 120 80 ,C C> so the maximum value of R occurs at (15, 120). The maximum revenue occurs when the aircraft is configured with 15 first class seats and 120 coach seats.
c. Answers will vary.
32. Answers will vary.
Chapter 8 Review Exercises
1. 2 55 2 8
x yx y
− =⎧⎨ + =⎩
Solve the first equation for y: 2 5y x= − . Substitute and solve: 5 2(2 5) 8
5 4 10 89 18
2
x xx x
xx
+ − =+ − =
==
2(2) 5 4 5 1y = − = − = − The solution is 2, 1x y= = − or (2, 1)− .
2. 2 3 27 3x yx y
+ =⎧⎨ − =⎩
Solve the second equation for y: 7 3y x= − Substitute into the first equation and solve: 2 3(7 3) 2
2 21 9 223 11
1123
x xx x
x
x
+ − =+ − =
=
=
11 77 69 87 323 23 23 23
y ⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
The solution is 11 8,23 23
x y= = or 11 8,23 23
⎛ ⎞⎜ ⎟⎝ ⎠
.
3. 3 4 4
132
x y
x y
− =⎧⎪⎨
− =⎪⎩
Solve the second equation for x: 132
x y= +
Substitute into the first equation and solve: 13 3 4 4239 4 42
55212
y y
y y
y
y
⎛ ⎞+ − =⎜ ⎟⎝ ⎠
+ − =
=
=
1 13 22 2
x ⎛ ⎞= + =⎜ ⎟⎝ ⎠
The solution is 12,2
x y= = or 12,2
⎛ ⎞⎜ ⎟⎝ ⎠
.
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Chapter 8 Review Exercises
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4. 2 0
135 42
x y
x y
+ =⎧⎪⎨
− = −⎪⎩
Solve the first equation for y: 2y x= − Substitute into the second equation and solve:
135 4( 2 )2
135 82
1313212
x x
x x
x
x
− − = −
+ = −
= −
= −
12 12
y ⎛ ⎞= − − =⎜ ⎟⎝ ⎠
The solution is 1 , 12
x y= − = or 1 , 12
⎛ ⎞−⎜ ⎟⎝ ⎠
.
5. 2 4 0
3 2 4 0x yx y
− − =⎧⎨ + − =⎩
Solve the first equation for x: 2 4x y= + Substitute into the second equation and solve: 3(2 4) 2 4 0
6 12 2 4 08 8
1
y yy y
yy
+ + − =+ + − =
= −= −
2( 1) 4 2x = − + = The solution is 2, 1x y= = − or (2, 1)− .
6. 3 5 0
2 3 5 0x yx y
− + =⎧⎨ + − =⎩
Solve the first equation for x: 3 5x y= −
Substitute into the second equation and solve: 2(3 5) 3 5 0
6 10 3 5 09 15
53
y yy y
y
y
− + − =− + − =
=
=
53 5 03
x ⎛ ⎞= − =⎜ ⎟⎝ ⎠
The solution is 50,3
x y= = or 50,3
⎛ ⎞⎜ ⎟⎝ ⎠
.
7. 2 53 4
y xx y
= −⎧⎨ = +⎩
Substitute the first equation into the second equation and solve:
3(2 5) 46 15 4
5 11115
x xx xx
x
= − += − +
− = −
=
11 32 55 5
y ⎛ ⎞= − = −⎜ ⎟⎝ ⎠
The solution is 11 3,5 5
x y= = − or 11 3,5 5
⎛ ⎞−⎜ ⎟⎝ ⎠
.
8. 5 25 2
x yy x
= +⎧⎨ = +⎩
Substitute the first equation into the second equation and solve:
5(5 2) 225 10 2
24 1212
y yy yy
y
= + += + +
− =
= −
1 5 15 2 22 2 2
x ⎛ ⎞= − + = − + = −⎜ ⎟⎝ ⎠
The solution is 1 1,2 2
x y= − = − or 1 1,2 2
⎛ ⎞− −⎜ ⎟⎝ ⎠
.
9. 3 4 0
1 3 4 02 2 3
x y
x y
− + =⎧⎪⎨
− + =⎪⎩
Multiply each side of the first equation by 3 and each side of the second equation by 6− and add:
3 9 12 03 9 8 0
4 0
x yx y
⎧ − + =⎪⎨− + − =⎪⎩
=
There is no solution to the system. The system is inconsistent.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
906 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10. 1 24
4 2 0
x y
y x
⎧ + =⎪⎨⎪ + + =⎩
Solve the second equation for y: 4 2y x= − − . Substitute into the first equation and solve:
1 ( 4 2) 24
1 22
502
x x
x x
+ − − =
− − =
=
There is no solution to the system. The system of equations is inconsistent.
11. 2 3 13 03 2 0x yx y+ − =⎧
⎨ − =⎩
Multiply each side of the first equation by 2 and each side of the second equation by 3, and add to eliminate y:
4 6 26 09 6 0
13 26 0 13 26
2
x yx y
xxx
⎧ + − =⎪⎨ − =⎪⎩
− ===
Substitute and solve for y: 3(2) 2 0
2 63
yyy
− =− = −
=
The solution is 2, 3x y= = or (2, 3).
12. 4 5 215 6 42
x yx y
+ =⎧⎨ + =⎩
Multiply each side of the first equation by 5 and each side of the second equation by –4 and add to eliminate x:
20 25 10520 24 168
63
x yx y
y
⎧ + =⎪⎨ − − = −⎪⎩
= −
Substitute and solve for x: 4 5( 63) 21
4 315 214 336
84
xx
xx
+ − =− =
==
The solution is 84, 63x y= = − or (84, 63)− .
13. 3 2 8
2 123
x y
x y
− =⎧⎪⎨
− =⎪⎩
Multiply each side of the second equation by –3 and add to eliminate x:
3 2 83 2 36
0 28
x yx y
⎧ − =⎪⎨− + = −⎪⎩
= −
There is no solution to the system. The system of equations is inconsistent.
14. 2 5 104 10 20
x yx y
+ =⎧⎨ + =⎩
Multiply each side of the first equation by –2 and add to eliminate x:
4 10 204 10 20
0 0
x yx y
⎧− − = −⎪⎨ + =⎪⎩
=
The system is dependent. 2 5 10
5 2 102 25
x yy x
y x
+ == − +
= − +
The solution is 2 25
y x= − + , x is any real number
or 2( , ) 2, is any real number5
x y y x x⎧ ⎫= − +⎨ ⎬
⎩ ⎭.
15. 2 6
2 3 133 2 3 16
x y zx y zx y z
+ − =⎧⎪ − + = −⎨⎪ − + = −⎩
Multiply each side of the first equation by –2 and add to the second equation to eliminate x;
2 4 2 122 3 13
5 5 25 5
x y zx y z
y zy z
⎧ − − + = −⎪⎨ − + = −⎪⎩
− + = −− =
Multiply each side of the first equation by –3 and add to the third equation to eliminate x:
3 6 3 183 2 3 16
8 6 34
x y zx y z
y z
− − + = −− + = −
− + = −
Multiply each side of the first result by 8 and add to the second result to eliminate y:
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Chapter 8 Review Exercises
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8 8 408 6 34
2 63
y zy z
zz
− =− + = −
− == −
Substituting and solving for the other variables: ( 3) 5
2y
y− − =
= 2(2) ( 3) 6
4 3 61
xx
x
+ − − =+ + =
= −
The solution is 1, 2, 3x y z= − = = − or ( 1, 2, 3)− − .
16. 5 2
2 72 11
x y zx y zx y z
+ − =⎧⎪ + + =⎨⎪ − + =⎩
Add the first equation and the second equation to eliminate z:
5 22 7
3 6 9 2 3
x y zx y z
x yx y
⎧ + − =⎪⎨ + + =⎪⎩
+ =− − = −
Multiply each side of the first equation by 2 and add to the third equation to eliminate z:
2 10 2 4 2 11
3 9 15 3 5
x y zx y z
x yx y
+ − =− + =
+ =+ =
Add the two results to eliminate x: 2 33 5
2
x yx y
y
− − = −+ =
=
Substituting and solving for the other variables: 3(2) 5
6 51
xx
x
+ =+ =
= −
2( 1) 2 72 2 7
7
zzz
− + + =− + + =
=
The solution is 1, 2, 7x y z= − = = or ( 1, 2, 7)− .
17. 2 4 15
2 4 275 6 2 3
x y zx y zx y z
− + = −⎧⎪ + − =⎨⎪ − − = −⎩
Multiply the first equation by 1− and the second equation by 2, and then add to eliminate x:
2 4 152 4 8 54
8 9 69
x y zx y z
y z
− + − =⎧⎪⎨ + − =⎪⎩
− =
Multiply the second equation by 5− and add to the third equation to eliminate x:
5 10 20 1355 6 2 3
16 18 138
x y zx y z
y z
− − + = −− − = −
− + = −
Multiply both sides of the first result by 2 and add to the second result to eliminate y: 16 18 13816 18 138
0 0
y zy z
− =− + = −
=
The system is dependent. 16 18 13818 138 16
9 698 8
y zz y
y z
− + = −+ =
= +
Substituting into the second equation and solving for x:
9 692 4 278 89 69 4 274 4
7 394 4
x z z
x z z
x z
⎛ ⎞+ + − =⎜ ⎟⎝ ⎠
+ + − =
= +
The solution is 7 394 4
x z= + , 9 698 8
y z= + , z is
any real number or 7 39( , , ) ,4 4
x y z x z⎧= +⎨
⎩
9 69 , is any real number8 8
y z z ⎫= + ⎬⎭
.
18. 4 3 15
3 5 57 5 9 10
x y zx y zx y z
− + =⎧⎪− + − = −⎨⎪− − − =⎩
Multiply the first equation by 3 and then add the second equation to eliminate x:
3 12 9 453 5 5
11 4 40
x y zx y z
y z
− + =⎧⎪⎨− + − = −⎪⎩
− + =
Multiply the first equation by 7 and add to the third equation to eliminate x: 7 28 21 105
7 5 9 10
33 12 115115 11 4
3
x y zx y z
y z
y z
− + =− − − =
− + =
− + =
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
908 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Multiply the first result by 1− and adding it to the second result:
11 4 4011511 4
35 03
y z
y z
− = −⎧⎪⎨− + =⎪⎩
= −
The system has no solution. The system is inconsistent.
19. 3 2 8
4 1x y
x y+ =⎧
⎨ + = −⎩
20. 2 5 2
5 3 82 0
x y zx zx y
+ + = −⎧⎪ − =⎨⎪ − =⎩
21. 1 0 3 42 4 1 51 2 5 2
1 3 0 ( 4) 4 42 1 4 5 3 91 5 2 2 4 4
A C−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥+ = +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
+ + − −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= + + =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− + +⎣ ⎦ ⎣ ⎦
22. 1 0 3 42 4 1 51 2 5 2
1 3 0 ( 4) 2 42 1 4 5 1 11 5 2 2 6 0
A C−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥− = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
− − − −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − − −⎣ ⎦ ⎣ ⎦
23. 1 0 6 1 6 0 6 0
6 6 2 4 6 2 6 4 12 241 2 6( 1) 6 2 6 12
A⋅ ⋅⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⋅ = ⋅ ⋅ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − ⋅ −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
24. 4 3 0
4 41 1 2
4 4 4( 3) 4 04 1 4 1 4( 2)
16 12 04 4 8
B−⎡ ⎤
− = − ⋅ ⎢ ⎥−⎣ ⎦− ⋅ − − − ⋅⎡ ⎤
= ⎢ ⎥− ⋅ − ⋅ − −⎣ ⎦−⎡ ⎤
= ⎢ ⎥− −⎣ ⎦
25. 1 0
4 3 02 4
1 1 21 2
1(4) 0(1) 1( 3) 0(1) 1(0) 0( 2)2(4) 4(1) 2( 3) 4(1) 2(0) 4( 2)1(4) 2(1) 1( 3) 2(1) 1(0) 2( 2)
4 3 012 2 82 5 4
AB⎡ ⎤
−⎡ ⎤⎢ ⎥= ⋅ ⎢ ⎥⎢ ⎥ −⎣ ⎦⎢ ⎥−⎣ ⎦+ − + + −⎡ ⎤
⎢ ⎥= + − + + −⎢ ⎥⎢ ⎥− + − − + − + −⎣ ⎦
−⎡ ⎤⎢ ⎥= − −⎢ ⎥⎢ ⎥− −⎣ ⎦
26. 1 0
4 3 02 4
1 1 21 2
4(1) 3(2) 0( 1) 4(0) 3(4) 0(2)1(1) 1(2) 2( 1) 1(0) 1(4) 2(2)
2 125 0
BA⎡ ⎤
−⎡ ⎤ ⎢ ⎥= ⋅⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎢ ⎥−⎣ ⎦− + − − +⎡ ⎤
= ⎢ ⎥+ − − + −⎣ ⎦− −⎡ ⎤
= ⎢ ⎥⎣ ⎦
27. 3 4
4 3 01 5
1 1 25 2
3(4) 4(1) 3( 3) 4(1) 3(0) 4( 2)1(4) 5(1) 1( 3) 5(1) 1(0) 5( 2)5(4) 2(1) 5( 3) 2(1) 5(0) 2( 2)
8 13 89 2 10
22 13 4
CB−⎡ ⎤
−⎡ ⎤⎢ ⎥= ⋅ ⎢ ⎥⎢ ⎥ −⎣ ⎦⎢ ⎥⎣ ⎦− − − − −⎡ ⎤
⎢ ⎥= + − + + −⎢ ⎥⎢ ⎥+ − + + −⎣ ⎦
−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥− −⎣ ⎦
28. 3 4
4 3 01 5
1 1 25 2
4(3) 3(1) 0(5) 4( 4) 3(5) 0(2)1(3) 1(1) 2(5) 1( 4) 1(5) 2(2)
9 316 3
BC−⎡ ⎤
−⎡ ⎤ ⎢ ⎥= ⋅⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎢ ⎥⎣ ⎦− + − − +⎡ ⎤
= ⎢ ⎥+ − − + −⎣ ⎦−⎡ ⎤
= ⎢ ⎥− −⎣ ⎦
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Chapter 8 Review Exercises
909 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
29. 4 61 3
A ⎡ ⎤= ⎢ ⎥
⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
( )
1 2
2 1 2
12 21 2 6
6 3
12
1 2 11 26 3
4 6 1 01 3 0 1
Interchange1 3 0 1
and 4 6 1 0
1 3 0 1 4
0 6 1 4
1 3 0 1
0 1
1 0 1 3
0 1
r r
R r r
R r
R r r
⎡ ⎤⎢ ⎥⎣ ⎦
⎛ ⎞⎡ ⎤→ ⎜ ⎟⎢ ⎥
⎣ ⎦ ⎝ ⎠⎡ ⎤
→ = − +⎢ ⎥− −⎣ ⎦⎡ ⎤
→ = −⎢ ⎥−⎣ ⎦⎡ ⎤−
→ = − +⎢ ⎥−⎢ ⎥⎣ ⎦
Thus, 1
1 21 26 3
1A− ⎡ ⎤−
= ⎢ ⎥−⎢ ⎥⎣ ⎦.
30. 3 21 2
A−⎡ ⎤
= ⎢ ⎥−⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
( )
1 2
2 1 2
12 24
1 2 1314 4
1 12 2
314 4
Interchange3 2 1 0
and 1 2 0 1
1 2 0 1 3
3 2 1 0
1 2 0 1
0 4 1 3
1 2 0 1 2
0 1
1 00 1
r r
R r r
R r
R r r
− ⎛ ⎞⎡ ⎤⎜ ⎟⎢ ⎥−⎣ ⎦ ⎝ ⎠
−⎡ ⎤→ = +⎢ ⎥−⎣ ⎦
−⎡ ⎤→ = −⎢ ⎥−⎣ ⎦
−⎡ ⎤→ = +⎢ ⎥− −⎣ ⎦
⎡ ⎤− −→ ⎢ ⎥− −⎢ ⎥⎣ ⎦
Thus, 1 1
1 2 231
4 4A− ⎡ ⎤− −
= ⎢ ⎥− −⎢ ⎥⎣ ⎦.
31. 1 3 31 2 11 1 2
A⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥−⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
2 1 2
3 1 3
2 2
1 2 1
3 2 3
1 3 3 1 0 01 2 1 0 1 01 1 2 0 0 1
1 3 3 1 0 00 1 2 1 1 0 0 4 1 1 0 1
1 3 3 1 0 00 1 2 1 1 0 0 4 1 1 0 1
1 0 3 2 3 03
0 1 2 1 1 0 4
0 0 7 3 4 1
R r rR r r
R r
R r rR r r
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦
⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− − −⎣ ⎦
⎡ ⎤⎢ ⎥→ − = −⎢ ⎥⎢ ⎥− − −⎣ ⎦
− −⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥−⎣ ⎦
( )13 37
3 4 17 7 7
5 9 37 7 7
1 3 11 1 27 7 7
2 3 23 4 17 7 7
1 0 3 2 3 00 1 2 1 1 0
0 0 1
1 0 03
0 1 0 2
0 0 1
R r
R r rR r r
⎡ ⎤− −⎢ ⎥
→ − =⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥ = +⎛ ⎞⎢ ⎥→ − ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠⎢ ⎥−⎣ ⎦
Thus,
5 9 37 7 7
1 1 1 27 7 73 4 17 7 7
A−
⎡ ⎤−⎢ ⎥
= −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
.
32. 3 1 23 2 11 1 1
A⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
1 3
2 1 2
3 1 3
3 1 2 1 0 03 2 1 0 1 01 1 1 0 0 1
1 1 1 0 0 1Interchange
3 2 1 0 1 0 and
3 1 2 1 0 0
1 1 1 0 0 13
0 1 4 0 1 3 3
0 2 1 1 0 3
r r
R r rR r r
⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦
⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− − −⎣ ⎦
( )2 2
1 1 1 0 0 10 1 4 0 1 3 0 2 1 1 0 3
R r⎡ ⎤⎢ ⎥→ − = −⎢ ⎥⎢ ⎥− − −⎣ ⎦
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Chapter 8: Systems of Equations and Inequalities
910 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
( )
1 2 1
3 2 3
13 37
31 27 7 7
3 517 7 7
1 3 194 17 7 7
2 3 231 27 7 7
1 0 3 0 1 20 1 4 0 1 3
20 0 7 1 2 3
1 0 3 0 1 20 1 4 0 1 3
0 0 1
1 0 03
0 1 0 4
0 0 1
R r rR r r
R r
R r rR r r
− −⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥−⎣ ⎦
⎡ ⎤− −⎢ ⎥
→ − =⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥ = +⎛ ⎞⎢ ⎥→ − ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠⎢ ⎥−⎣ ⎦
Thus,
3 517 7 7
1 94 17 7 7
31 27 7 7
A−
⎡ ⎤−⎢ ⎥
= −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
.
33. 4 81 2
A−⎡ ⎤
= ⎢ ⎥−⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
1 2
4 8 1 01 2 0 1
Interchange1 2 0 1
and 4 8 1 0 r r
−⎡ ⎤⎢ ⎥−⎣ ⎦
− ⎛ ⎞⎡ ⎤→ ⎜ ⎟⎢ ⎥−⎣ ⎦ ⎝ ⎠
( )
( )
2 1 2
1 1
1 2 0 1 4
0 0 1 4
1 2 0 1
0 0 1 4
R r r
R r
−⎡ ⎤→ = +⎢ ⎥
⎣ ⎦− −⎡ ⎤
→ = −⎢ ⎥⎣ ⎦
There is no inverse because there is no way to obtain the identity on the left side. The matrix is singular.
34. 3 16 2
A−⎡ ⎤
= ⎢ ⎥−⎣ ⎦
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
1 113 3
1 13
1 13 3
2 1 2
3 1 1 06 2 0 1
1 0
6 2 0 1
1 0 6
0 0 2 1
R r
R r r
−⎡ ⎤⎢ ⎥−⎣ ⎦
⎡ ⎤− −→ = −⎢ ⎥
−⎢ ⎥⎣ ⎦⎡ ⎤− −
→ = +⎢ ⎥−⎢ ⎥⎣ ⎦
There is no inverse because there is no way to obtain the identity on the left side. The matrix is singular.
35. 3 2 1
10 10 5x yx y
− =⎧⎨ + =⎩
Write the augmented matrix:
( )
( )
( )
2 1 2
1 2
2 1 2
12 2501
10
3 2 110 10 5
23 1 3161 2
Interchange161 2 and 23 1
161 2 30 50 5
161 2 0 1
R r r
r r
R r r
R r
−⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤−→ = − +⎢ ⎥⎣ ⎦
⎛ ⎞⎡ ⎤→ ⎜ ⎟⎢ ⎥
−⎣ ⎦ ⎝ ⎠⎡ ⎤
→ = − +⎢ ⎥− −⎣ ⎦
⎡ ⎤→ = −⎢ ⎥
⎢ ⎥⎣ ⎦
( )25
1 2 11
10
01 160 1
R r r⎡ ⎤⎢ ⎥→ = − +⎢ ⎥⎣ ⎦
The solution is 2 1,5 10
x y= = or 2 1,5 10
⎛ ⎞⎜ ⎟⎝ ⎠
.
36. 3 2 6
12
x y
x y
+ =⎧⎪⎨
− = −⎪⎩
Write the augmented matrix:
12
3 2 61 1
⎡ ⎤⎢ ⎥− −⎣ ⎦
( )
( )
( )
12
1 2
12
2 1 2152
112
2 23 52
1 2 132
Interchange1 1
and 3 2 6
1 1 3
0 5
1 1
0 1
1 0 1
0 1
r r
R r r
R r
R r r
⎡ ⎤− − ⎛ ⎞→ ⎢ ⎥ ⎜ ⎟
⎝ ⎠⎣ ⎦⎡ ⎤− −
→ = − +⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤− −
→ =⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
→ = +⎢ ⎥⎣ ⎦
The solution is 31,2
x y= = or 31,2
⎛ ⎞⎜ ⎟⎝ ⎠
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8 Review Exercises
911 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
37. 5 6 3 64 7 2 33 7 1
x y zx y zx y z
− − =⎧⎪ − − = −⎨⎪ + − =⎩
Write the augmented matrix:
( )1 2 1
2 1 2
3 1 3
1392 2 211
11 111
3 32
5 6 3 627 34
3 71 1
91 1 127 34
3 71 1
91 1 14
0 39 11 23
4 260 2
91 1 1
0 10 131 2
R r r
R r rR r r
R r
R r
⎡ ⎤− −⎢ ⎥
−− −⎢ ⎥⎢ ⎥−⎣ ⎦
⎡ ⎤−⎢ ⎥
−− −→ = − +⎢ ⎥⎢ ⎥−⎣ ⎦⎡ ⎤−
= − +⎢ ⎥ ⎛ ⎞→ −− ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −−⎣ ⎦
⎡ ⎤−⎢ ⎥ ⎛ ⎞= −− ⎜ ⎟⎢ ⎥→
⎜ = −⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦
⎟
9 6011 11
1 2 139211 11
3 2 31042411 11
1 00 1 0 0
R r rR r r
⎡ ⎤−= − +⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥
⎣ ⎦
( )9 6011 11
392 113 311 11 24
133
91 3 11113
3 22 3 213 11
3
1 00 1 0 0 1
1 0 0 90 1 0
0 0 1
R r
R r r
R r r
⎡ ⎤−⎢ ⎥→ − =⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
⎛ ⎞= +⎢ ⎥⎜ ⎟→ ⎢ ⎥ ⎜ ⎟= +⎢ ⎥ ⎝ ⎠⎢ ⎥⎣ ⎦
The solution is 13 139, ,3 3
x y z= = = or
13 139, ,3 3
⎛ ⎞⎜ ⎟⎝ ⎠
.
38. 2 54 3 18 5
x y zx y zx y z
+ + =⎧⎪ − − =⎨⎪ + − =⎩
Write the augmented matrix: 52 1 1
34 1 18 51 1
⎡ ⎤⎢ ⎥
−−⎢ ⎥⎢ ⎥−⎣ ⎦
2 1 2
3 1 3
51 112 2 2
1 1253 1
2 23
13 1
1 2 15 23
3 2 3
52 1 12
0 3 5 9 4
0 3 5 15
10 1 3
0 3 5 15
1 0 1
0 1 3 3
0 0 0 6
R r rR r r
R r
R r
R r r
R r r
⎡ ⎤= − +⎢ ⎥ ⎛ ⎞
→ − − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− − −⎣ ⎦⎡ ⎤ ⎛ ⎞=⎢ ⎥ ⎜ ⎟→ ⎢ ⎥ ⎜ ⎟= −⎢ ⎥ ⎝ ⎠− − −⎣ ⎦⎡ ⎤−
⎛ ⎞⎢ ⎥ = − +→ ⎜ ⎟⎢ ⎥ ⎜ ⎟= +⎢ ⎥ ⎝ ⎠−⎣ ⎦
There is no solution; the system is inconsistent.
39. 2 1
2 3 34 3 4 3
x zx yx y z
− =⎧⎪ + = −⎨⎪ − − =⎩
Write the augmented matrix:
2 1 2
3 1 3
1 0 2 12 3 0 34 3 4 3
1 0 2 12
0 3 4 5 4
0 3 4 1
R r rR r r
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− −⎣ ⎦
−⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −⎣ ⎦
( )54 12 23 3 3
1 0 2 10 1
0 3 4 1
R r⎡ ⎤−⎢ ⎥→ − =⎢ ⎥⎢ ⎥− −⎣ ⎦
( )543 2 33 3
1 0 2 10 1 3
0 0 8 6
R r r⎡ ⎤−⎢ ⎥→ − = +⎢ ⎥⎢ ⎥−⎣ ⎦
( )54 13 33 3 8
34
12
1 3 123 4
2 3 2334
1 0 2 10 1
0 0 1
1 0 0 20 1 0
0 0 1
R r
R r r
R r r
⎡ ⎤−⎢ ⎥
→ − =⎢ ⎥⎢ ⎥−⎣ ⎦⎡ ⎤− = +⎛ ⎞⎢ ⎥
→ − ⎜ ⎟⎢ ⎥ ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠−⎣ ⎦
The solution is 1 2 3, ,2 3 4
x y z= − = − = − or
1 2 3, ,2 3 4
⎛ ⎞− − −⎜ ⎟⎝ ⎠
.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8: Systems of Equations and Inequalities
912 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
40. 2 2
2 2 16 4 3 5
x y zx y zx y z
+ − =⎧⎪ − + = −⎨⎪ + + =⎩
Write the augmented matrix: 1 2 1 22 2 1 16 4 3 5
−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥⎣ ⎦
( )
2 1 2
3 1 3
51 12 22 6 6
13
1 2 1512 6
3 2 313
1351
2 61
15
1 2 1 22
0 6 3 5 6
0 8 9 7
1 2 1 20 1
0 8 9 7
1 0 02
0 1 8
0 0 5
1 0 0
0 1
0 0 1
R r rR r r
R r
R r rR r r
R
−⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −⎣ ⎦
⎡ ⎤−⎢ ⎥→ − = −⎢ ⎥⎢ ⎥− −⎣ ⎦⎡ ⎤⎢ ⎥ = − +⎛ ⎞
→ −⎢ ⎥ ⎜ ⎟= +⎝ ⎠⎢ ⎥−⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥
→ −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
( )
( )
13 35
134 1
2 3 25 21
15
1 0 0
0 1 0
0 0 1
r
R r r
=
⎡ ⎤⎢ ⎥
→ = +⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
The solution is 1 4 1, ,3 5 15
x y z= = = − or
1 4 1, ,3 5 15
⎛ ⎞−⎜ ⎟⎝ ⎠
.
41. 0 0
5 6 0 5 62 2 1 0 2 2 1
x y z x y zx y z x y zx y z x y z
− + = − + =⎧ ⎧⎪ ⎪− − − = → − − =⎨ ⎨⎪ ⎪− + − = − + =⎩ ⎩
Write the augmented matrix: 1 1 1 01 1 5 62 2 1 1
−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥−⎣ ⎦
2 1 2
3 1 3
1 1 1 00 0 6 6
20 0 1 1
R r rR r r
−⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦
( )12 26
1 1 1 00 0 1 1 0 0 1 1
R r−⎡ ⎤
⎢ ⎥→ − = −⎢ ⎥⎢ ⎥−⎣ ⎦
1 2 1
3 2 3
1 1 0 10 0 1 1 0 0 0 0
R r rR r r
−⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥⎣ ⎦
The system is dependent. 1
1x yz
= +⎧⎨ = −⎩
The solution is 1x y= + , 1z = − , y is any real
number or {( , , ) 1, 1, is anyx y z x y z y= + = −
}real number .
42. 4 3 5 02 4 3 06 2 0
x y zx y zx y z
− + =⎧⎪ + − =⎨⎪ + + =⎩
Write the augmented matrix:
( )3 54 4
11 14
4 3 5 0 1 02 4 3 0 2 4 3 0 6 2 1 0 6 2 1 0
R r⎡ ⎤− −⎡ ⎤⎢ ⎥⎢ ⎥− → − =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
3 54 4
2 1 211 112 2
3 1 313 132 2
3 524 4
2 2112
3 313
132
1 2 14
3 2 3
1 02
0 0 6
0 0
1 00 1 1 0 0 1 1 0
1 0 00 1 1 0 0 0 0 0
R r rR r r
R r
R r
R r r
R r r
⎡ ⎤−= − +⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦
⎡ ⎤− ⎛ ⎞=⎢ ⎥ ⎜ ⎟→ −⎢ ⎥ ⎜ ⎟=⎢ ⎥ ⎝ ⎠−⎣ ⎦⎡ ⎤
⎛ ⎞= +⎢ ⎥→ − ⎜ ⎟⎢ ⎥ ⎜ ⎟= − +⎝ ⎠⎢ ⎥⎣ ⎦
The solution is 12
x z= − , y z= , z is any real
number or 1( , , ) , , is any real number2
x y z x z y z z⎧ ⎫= − =⎨ ⎬
⎩ ⎭.
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Chapter 8 Review Exercises
913 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
43.
12 2 3
2 2 3 03 4 5 3
x y z tx y z tx y z tx y z t
− − − =⎧⎪ + − + =⎪⎨ − − − =⎪⎪ − + + = −⎩
Write the augmented matrix: 1 1 1 1 12 1 1 2 31 2 2 3 03 4 1 5 3
− − −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− − −⎢ ⎥
− −⎣ ⎦
2 1 2
3 1 3
4 1 4
2 3
2
1 1 1 1 12
0 3 1 4 1
0 1 1 2 13
0 1 4 8 6
1 1 1 1 1Interchange0 1 1 2 1
and 0 3 1 4 1
0 1 4 8 6
1 1 1 1 10 1 1 2 1
0 3 1 4 10 1 4 8 6
R r rR r rR r r
r r
R
− − −⎡ ⎤= − +⎛ ⎞⎢ ⎥
⎜ ⎟⎢ ⎥→ = − +⎜ ⎟⎢ ⎥− − − − ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠− −⎣ ⎦− − −⎡ ⎤
⎢ ⎥− − − − ⎛ ⎞⎢ ⎥→ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥
− −⎣ ⎦− − −⎡ ⎤
⎢ ⎥⎢ ⎥→ = −⎢ ⎥⎢ ⎥
− −⎣ ⎦
( )2r
1 2 1
3 2 3
4 2 4
13 32
14 45
2 3 2
4
1 0 0 1 20 1 1 2 1
30 0 2 2 20 0 5 10 5
1 0 0 1 20 1 1 2 1
0 0 1 1 10 0 1 2 1
1 0 0 1 2=0 1 0 1 0
0 0 1 1 1 =0 0 0 1 2
R r rR r rR r r
R r
R r
R r rR r
⎡ ⎤= +⎛ ⎞⎢ ⎥
⎜ ⎟⎢ ⎥→ = − +⎜ ⎟⎢ ⎥− − − ⎜ ⎟= +⎢ ⎥ ⎝ ⎠−⎣ ⎦⎡ ⎤⎢ ⎥ ⎛ ⎞= −⎢ ⎥ ⎜ ⎟→
⎜ ⎟⎢ ⎥ =⎝ ⎠⎢ ⎥−⎣ ⎦
⎡ ⎤⎢ ⎥ − +⎢ ⎥→⎢ ⎥ −⎢ ⎥
−⎣ ⎦3 4r
⎛ ⎞⎜ ⎟+⎝ ⎠
1 4 1
2 4 2
3 4 3
1 0 0 0 40 1 0 0 2
0 0 1 0 30 0 0 1 2
R r rR r rR r r
⎡ ⎤= − +⎛ ⎞⎢ ⎥
⎜ ⎟⎢ ⎥→ = − +⎜ ⎟⎢ ⎥ ⎜ ⎟= − +⎢ ⎥ ⎝ ⎠−⎣ ⎦
The solution is 4, 2, 3, 2x y z t= = = = − or (4, 2, 3, 2)− .
44.
3 3 42 3
3 2 35 6
x y z tx y zx z tx y z
− + − =⎧⎪ + − = −⎪⎨ + + =⎪
+ + =⎪⎩
Write the augmented matrix: 1 3 3 1 41 2 1 0 31 0 3 2 31 1 5 0 6
− −⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
( )
2 1 2
3 1 3
4 1 4
74 115 5 5
2 25
3 2 15 5 5
74 1 15 5 5
1612 125 5 5
26 3815 5 5
1 3 3 1 40 5 4 1 7
0 3 0 3 10 4 2 1 21 3 3 1 40 1
0 3 0 3 10 4 2 1 2
1 0
0 1
0 0
0 0
R r rR r rR r r
R r
R
− −⎡ ⎤ = − +⎛ ⎞⎢ ⎥− − ⎜ ⎟⎢ ⎥→ = − +⎜ ⎟−⎢ ⎥ ⎜ ⎟= − +⎝ ⎠⎢ ⎥⎣ ⎦− −⎡ ⎤
⎢ ⎥− −⎢ ⎥→ =⎢ ⎥−⎢ ⎥⎣ ⎦⎡ ⎤− −⎢ ⎥ =− −⎢ ⎥
→ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
2 1
3 2 3
4 2 4
334
r rR r rR r r
+⎛ ⎞⎜ ⎟= − +⎜ ⎟⎜ ⎟= − +⎝ ⎠
( )
3 2 15 5 5
74 15 5 5 5
3 31243
26 3815 5 5
31 3 11 5
3 42 3 24 5
3 264 3 42 5
3
13432
15
1 0
0 1
0 0 1 1
0 0
1 0 0 1 10 1 0 1
0 0 1 1
0 0 0 5
1 0 0 1 10 1 0 1
0 0 1 1
0 0 0 1
R r
R r r
R r r
R r r
⎡ ⎤− −⎢ ⎥
− −⎢ ⎥→ =⎢ ⎥
⎢ ⎥⎢ ⎥⎣ ⎦
− −⎡ ⎤⎛ ⎞= − +⎢ ⎥− ⎜ ⎟⎢ ⎥→ = +⎜ ⎟⎢ ⎥ ⎜ ⎟⎢ ⎥ ⎜ ⎟= − +⎝ ⎠−⎢ ⎥⎣ ⎦
− −⎡ ⎤⎢ ⎥−⎢ ⎥→ ⎢ ⎥⎢ ⎥
−⎢ ⎥⎣ ⎦
( )14 45 R r= −
1715
1 1 4 15
2 4 22215 3 4 32
15
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
R r rR r rR r r
⎡ ⎤−⎢ ⎥ = +⎛ ⎞−⎢ ⎥ ⎜ ⎟→ = − +⎢ ⎥ ⎜ ⎟⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−⎣ ⎦
The solution is 17 1 22 215 5 15 15
, , ,x y z t= − = − = = −
or 17 1 22 2, , ,15 5 15 15
⎛ ⎞− − −⎜ ⎟⎝ ⎠
.
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Chapter 8: Systems of Equations and Inequalities
914 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
45. 3 4
3(3) 4(1) 9 4 51 3
= − = − =
46. 4 0
4(3) 1(0) 12 0 121 3
−= − − = − − = −
47. 01 4
6 62 1 1 261 2 1 4 03 31 4 4 134 1
1(6 6) 4( 3 24) 0( 1 8)1(0) 4( 27) 0( 9) 0 108 0108
− −− = − +
= − − − − + − −= − − + − = + +=
48. 3 102
5 0 5 01 10 5 2 3 1013 32 1 1 231 2
2(3 10) 3(0 5) 10(0 1)2( 7) 3(5) 10(1)
14 15 1019
= − +− −−
= − − + + += − − += − − += −
49. 32 1
0 5 5 01 15 0 2 1 ( 3)16 0 0 62 26 02
2(0 6) 1(0 2) 3(30 0)2( 6) 1( 2) 3(30)
12 2 90100
−= − + −
= − − − − −= − − − −= − + −= −
50. 012
3 32 1 1 231 2 2 1 04 2 1 2 1 41 4 2
2(4 12) 1(2 3) 0(4 2)2( 8) 1(5) 0(6)
16 5 011
−= − − +
− −−
= − − − + + += − − − += − +=
51. 2 4
3 2 4x yx y
− =⎧⎨ + =⎩
Set up and evaluate the determinants to use Cramer’s Rule:
1 23 2
1(2) 3( 2) 2 6 8D −= = − − = + =
4 24 2
4(2) 4( 2) 8 8 16xD −= = − − = + =
1 43 4
1(4) 3(4) 4 12 8yD = = − = − = −
The solution is 168
2xDD
x = = = , 88
1yDD
y −= = = −
or (2, 1)− .
52. 3 5
2 3 5x yx y
− = −⎧⎨ + =⎩
Set up and evaluate the determinants to use Cramer’s Rule:
1 31(3) 2( 3) 3 6 9
2 3D
−= = − − = + =
5 35(3) 5( 3) 15 15 0
5 3xD− −
= = − − − = − + =
1 51(5) 2( 5) 5 10 15
2 5yD−
= = − − = + =
The solution is 09
0xDD
x = = = , 15 59 3
yDD
y = = =
or 53
0,⎛ ⎞⎜ ⎟⎝ ⎠
.
53. 2 3 13 0
3 2 0x y
x y+ − =⎧
⎨ − =⎩
Write the system is standard form: 2 3 133 2 0
x yx y
+ =⎧⎨ − =⎩
Set up and evaluate the determinants to use Cramer’s Rule:
2 34 9 13
3 2D = = − − = −
−
13 326 0 26
0 2xD = = − − = −−
132 0 39 393 0
yD = = − = −
The solution is 2613
2xDD
x −−
= = = ,
3913
3yDD
y −−
= = = or (2, 3).
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 8 Review Exercises
915 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
54. 3 4 12 05 2 6 0x yx y− − =⎧
⎨ + + =⎩
Write the system in standard form: 3 4 125 2 6
x yx y
− =⎧⎨ + = −⎩
Set up and evaluate the determinants to use Cramer's Rule:
3 46 20 26
5 2D
−= = + =
12 424 24 0
6 2xD−
= = − =−
3 12 18 60 785 6
yD = = − − = −−
The solution is 0 026
xDxD
= = = ,
78 326
yDy
D−
= = = − or (0, 3)− .
55. 2 6
2 3 133 2 3 16
x y zx y zx y z
+ − =⎧⎪ − + = −⎨⎪ − + = −⎩
Set up and evaluate the determinants to use Cramer’s Rule:
( )
1 2 12 1 33 2 3
1 3 1 3 2 11 2 ( 1)
2 3 2 3 3 21 3 6 2( 3 6) ( 1)( 4 3)3 6 1 10
D−
= −−
− − −= − + −
− − −
= − + − − + + − − +
= + + =
( )
6 2 113 1 316 2 3
1 3 13 3 13 16 2 ( 1)
2 3 16 3 16 26 3 6 2( 39 48) ( 1)(26 16)18 18 10 10
xD−
= − −− −
− − − −= − + −
− − − −
= − + − − + + − −
= − − = −
( )
1 6 12 13 33 16 3
13 3 2 3 2 131 6 ( 1)
16 3 3 3 3 161 39 48 6(6 9) ( 1)( 32 39)9 18 7 20
yD−
= −−
− −= − + −
− −
= − + − − + − − +
= + − =
( )
1 2 62 1 133 2 16
1 13 2 13 2 11 2 6
2 16 3 16 3 21 16 26 2( 32 39) 6( 4 3)
10 14 6 30
zD = − −− −
− − − −= − +
− − − −
= − − − + + − +
= − − − = −
The solution is 10 110
xDxD
−= = = − ,
20 210
yDy
D= = = , 30 3
10zD
zD
−= = = − or
( 1, 2, 3)− − .
56. 8
2 3 23 9 9
x y zx y zx y z
− + =⎧⎪ + − = −⎨⎪ − − =⎩
Set up and evaluate the determinants to use Cramer’s Rule:
1 1 12 3 13 1 9
3 1 2 32 11 ( 1) 11 9 3 13 9
1( 27 1) 1( 18 3) 1( 2 9)28 15 1154
D−
= −− −
− −= − − +− − −−
= − − + − + + − −= − − −= −
8 1 12 3 19 1 9
3 1 2 3128 ( 1) 11 9 9 19 9
8( 27 1) 1(18 9) 1(2 27)224 27 25222
xD−
= − −− −
− −−−= − − +− − −−
= − − + + + −= − + −= −
1 8 12 2 13 9 9
1 2 1 22 21 8 19 9 3 9 3 9
1(18 9) 8( 18 3) 1(18 6)27 120 24171
yD = − −−
− −− −= − +− −
= + − − + + += + +=
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Chapter 8: Systems of Equations and Inequalities
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1 1 82 3 23 1 9
3 2 2 32 21 ( 1) 81 9 3 13 9
1(27 2) 1(18 6) 8( 2 9)25 24 88
39
zD−
= −−
− −= − − +− −
= − + + + − −= + −= −
The solution is 222 3754 9
xDx
D−
= = =−
,
171 1954 6
yDy
D= = = −
−, 39 13
54 18zD
zD
−= = =
− or
37 19 13, ,9 6 18
⎛ ⎞−⎜ ⎟⎝ ⎠
.
57. Let 8x ya b
= .
Then ( )2
2 8 162
x ya b
= = by Theorem (14).
The value of the determinant is multiplied by k when the elements of a column are multiplied by k.
58. Let 8x ya b
= .
Then 8y xb a
= − by Theorem (11). The
value of the determinant changes sign when any 2 columns are interchanged.
59. Find the partial fraction decomposition:
6( 4) ( 4)
( 4) 46 ( 4)
A Bx x x xx x x x
A x Bx
⎛ ⎞ ⎛ ⎞− = − +⎜ ⎟ ⎜ ⎟− −⎝ ⎠⎝ ⎠= − +
Let x =4, then 6 (4 4) (4)4 6
32
A BB
B
= − +=
=
Let x = 0, then 6 (0 4) (0)4 6
32
A BA
A
= − +− =
= −
3 36 2 2
( 4) 4x x x x
−= +
− −
60. Find the partial fraction decomposition:
( 2)( 3) 2 3x A B
x x x x= +
+ − + −
Multiply both sides by ( 2)( 3)x x+ − . ( 3) ( 2)x A x B x= − + +
Let 2x = − , then 2 ( 2 3) ( 2 2)5 2
25
A BA
A
− = − − + − +− = −
=
Let 3x = , then 3 (3 3) (3 2)5 3
35
A BB
B
= − + +=
=
2 35 5
( 2)( 3) 2 3x
x x x x= +
+ − + −
61. Find the partial fraction decomposition:
2 24
1( 1)x A B C
x xx x x−
= + +−−
Multiply both sides by 2 ( 1)x x − 24 ( 1) ( 1)x Ax x B x Cx− = − + − +
Let 1x = , then 21 4 (1)(1 1) (1 1) (1)
33
A B CC
C
− = − + − +− =
= −
Let 0x = , then 20 4 (0)(0 1) (0 1) (0)
44
A B CB
B
− = − + − +− = −
=
Let 2x = , then 22 4 (2)(2 1) (2 1) (2)
2 2 42 2 4 4( 3)2 6
3
A B CA B C
AAA
− = − + − +− = + +
= − − − −==
2 24 3 4 3
1( 1)x
x xx x x− −
= + +−−
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Chapter 8 Review Exercises
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62. Find the partial fraction decomposition:
2 22 6
2 1( 2) ( 1) ( 2)x A B C
x xx x x−
= + +− −− − −
Multiply both sides by 2( 2) ( 1)x x− − . 22 6 ( 2)( 1) ( 1) ( 2)x A x x B x C x− = − − + − + −
Let 1x = , then 22(1) 6 (1 2)(1 1) (1 1) (1 2)
44
A B CC
C
− = − − + − + −− =
= −
Let 2x = , then 22(2) 6 (2 2)(2 1) (2 1) (2 2)
2A B CB
− = − − + − + −− =
Let 0x = , then 22(0) 6 (0 2)(0 1) (0 1) (0 2)
6 2 46 2 ( 2) 4( 4)
2 84
A B CA B CA
AA
− = − − + − + −− = − +− = − − + −
==
2 22 6 4 2 4
2 1( 2) ( 1) ( 2)x
x xx x x− − −
= + +− −− − −
63. Find the partial fraction decomposition:
2 21( 9)( 1) 9x A Bx C
xx x x+
= +++ + +
Multiply both sides by 2( 1)( 9)x x+ + . 2( 9) ( )( 1)x A x Bx C x= + + + +
Let 1x = − , then
( )( ) ( )( )( )21 1 9 1 1 1
1 (10) ( )(0)1 10
110
A B C
A B CA
A
− = − + + − + − +
− = + − +− =
= −
Let 0x = , then
( ) ( )( )( )20 0 9 0 0 1
0 910 9
109
10
A B C
A C
C
C
= + + + +
= +
⎛ ⎞= − +⎜ ⎟⎝ ⎠
=
Let 1x = , then ( ) ( )( )( )21 1 9 1 1 1
1 (10) ( )(2)1 10 2 2
1 91 10 2 210 10
91 1 25
A B C
A B CA B C
B
B
= + + + +
= + += + +
⎛ ⎞ ⎛ ⎞= − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= − + +
1251
10
B
B
=
=
2 2
1 1 910 10 10
1( 9)( 1) 9
xxxx x x
− += +
++ + +
64. Find the partial fraction decomposition:
2 23
2( 2)( 1) 1x A Bx C
xx x x+
= +−− + +
Multiply both sides by 2( 2)( 1)x x− + .
23 ( 1) ( )( 2)x A x Bx C x= + + + − Let 2x = , then
23(2) ((2) 1) ( (2) )(2 2)6 5
65
A B CA
A
= + + + −=
=
Let 0x = , then 23(0) (0 1) ( (0) )(0 2)
0= 260 2562535
A B CA C
C
C
C
= + + + −−
= −
=
=
Let 1x = , then 23(1) (1 1) ( (1) )(1 2)
3 26 33 25 5
65
A B CA B C
B
B
= + + + −= − −
⎛ ⎞= − −⎜ ⎟⎝ ⎠
= −
2 2
6 6 33 5 5 5
2( 2)( 1) 1
xxxx x x
− += +
−− + +
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Chapter 8: Systems of Equations and Inequalities
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65. Find the partial fraction decomposition: 3
2 2 2 2 2( 4) 4 ( 4)x Ax B Cx D
x x x+ +
= ++ + +
Multiply both sides by 2 2( 4)x + . 3 2
3 3 2
3 3 2
( )( 4)
4 4
(4 ) 4
x Ax B x Cx D
x Ax Bx Ax B Cx D
x Ax Bx A C x B D
= + + + +
= + + + + +
= + + + + +
1; 0A B= = 4 0
4(1) 04
A CCC
+ =+ =
= −
4 04(0) 0
0
B DDD
+ =+ =
=
3
2 2 2 2 24
( 4) 4 ( 4)x x x
x x x−
= ++ + +
66. Find the partial fraction decomposition: 3
2 2 2 2 21
( 16) 16 ( 16)x Ax B Cx D
x x x+ + +
= ++ + +
Multiply both sides by 2 2( 16)x + . 3 2
3 3 2
3 3 2
1 ( )( 16)
1 16 16
1 (16 ) 16
x Ax B x Cx D
x Ax Bx Ax B Cx D
x Ax Bx A C x B D
+ = + + + +
+ = + + + + +
+ = + + + + +
1; 0A B= = 16 0
16(1) 016
A CCC
+ =+ =
= −
16 116(0) 1
1
B DDD
+ =+ =
=
3
2 2 2 2 21 16 1
( 16) 16 ( 16)x x x
x x x+ − +
= ++ + +
67. Find the partial fraction decomposition: 2 2
2 2 2
2
( 1)( 1) ( 1)( 1)( 1)
1 1 1
x xx x x x x
A B Cx Dx x x
=+ − + − +
+= + +
− + +
Multiply both sides by 2( 1)( 1)( 1)x x x− + + . 2 2 2( 1)( 1) ( 1)( 1)
( )( 1)( 1)x A x x B x x
Cx D x x= + + + − +
+ + − +
Let 1x = , then 2 2 21 (1 1)(1 1) (1 1)(1 1)
( (1) )(1 1)(1 1)1 4
14
A BC D
A
A
= + + + − ++ + − +
=
=
Let 1x = − , then 2 2
2
( 1) ( 1 1)(( 1) 1)
( 1 1)(( 1) 1) ( ( 1) )( 1 1)( 1 1)
1 414
A
BC D
B
B
− = − + − +
+ − − − ++ − + − − − +
= −
= −
Let 0x = , then 2 2 20 (0 1)(0 1) (0 1)(0 1)
( (0) )(0 1)(0 1)0
1 104 412
A BC D
A B D
D
D
= + + + − ++ + − +
= − −
⎛ ⎞= − − −⎜ ⎟⎝ ⎠
=
Let 2x = , then 2 2 22 (2 1)(2 1) (2 1)(2 1)
( (2) )(2 1)(2 1)4 15 5 6 3
1 1 14 15 5 6 34 4 215 5 36 44 4 2
6 00
A BC D
A B C D
C
C
CC
= + + + − ++ + − +
= + + +
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + −
==
( )( )2
22 2
1 1 14 4 2
1 1 11 1x
x x xx x
−= + +
− + ++ −
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Chapter 8 Review Exercises
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68. Find the partial fraction decomposition:
( )( ) ( )2 2 2
2
4 44 1 4 ( 1)( 1)
1 1 4
x x x x xA B Cx D
x x x
=+ − + − +
+= + +
− + +
Multiply both sides by ( )2 4 ( 1)( 1)x x x+ − + .
( ) ( )2 24 ( 1) 4 ( 1) 4 ( )( 1)( 1)
A x x B x xCx D x x
= + + + − ++ + − +
Let 1x = , then ( ) ( )
( )
2 24 (1 1) 1 4 (1 1) 1 4 (1) (1 1)(1 1)4 10
4 210 5
A BC D
A
A
= + + + − +
+ + − +
=
= =
Let 1x = − , then ( ) ( )
( )
2 24 ( 1 1) ( 1) 4 ( 1 1) ( 1) 4 ( 1) ( 1 1)( 1 1)4 10
4 210 5
A BC D
B
B
= − + − + + − − − +
+ − + − − − +
= −
= − = −
Let 0x = , then ( ) ( )
( )
2 24 (0 1) 0 4 (0 1) 0 4 (0) (0 1)(0 1)4 4 4
2 24 4 45 5
8 845 5
45
A BC D
A B D
D
D
D
= + + + − +
+ + − +
= − −
⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= + −
= −
Let 2x = , then ( ) ( )
( )
2 24 (2 1) 2 4 (2 1) 2 4 (2) (2 1)(2 1)
4 24 8 6 32 2 44 24 8 6 35 5 5
48 16 124 65 5 5
6 00
A BC D
A B C D
C
C
CC
= + + + − +
+ + − +
= + + +
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + −
==
( )( ) 22 2
2 2 45 5 54
1 1 44 1 x x xx x
− −= + +
− + ++ −
69. Solve the first equation for y, substitute into the second equation and solve:
2 2
2 3 0 2 3
5
x y y x
x y
+ + = → = − −⎧⎪⎨
+ =⎪⎩
2 2 2 2
2
( 2 3) 5 4 12 9 5
5 12 4 0 (5 2)( 2) 02 or 2511 15
x x x x x
x x x x
x x
y y
+ − − = → + + + =
+ + = → + + =
⇒ = − = −
= − =
Solutions: 2 11, , ( 2, 1)5 5
⎛ ⎞− − −⎜ ⎟⎝ ⎠
.
70. Add the equations to eliminate y, and solve: 2 2
2
2
16
2 8
2 8
x y
x y
x x
⎧ + =⎪⎨
− = −⎪⎩
+ =
2 2 8 0 ( 4)( 2) 04 or 2
x x x xx x
+ − = → + − == − =
2 2 2
2 2 2
If 4 :
( 4) 16 0 0
If 2 :
(2) 16 12 2 3
x
y y y
x
y y y
= −
− + = → = → =
=
+ = → = → = ±
Solutions: ( ) ( ) ( )– 4, 0 , 2, 2 3 , 2, 2 3− .
71. Multiply each side of the second equation by 2 and add the equations to eliminate xy:
2 2
22 2
2
2
2 10 2 10
3 2 2 6 4
7 14
2
2
xy y xy y
xy y xy y
y
y
y
⎧ + = ⎯⎯→ + =⎪⎨
− + = ⎯⎯→ − + =⎪⎩
=
=
= ±
( ) ( )
( ) ( )
2
2
If 2 :
2 2 2 10 2 2 8 2 2
If 2 :
2 2 2 10 2 2 8
2 2
y
x x x
y
x x
x
=
+ = → = → =
= −
− + − = → − =
→ = −
Solutions: ( ) ( )2 2, 2 , 2 2, 2− −
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Chapter 8: Systems of Equations and Inequalities
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72. Multiply each side of the first equation by –2 and add the equations to eliminate y:
22 2 2 2
2 2 2 2
2
3 1 6 2 2
7 2 5 0 7 2 5
3
3
x y x y
x y x y
x
x
−⎧ − = ⎯⎯→ − + = −⎪⎨
− − = ⎯⎯→ − =⎪⎩
=
= ±
( )
( )
2 2 2
2 2 2
If 3 :
3 3 1 8
8 2 2
If 3 :
3 3 1 8
8 2 2
x
y y
y
x
y y
y
=
− = → − = −
→ = ± = ±
= −
− − = → − = −
→ = ± = ±
Solutions:
( ) ( ) ( )( )
3, 2 2 , 3, 2 2 , 3, 2 2 ,
3, 2 2
− − − −
73. Substitute into the second equation into the first equation and solve:
2 2
2
6
3
x y y
x y
⎧ + =⎪⎨
=⎪⎩
2
2
3 6
3 0( 3) 0 0 or 3
y y y
y yy y y y
+ =
− =− = → = =
2 2
2 2
If 0 : 3(0) 0 0
If 3 : 3(3) 9 3
y x x x
y x x x
= = → = → =
= = → = → = ±
Solutions: (0, 0), (–3, 3), (3, 3).
74. Multiply each side of the second equation by –1 and add the equations to eliminate y:
2 2 2 2
12 2 2 2
2
2 9 2 9
9 9
0 0
x y x y
x y x y
x x
−
⎧ + = ⎯⎯→ + =⎪⎨
+ = ⎯⎯→ − − = −⎪⎩
= ⇒ =
2 2 2If 0 : 0 9 9 3x y y y= + = → = → = ± Solutions: (0, 3), (0, –3)
75. Factor the second equation, solve for x, substitute into the first equation and solve:
2 2
2 2
3 4 5 8
3 2 0
x xy y
x xy y
⎧ + + =⎪⎨
+ + =⎪⎩
2 23 2 0( 2 )( ) 0 2 or x xy yx y x y x y x y
+ + =+ + = → = − = −
Substitute 2x y= − and solve: 2 2
2 2
2 2 2
2
2
3 4 5 8
3( 2 ) 4( 2 ) 5 8
12 8 5 8
9 8
8 2 29 3
x xy y
y y y y
y y y
y
y y
+ + =
− + − + =
− + =
=
= ⇒ = ±
Substitute x y= − and solve:
2 2
2 2
2 2 2
2
2
3 4 5 8
3( ) 4( ) 5 8
3 4 5 8
4 8
2 2
x xy y
y y y y
y y y
y
y y
+ + =
− + − + =
− + =
=
= ⇒ = ±
2 2 2 2 4 2If : 2
3 3 3
2 2 2 2 4 2If : 23 3 3
y x
y x
⎛ ⎞ −= = − =⎜ ⎟⎜ ⎟
⎝ ⎠⎛ ⎞− −
= = − =⎜ ⎟⎜ ⎟⎝ ⎠
If 2 : 2
If 2 : 2
y x
y x
= = −
= − =
Solutions:
( )
( )
4 2 2 2 4 2 2 2, , , , 2, 2 ,3 3 3 3
2, 2
⎛ ⎞ ⎛ ⎞− −−⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
−
76. 2 2
2
3 2 2 6
2 4 0
x xy y
xy y
⎧ + − =⎪⎨
− + =⎪⎩
→ 2 2
2
3 2 2 6
2 4
x xy y
xy y
⎧ + − =⎪⎨
− = −⎪⎩
Multiply each side of the first equation by 2 and each side of the second equation by 3 and add to eliminate the constant:
2 2
2
2 2
6 4 4 12
3 6 12
6 7 10 0
x xy y
xy y
x xy y
+ − =
− = −
+ − =
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Chapter 8 Review Exercises
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( )( )6 5 2 05 or 26
x y x y
x y x y
− + =
= = −
Substitute 56x y= and solve:
2
2 2
5 2 46
7 24 2 4246 7 7
y y y
y y y
⋅ − = −
− = − → = → = ±
Substitute 2x y= − and solve:
2
2
2
2 2 4
4 4
11
y y y
y
yy
− ⋅ − = −
− = −
== ±
2 42 5 2 42 5 42If :
7 6 7 212 42 5 2 42 5 42If :
7 6 7 21
y x
y x
= = ⋅ =
−= − = ⋅ = −
If 1: 2; If 1: 2y x y x= = − = − = Solutions:
( )
( )
5 42 2 42 5 42 2 42, , , , 2,1 ,21 7 21 7
2, 1
⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠−
77.
2 2
2
3 2
1 0
x x y y
x x yy
⎧ − + + = −⎪⎨ −
+ + =⎪⎩
Multiply each side of the second equation by –y and add the equations to eliminate y:
2 2
2 2
3 2
0
2 2 1
x x y y
x x y y
xx
− + + = −
− + − − =
− = −=
If 1:x = 2 2
2
1 3(1) 2
0( 1) 0
0 or 1
y y
y yy y
y y
− + + = −
+ =+ =
= = − Note that 0y ≠ because that would cause division by zero in the original system. Solution: (1, –1)
78. Multiply each side of the second equation by –x and add the equations to eliminate x:
2 2 2 2
2
2
2
2 221 2
2
2 0 ( 2) 0
0 or 2
x x y y x x y yyx x x y
xy y
y yy yy y
⎧ + + = + → + + = +⎪⎨ −
+ = → − − = −⎪⎩
=
− =− =
= =
2 2 2
2 2 2
If 0 :
0 0 2 2 0( 1)( 2) 0 1 or 2
If 2 :
2 2 2 0( 1) 0 0 or 1
y
x x x xx x x x
y
x x x xx x x x
=
+ + = + → + − =→ − + = → = = −
=
+ + = + → + =→ + = → = = −
Note that 0x ≠ because that would cause division by zero in the original system. Solutions: (1, 0), (–2, 0), (–1, 2)
79. 3 4 12x y+ ≤ Graph the line 3 4 12x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since
( ) ( )3 0 4 0 12+ ≤ is true, shade the side of the line containing (0, 0).
y
x−5
3 + 4 x y 12≤
5
5
−5
80. 2 3 6x y− ≥ Graph the line 2 3 6x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since ( ) ( )2 0 3 0 6− ≥ is false, shade the side of the line opposite (0, 0).
y
x−5 5
5
−5
62 − 3x y ≥
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Chapter 8: Systems of Equations and Inequalities
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81. 2y x≤
Graph the parabola 2y x= . Use a solid curve since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 1). Since 20 1≤ is false, shade the opposite side of the parabola from (0, 1).
y
x−5 5
5
−5
2y x≤
82. 2x y≥
Graph the circle 2x y= . Use a solid curve since the inequality uses ≥ . Choose a test point not on the parabola, such as (1, 0). Since 21 0≥ is true, shade the same side of the parabola as (1, 0).
y
x−5 5
5
−5
2x y≥
83. 2 2
2x yx y
− + ≤⎧⎨ + ≥⎩
Graph the line 2 2x y− + = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since
2(0) 0 2− + ≤ is true, shade the side of the line containing (0, 0). Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
y
–5
5
x–5 5
–2x + y = 2
x + y = 2
The graph is unbounded. Find the vertices: To find the intersection of 2x y+ = and
2 2x y− + = , solve the system: 2
2 2x yx y
+ =⎧⎨− + =⎩
Solve the first equation for x: 2x y= − . Substitute and solve:
2(2 ) 24 2 2
3 62
y yy y
yy
− − + =− + + =
==
2 2 0x = − = The point of intersection is (0, 2). The corner point is (0, 2).
84. 2 6
2 2x yx y
− ≤⎧⎨ + ≥⎩
Graph the line 2 6x y− = . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6− ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
y
–5
5
x–2 8
x – 2y = 6
(2, –2)
2x + y = 2 The graph is unbounded. Find the vertices: To find the intersection of 2 6x y− = and 2 2x y+ = , solve the system:
2 62 2x y
x y− =⎧
⎨ + =⎩
Solve the first equation for x: 2 6x y= + . Substitute and solve: 2(2 6) 2
4 12 25 10
2
y yy y
yy
+ + =+ + =
= −= −
2( 2) 6 2x = − + = The point of intersection is (2, 2)− . The corner point is (2, 2)− .
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Chapter 8 Review Exercises
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85.
004
2 3 6
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≤⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 4x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 4+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 3 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≤ is true, shade the side of the line containing (0, 0).
y
–2
8
x8
(0, 2)
(3, 0)(0, 0)
x + y = 4
2x + 3y = 6
The overlapping region is the solution. The graph is bounded. Find the vertices: The x-axis and y-axis intersect at (0, 0). The intersection of 2 3 6x y+ = and the y-axis is (0, 2). The intersection of 2 3 6x y+ = and the x-axis is (3, 0). The three corner points are (0, 0), (0, 2), and (3, 0).
86.
00
3 62 2
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≥⎪⎪ + ≥⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 3 6x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) + 0 ≥ 6 is false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0).
y
–2
8
x–2 8
(0, 6)
(2, 0)
3x + y = 6
2x + y = 2
The overlapping region is the solution. The graph
is unbounded. Find the vertices: The intersection of 3 6x y+ = and the y-axis is (0, 6). The intersection of 3 6x y+ = and the x-axis is (2, 0). The two corner points are (0, 6), and (2, 0).
87.
00
2 82 2
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≥⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 8x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≤ 8 is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 + 2(0) ≥ 2 is false, shade the opposite side of the line from (0, 0).
y
–1
9
x–1 9
2x + y = 8
x + 2y = 2
(2, 0)
(4, 0)(0, 1)
(0, 8)
The overlapping region is the solution. The graph
is bounded. Find the vertices: The intersection of 2 2x y+ = and the y-axis is (0, 1). The
intersection of 2 2x y+ = and the x-axis is (2, 0). The intersection of 2 8x y+ = and the y-axis is (0, 8). The intersection of 2 8x y+ = and the x-axis is (4, 0). The four corner points are (0, 1), (0, 8), (2, 0), and (4, 0).
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88.
00
3 92 3 6
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≥⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 3 9x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 9+ ≤ is true, shade the side of the line containing (0, 0).
y
–1
9
x–1 9
(0, 2)
(0, 9)
(3, 0)
2x + 3y = 6
3x + y = 9
Graph the line 2 3 6x y+ = . Use a solid line since
the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices: The intersection of 2 3 6x y+ = and the y-axis is (0, 2). The intersection of 2 3 6x y+ = and the x-axis is (3, 0). The intersection of 3 9x y+ = and the y-axis is (0, 9). The intersection of 3 9x y+ = and the x-axis is (3, 0). The three corner points are (0, 2), (0, 9), and (3, 0).
89. Graph the system of inequalities: 2 2 16
2x y
x y⎧ + ≤⎪⎨
+ ≥⎪⎩
Graph the circle 2 2 16x y+ = .Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 16+ ≤ is true, shade the side of the circle containing (0, 0). Graph the line 2x y+ = . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
y
–5
5
x–5 5
x + y = 2
x2+ y2 = 16
90. Graph the system of inequalities: 2 1
3y x
x y⎧ ≤ −⎪⎨
− ≤⎪⎩
Graph the parabola 2 1y x= − . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since
20 0 1≤ − is false, shade the opposite side of the parabola from (0, 0). Graph the line 3x y− = . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 3− ≤ is true, shade the same side of the line as (0, 0). The overlapping region is the solution.
y
–5
5
x–2 8
x – y = 3
y2 = x – 1
91. Graph the system of inequalities: 2
4y x
xy⎧ ≤⎪⎨
≤⎪⎩
Graph the parabola 2y x= . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (1, 2). Since 22 1≤ is false, shade the opposite side of the parabola from (1, 2). Graph the hyperbola 4xy = . Use a solid line since the inequality uses ≤ . Choose a test point not on the hyperbola, such as (1, 2). Since 1 2 4⋅ ≤ is true, shade the same side of the hyperbola as (1, 2). The overlapping region is the solution.
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y
–5
5
x–5
5
xy = 4
y = x2
92. Graph the system of inequalities: 2 2
2 2
1
4
x y
x y
⎧ + ≥⎪⎨
+ ≤⎪⎩
Graph the circle 2 2 1x y+ = . Use a solid line since the inequality uses ≥ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 1+ ≥ is false, shade the opposite side of the circle from (0, 0). Graph the circle 2 2 4x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 4+ ≤ is true, shade the same side of the circle as (0, 0). The overlapping region is the solution.
y
–5
5
x–5 5
x2 + y2 = 4
x2 + y2 = 1
93. Maximize 3 4z x y= + subject to 0x ≥ , 0y ≥ , 3 2 6x y+ ≥ , 8x y+ ≤ . Graph the constraints.
y
x
(0,8)
(8,0)(2,0)
(0,3)
The corner points are (0, 3), (2, 0), (0, 8), (8, 0).
Evaluate the objective function:
Vertex Value of 3 4(0, 3) 3(0) 4(3) 12(0, 8) 3(0) 4(8) 32(2, 0) 3(2) 4(0) 6(8, 0) 3(8) 4(0) 24
z x yzzzz
= += + == + == + == + =
The maximum value is 32 at (0, 8).
94. Maximize 2 4z x y= + subject to 0x ≥ , 0y ≥ , 6x y+ ≤ , 2x ≥ . Graph the constraints.
y
x
(2,4)
(6,0)(2,0)
The corner points are (2, 4), (2, 0), (6, 0).
Evaluate the objective function: Vertex Value of 2 4(2, 4) 2(2) 4(4) 20(2, 0) 2(2) 4(0) 4(6, 0) 2(6) 4(0) 12
z x yzzz
= += + == + == + =
The maximum value is 20 at (2, 4).
95. Minimize 3 5z x y= + subject to 0x ≥ , 0y ≥ , 1x y+ ≥ , 3 2 12x y+ ≤ , 3 12x y+ ≤ .
Graph the constraints.
y
x(4,0)(1,0)
(0,1)
(0,4)127 , 24
7( )
To find the intersection of 3 2 12 and 3 12x y x y+ = + = , solve the system:
3 2 123 12
x yx y
+ =⎧⎨ + =⎩
Solve the second equation for x: 12 3x y= −
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Chapter 8: Systems of Equations and Inequalities
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Substitute and solve: 3(12 3 ) 2 12
36 9 2 127 24
247
y yy y
y
y
− + =− + =
− = −
=
24 72 1212 3 127 7 7
x ⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
The point of intersection is 12 24,7 7
⎛ ⎞⎜ ⎟⎝ ⎠
.
The corner points are (0, 1), (1, 0), (0, 4), (4, 0), 12 24,7 7
⎛ ⎞⎜ ⎟⎝ ⎠
.
Evaluate the objective function: Vertex Value of 3 5(0, 1) 3(0) 5(1) 5(0, 4) 3(0) 5(4) 20(1, 0) 3(1) 5(0) 3(4, 0) 3(4) 5(0) 1212 24 12 24 156, 3 57 7 7 7 7
z x yz
zz
z
z
= += + =
= + == + =
= + =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
The minimum value is 3 at (1, 0).
96. Minimize 3z x y= + subject to 0x ≥ , 0y ≥ , 8x ≤ , 6y ≤ , 2 4x y+ ≥ . Graph the
constraints.
y
x
(0,6)
(8,0)(2,0)
(0,4)
(8,6)
The corner points are (0, 4), (2, 0), (0, 6), (8, 0),
(8, 6). Evaluate the objective function:
( )
Vertex Value of 3(0, 6) 3(0) 6 6(0, 4) 3(0) 4 4(2, 0) 3(2) 0 6(8, 0) 3(8) 0 248, 6 3(8) 6 30
z x yzzzzz
= += + == + == + == + == + =
The minimum value is 4 at (0, 4).
97. 2 5 54 10
x yx y A
+ =⎧⎨ + =⎩
Multiply each side of the first equation by –2 and eliminate x:
4 10 104 10
0 10
x yx y A
A
− − = −⎧⎪⎨ + =⎪⎩
= −
If there are to be infinitely many solutions, the result of elimination should be 0 = 0. Therefore,
10 0 or 10A A− = = .
98. 2 5 54 10
x yx y A
+ =⎧⎨ + =⎩
Multiply each side of the first equation by –2 and eliminate x:
4 10 104 10
0 10
x yx y A
A
− − = −⎧⎪⎨ + =⎪⎩
= −
If the system is to be inconsistent, the result of elimination should be 0 = any number except 0. Therefore, 10 0 or 10A A− ≠ ≠ .
99. 2y ax bx c= + + At (0, 1) the equation becomes:
21 (0) (0)1a b c
c= + +=
At (1, 0) the equation becomes: 20 (1) (1)
00
a b ca b c
a b c
= + += + +
+ + =
At (–2, 1) the equation becomes: 21 ( 2) ( 2)
1 4 24 2 1
a b ca b c
a b c
= − + − += − +
− + =
The system of equations is: 0
4 2 11
a b ca b c
c
+ + =⎧⎪ − + =⎨⎪ =⎩
Substitute 1c = into the first and second equations and simplify:
1 01
1
a ba b
a b
+ + =+ = −
= − −
4 2 1 14 2 0
a ba b
− + =− =
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Solve the first equation for a, substitute into the second equation and solve:
4( 1) 2 04 4 2 0
6 423
b bb b
b
b
− − − =− − − =
− =
= −
2 113 3
a = − = −
The quadratic function is 21 2 13 3
y x x= − − + .
100. 2 2 0x y Dx Ey F+ + + + = At (0, 1) the equation becomes:
2 20 1 (0) (1) 01
D E FE F
+ + + + =+ = −
At (1, 0) the equation becomes: 2 21 0 (1) (0) 0
1D E F
D F+ + + + =
+ = −
At (–2, 1) the equation becomes: 2 2( 2) 1 ( 2) (1) 0
2 5D E F
D E F− + + − + + =
− + + = −
The system of equations is: 11
2 5
E FD F
D E F
+ = −⎧⎪ + = −⎨⎪− + + = −⎩
Substitute 1E F+ = − into the third equation and solve for D:
2 ( 1) 52 4
2
DDD
− + − = −− = −
=
Substitute and solve: 2 1
3FF
+ = −= −
( 3) 12
EE
+ − = −=
The equation of the circle is 2 2 2 2 3 0x y x y+ + + − = .
101. Let x = the number of pounds of coffee that costs $6.00 per pound, and let y = the number of pounds of coffee that costs $9.00 per pound. Then 100x y+ = represents the total amount of coffee in the blend. The value of the blend will be represented by the equation: 6 9 6.90(100)x y+ = . Solve the system of equations:
1006 9 690
x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 100y x= − . Solve by substitution: 6 9(100 ) 690
6 900 9 6903 210
70
x xx x
xx
+ − =+ − =
− = −=
100 70 30y = − = The blend is made up of 70 pounds of the $6.00-per-pound coffee and 30 pounds of the $9.00-per-pound coffee.
102. Let x = the number of acres of corn, and let y = the number of acres of soybeans. Then
1000x y+ = represents the total acreage on the farm. The total cost will be represented by the equation: 65 45 54,325x y+ = . Solve the system of equations:
100065 45 54,325
x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 1000y x= − Solve by substitution: 65 45(1000 ) 54,325
65 45,000 45 54,32520 9325
466.25
x xx x
xx
+ − =+ − =
==
1000 466.25 533.75y = − = Corn should be planted on 466.25 acres and soybeans should be planted on 533.75 acres.
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103. Let x = the number of small boxes, let y = the number of medium boxes, and let z = the number of large boxes. Oatmeal raisin equation: 2 2 15x y z+ + = Chocolate chip equation: 2 10x y z+ + = Shortbread equation: 3 11y z+ =
2 2 152 103 11
x y zx y z
y z
+ + =⎧⎪ + + =⎨⎪ + =⎩
Multiply each side of the second equation by –1 and add to the first equation to eliminate x:
2 2 152 10
3 11 5
x y zx y z
y zy
+ + =⎧⎪⎨− − − = −⎪⎩
+ ==
Substituting and solving for the other variables: 5 3 11
3 62
zzz
+ ===
5 2(2) 10
9 101
xx
x
+ + =+ =
=
Thus, 1 small box, 5 medium boxes, and 2 large boxes of cookies should be purchased.
104. a. Let x = the number of lower-priced packages, and let y = the number of quality packages. Peanut inequality: 8 6 120(16)
4 3 960x yx y
+ ≤+ ≤
Cashew inequality: 4 6 72(16)2 3 576
x yx y
+ ≤+ ≤
The system of inequalities is:
00
4 3 9602 3 576
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≤⎩
b. Graphing:
To find the intersection of 2 3 576x y+ = and 4 3 960x y+ = , solve the system:
4 3 9602 3 576
x yx y
+ =⎧⎨ + =⎩
Subtract the second equation from the first:
4 3 9602 3 576
2 384 192
x yx y
xx
+ =− − =
==
Substitute and solve: 2(192) 3 576
3 19264
yyy
+ ===
The corner points are (0, 0), (0, 192), (240, 0), and (192, 64).
105. Let x = the speed of the boat in still water, and let y = the speed of the river current. The distance from Chiritza to the Flotel Orellana is 100 kilometers.
Rate Time Distancetrip downstream 5 / 2 100trip downstream 3 100
x yx y
+−
The system of equations is: 5 ( ) 10023( ) 100
x y
x y
⎧ + =⎪⎨⎪ − =⎩
Multiply both sides of the first equation by 6, multiply both sides of the second equation by 5, and add the results.
15 15 60015 15 500
30 1100
x yx y
x
+ =− =
=
1100 11030 3
x = =
1103 3 1003110 3 100
10 3103
y
yy
y
⎛ ⎞ − =⎜ ⎟⎝ ⎠
− ==
=
The speed of the boat is 110 / 3 36.67 km/hr≈ ; the speed of the current is 10 / 3 3.33 km/hr≈ .
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106. Let x = the speed of the jet stream, and let d = the distance from Chicago to Ft. Lauderdale. The jet stream flows from Chicago to Ft. Lauderdale because the time is shorter in that direction.
Rate Time DistanceChicago to
475 5 / 2Ft. LauderdaleFt. Lauderdale
475 17 / 6to Chicago
x d
x d
+
−
The system of equations is: ( )( )
( )( )475 5 / 2
475 17 / 6
y d
y d
+ =⎧⎪⎨
− =⎪⎩
Simplifying the system, we obtain: 2 5 2375
6 17 8075d x
d x− =⎧
⎨ + =⎩
Multiply the first equation by 3− and add the result to the second equation:
6 15 71256 17 8075
d xd x
− + = −+ =
32 950950 47532 16
x
x
=
= =
The speed of the jet stream is approximately 475 /16 29.69≈ miles per hour.
107. Let x = the number of hours for Bruce to do the job alone, let y = the number of hours for Bryce to do the job alone, and let z = the number of hours for Marty to do the job alone. Then 1/x represents the fraction of the job that Bruce does in one hour. 1/y represents the fraction of the job that Bryce does in one hour. 1/z represents the fraction of the job that Marty does in one hour. The equation representing Bruce and Bryce working together is:
( )1 1 1 3 0.75
4 / 3 4x y+ = = =
The equation representing Bryce and Marty working together is:
( )1 1 1 5 0.625
8 / 5 6y z+ = = =
The equation representing Bruce and Marty working together is:
( )1 1 1 3 0.375
8 / 3 8x z+ = = =
Solve the system of equations: 1 1
1 1
1 1
0.75
0.625
0.375
x y
y z
x z
− −
− −
− −
⎧ + =⎪⎪ + =⎨⎪ + =⎪⎩
Let 1 1 1, ,u x v y w z− − −= = = 0.750.6250.375
u vv wu w
+ =⎧⎪ + =⎨⎪ + =⎩
Solve the first equation for u: 0.75u v= − . Solve the second equation for w: 0.625w v= − . Substitute into the third equation and solve: (0.75 ) (0.625 ) 0.375
2 10.5
v vvv
− + − =− = −
=
0.75 0.5 0.25u = − = 0.625 0.5 0.125w = − =
Solve for , , andx y z : 4, 2, 8 (reciprocals)x y z= = =
Bruce can do the job in 4 hours, Bryce in 2 hours, and Marty in 8 hours.
108. Let x = the number of dancing girls produced, and let y = the number of mermaids produced. The total profit is: 25 30P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of figurines must be produced.
3 3 90x y+ ≤ 90 hours are available for molding.
6 4 120x y+ ≤ 120 hours are available for painting.
2 3 60x y+ ≤ 60 hours are available for glazing. Graph the constraints.
y
x
(0,20)
(20,0)(0,0)
(12,12)
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To find the intersection of 6 4 120x y+ = and 2 3 60x y+ = , solve the system:
6 4 1202 3 60
x yx y
+ =⎧⎨ + =⎩
Multiply the second equation by 3, and subtract from the first equation:
6 4 1206 9 180
x yx y
+ =− − = −
5 6012
yy
− = −=
Substitute and solve: 2 3(12) 60
2 2412
xxx
+ ===
The point of intersection is (12, 12). The corner points are (0, 0), (0, 20), (20, 0), (12, 12). Evaluate the objective function:
Vertex Value of 25 30(0, 0) 25(0) 30(0) 0
(0, 20) 25(0) 30(20) 600(20, 0) 25(20) 30(0) 500(12, 12) 25(12) 30(12) 660
P x yP
PPP
= += + =
= + == + == + =
The maximum profit is $660, when 12 dancing girl and 12 mermaid figurines are produced each day. To determine the excess, evaluate each constraint at 12x = and 12y = :
Molding: 3 3 3(12) 3(12) 36 36 72Painting: 6 4 6(12) 4(12) 72 48 120Glazing: 2 3 2(12) 3(12) 24 36 60
x yx yx y
+ = + = + =+ = + = + =+ = + = + =
Painting and glazing are at their capacity. Molding has 18 more hours available, since only 72 of the 90 hours are used.
109. Let x = the number of gasoline engines produced each week, and let y = the number of diesel engines produced each week. The total cost is:
450 550C x y= + . Cost is to be minimized; thus, this is the objective function. The constraints are:
20 60x≤ ≤ number of gasoline engines needed and capacity each week.
15 40y≤ ≤ number of diesel engines needed and capacity each week.
50x y+ ≥ number of engines produced to prevent layoffs.
Graph the constraints.
y
x
(60,40)
(60,15)
(20,40)
(35,15)
(20,30)
The corner points are (20, 30), (20, 40), (35, 15),
(60, 15), (60, 40) Evaluate the objective function:
( ) ( ) ( )
Vertex Value of 450 550(20, 30) 450(20) 550(30) 25,500(20, 40) 450(35) 550(40) 31,000(35, 15) 450(35) 550(15) 24,000(60, 15) 450(60) 550(15) 35, 25060, 40 450 60 550 40 49,000
C x yCCCC
C
= += + == + == + == + =
= + =
The minimum cost is $24,000, when 35 gasoline engines and 15 diesel engines are produced. The excess capacity is 15 gasoline engines, since only 20 gasoline engines had to be delivered.
110. Answers will vary.
Chapter 8 Test
1. 2 74 3 9
x yx y
− + = −⎧⎨ + =⎩
Substitution: We solve the first equation for y, obtaining
2 7y x= − Next we substitute this result for y in the second equation and solve for x.
( )4 3 9
4 3 2 7 94 6 21 9
10 3030 310
x yx x
x xx
x
+ =
+ − =
+ − ==
= =
We can now obtain the value for y by letting 3x = in our substitution for y.
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( )2 72 3 7 6 7 1
y xy
= −
= − = − = −
The solution of the system is 3x = , 1y = − or (3, 1)− .
Elimination: Multiply each side of the first equation by 2 so that the coefficients of x in the two equations are negatives of each other. The result is the equivalent system
4 2 144 3 9
x yx y
− + = −⎧⎨ + =⎩
We can replace the second equation of this system by the sum of the two equations. The result is the equivalent system
4 2 145 5
x yy
− + = −⎧⎨ = −⎩
Now we solve the second equation for y. 5 5
5 15
y
y
= −−
= = −
We back-substitute this value for y into the original first equation and solve for x.
( )2 7
2 1 72 6
6 32
x yx
x
x
− + = −
− + − = −
− = −−
= =−
The solution of the system is 3x = , 1y = − or (3, 1)− .
2. 1 2 13
5 30 18
x y
x y
⎧ − =⎪⎨⎪ − =⎩
We choose to use the method of elimination and multiply the first equation by 15− to obtain the equivalent system
5 30 155 30 18
x yx y
− + = −⎧⎨ − =⎩
We replace the second equation by the sum of the two equations to obtain the equivalent system
5 30 150 3
x y− + = −⎧⎨ =⎩
The second equation is a contradiction and has no solution. This means that the system itself has no solution and is therefore inconsistent.
3. 2 5 (1)
3 4 2 (2)5 2 3 8 (3)
x y zx y z
x y z
− + =⎧⎪ + − = −⎨⎪ + + =⎩
We use the method of elimination and begin by eliminating the variable y from equation (2). Multiply each side of equation (1) by 4 and add the result to equation (2). This result becomes our new equation (2). 2 5 4 4 8 20
3 4 2 3 4 2
7 7 18 (2)
x y z x y zx y z x y z
x z
− + = − + =+ − = − + − = −
+ =
We now eliminate the variable y from equation (3) by multiplying each side of equation (1) by 2 and adding the result to equation (3). The result becomes our new equation (3). 2 5 2 2 4 105 2 3 8 5 2 3 8
7 7 18 (3)
x y z x y zx y z x y z
x z
− + = − + =+ + = + + =
+ =
Our (equivalent) system now looks like 2 5 (1)
7 7 18 (2)7 7 18 (3)
x y zx zx z
− + =⎧⎪ + =⎨⎪ + =⎩
Treat equations (2) and (3) as a system of two equations containing two variables, and eliminate the x variable by multiplying each side of equation (2) by 1− and adding the result to equation (3). The result becomes our new equation (3). 7 7 18 7 7 187 7 18 7 7 18 0 0 (3)
x z x zx z x z
+ = − − = −+ = + =
=
We now have the equivalent system 2 5 (1)
7 7 18 (2)0 0 (3)
x y zx z
− + =⎧⎪ + =⎨⎪ =⎩
This is equivalent to a system of two equations with three variables. Since one of the equations contains three variables and one contains only two variables, the system will be dependent. There are infinitely many solutions. We solve equation (2) for x and determine that
187
x z= − + . Substitute this expression into
equation (1) to obtain y in terms of z.
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Chapter 8: Systems of Equations and Inequalities
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2 518 2 5718 2 57
177
177
x y z
z y z
z y z
y z
y z
− + =
⎛ ⎞− + − + =⎜ ⎟⎝ ⎠
− + − + =
− + =
= −
The solution is 18 17,7 7
x z y z= − + = − ,
z is any real number or 18( , , ) ,7
x y z x z⎧= − +⎨
⎩
17 , is any real number7
y z z ⎫= − ⎬⎭
.
4. 23
3 2 8 3 (1)
1 (2)
6 3 15 8 (3)
x y z
x y z
x y z
+ − = −⎧⎪
− − + =⎨⎪ − + =⎩
We start by clearing the fraction in equation (2) by multiplying both sides of the equation by 3.
3 2 8 3 (1)3 2 3 3 (2)
6 3 15 8 (3)
x y zx y z
x y z
+ − = −⎧⎪− − + =⎨⎪ − + =⎩
We use the method of elimination and begin by eliminating the variable x from equation (2). The coefficients on x in equations (1) and (2) are negatives of each other so we simply add the two equations together. This result becomes our new equation (2). 3 2 8 3
3 2 3 3
5 0 (2)
x y zx y z
z
+ − = −− − + =
− =
We now eliminate the variable x from equation (3) by multiplying each side of equation (1) by
2− and adding the result to equation (3). The result becomes our new equation (3). 3 2 8 3 6 4 16 66 3 15 8 6 3 15 8
7 31 14 (3)
x y z x y zx y z x y z
y z
+ − = − − − + =− + = − + =
− + =
Our (equivalent) system now looks like 3 2 8 3 (1)
5 0 (2)7 31 14 (3)
x y zz
y z
+ − = −⎧⎪ − =⎨⎪ − + =⎩
We solve equation (2) for z by dividing both
sides of the equation by 5− . 5 0
0zz
− ==
Back-substitute 0z = into equation (3) and solve for y.
7 31 147 31(0) 14
7 142
y zy
yy
− + =− + =
− == −
Finally, back-substitute 2y = − and 0z = into equation (1) and solve for x.
3 2 8 33 2( 2) 8(0) 3
3 4 33 1
13
x y zx
xx
x
+ − = −+ − − = −
− = −=
=
The solution of the original system is 13
x = , 2y = − , 0z = or 1 , 2, 03
⎛ ⎞−⎜ ⎟⎝ ⎠
.
5. 4 5 02 6 19
5 5 10
x y zx y
x y z
− + =⎧⎪− − + = −⎨⎪ + − =⎩
We first check the equations to make sure that all variable terms are on the left side of the equation and the constants are on the right side. If a variable is missing, we put it in with a coefficient of 0. Our system can be rewritten as
4 5 02 0 25
5 5 10
x y zx y z
x y z
− + =⎧⎪− − + = −⎨⎪ + − =⎩
The augmented matrix is 4 5 1 02 1 0 25
1 5 5 10
−⎡ ⎤⎢ ⎥− − −⎢ ⎥
−⎢ ⎥⎣ ⎦
6. The matrix has three rows and represents a system with three equations. The three columns to the left of the vertical bar indicate that the system has three variables. We can let x, y, and z denote these variables. The column to the right of the vertical bar represents the constants on the right side of the equations. The system is
3 2 4 61 0 8 22 1 3 11
x y zx y zx y z
+ + = −⎧⎪ + + =⎨⎪− + + = −⎩
or 3 2 4 6
8 22 3 11
x y zx z
x y z
+ + = −⎧⎪ + =⎨⎪− + + = −⎩
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7. 1 1 4 6
2 2 0 4 1 33 2 1 8
2 2 4 6 6 40 8 1 3 1 116 4 1 8 5 12
A C−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥+ = − + −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − + − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
8. 1 1 4 6
3 0 4 3 1 33 2 1 8
1 1 12 18 11 190 4 3 9 3 53 2 3 24 6 22
A C−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥− = − − −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − − − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
9. AC cannot be computed because the dimensions are mismatched. To multiply two matrices, we need the number of columns in the first matrix to be the same as the number of rows in the second matrix. Matrix A has 2 columns, but matrix C has 3 rows. Therefore, the operation cannot be performed.
10. Here we are taking the product of a 2 3× matrix and a 3 2× matrix. Since the number of columns in the first matrix is the same as the number of rows in the second matrix (3 in both cases), the operation can be performed and will result in a 2 2× matrix.
( ) ( ) ( ) ( )( ) ( )
1 1 2 0 5 3 1 1 2 4 5 2
0 1 3 0 1 3 0 1 3 4 1 2
1 11 2 5
0 40 3 1
3 2
16 173 10
BA
⋅ + − ⋅ + ⋅ ⋅ − + − ⋅ − + ⋅
⋅ + ⋅ + ⋅ ⋅ − + − + ⋅
−⎡ ⎤−⎡ ⎤ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦
= ⎡ ⎤⎢ ⎥⎣ ⎦⎡ ⎤
= ⎢ ⎥−⎣ ⎦
11. We first form the matrix
[ ]2
3 2 1 0|
5 4 0 1A I
⎡ ⎤= ⎢ ⎥
⎣ ⎦
Next we use row operations to transform [ ]2|A I into reduced row echelon form.
( )1 1
2 113 33
3 2 1 05 4 0 1
1 0
5 4 0 1R r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ =⎢ ⎥
⎢ ⎥⎣ ⎦
( )
( )
( )
2 1 2
2 2
1 2 1
2 13 3
523 32 13 3 3
25 32 2
25 3 32 2
1 0 5
0 1
1 0
0 1
1 0 2 1
0 1
R r r
R r
R rr
⎡ ⎤→ = − +⎢ ⎥
−⎢ ⎥⎣ ⎦⎡ ⎤
→ =⎢ ⎥−⎢ ⎥⎣ ⎦
−⎡ ⎤→ = − +⎢ ⎥−⎣ ⎦
Therefore, 15 32 2
2 1A−
−⎡ ⎤= ⎢ ⎥−⎣ ⎦
.
12. We first form the matrix
[ ]3
1 1 1 1 0 0| 2 5 1 0 1 0
2 3 0 0 0 1B I
−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
Next we use row operations to transform [ ]3|B I into reduced row echelon form.
( )
2 1 2
3 1 3
2 2
1 2 1
3
3 2 1 17 7 7 7
54 17 7 73 2 17 7 7
51 47 7 7
1 1 1 1 0 02 5 1 0 1 02 3 0 0 0 1
1 1 1 1 0 02
0 7 3 2 1 0 2
0 5 2 2 0 1
1 1 1 1 0 0
0 1 0
0 5 2 2 0 1
1 0 0
0 1 0
0 0 1
R r rR r r
R r
R r rR
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
−⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −⎣ ⎦
−⎡ ⎤⎢ ⎥
→ − − =⎢ ⎥⎢ ⎥− −⎣ ⎦⎡ ⎤⎢ ⎥ = +⎢ ⎥→ − −⎢ ⎥
− −⎢ ⎥⎣ ⎦2 35r r
⎛ ⎞⎜ ⎟= − +⎝ ⎠
( )3 3
1 3 1
32 3 27
54 17 7 73 2 17 7 7
47
1 0 0
0 1 0 7
0 0 1 4 5 7
1 0 0 3 3 40 1 0 2 2 3 0 0 1 4 5 7
R r
R r r
R r r
⎡ ⎤⎢ ⎥⎢ ⎥→ − − =⎢ ⎥
− −⎢ ⎥⎣ ⎦−⎡ ⎤ ⎛ ⎞= − +⎢ ⎥ ⎜ ⎟→ − −⎢ ⎥ ⎜ ⎟= +⎝ ⎠⎢ ⎥− −⎣ ⎦
Thus, 1
3 3 42 2 34 5 7
B−
−⎡ ⎤⎢ ⎥= − −⎢ ⎥⎢ ⎥− −⎣ ⎦
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Chapter 8: Systems of Equations and Inequalities
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13. 6 3 122 2x yx y+ =⎧
⎨ − = −⎩
We start by writing the augmented matrix for the system.
6 3 122 1 2
⎡ ⎤⎢ ⎥− −⎣ ⎦
Next we use row operations to transform the augmented matrix into row echelon form.
( )
( )
1 2
2 1
11 12
2 1 2
12
12
6 3 12 2 1 2
2 1 2 6 3 12
1 1
6 3 12
1 1 6
0 6 18
R rR r
R r
R r r
=− − ⎛ ⎞⎡ ⎤ ⎡ ⎤→ ⎜ ⎟⎢ ⎥ ⎢ ⎥− − =⎣ ⎦ ⎣ ⎦ ⎝ ⎠
⎡ ⎤− −→ =⎢ ⎥
⎢ ⎥⎣ ⎦⎡ ⎤− −
→ = − +⎢ ⎥⎢ ⎥⎣ ⎦
( )
( )
12 26
12 2 12
12
12
1 1
0 1 3
1 0
0 1 3
R r
R r r
⎡ ⎤− −→ =⎢ ⎥
⎢ ⎥⎣ ⎦⎡ ⎤
→ = +⎢ ⎥⎢ ⎥⎣ ⎦
The solution of the system is 12x = , 3y = or
( )12 , 3
14. 1 74
8 2 56
x y
x y
⎧ + =⎪⎨⎪ + =⎩
We start by writing the augmented matrix for the system.
141 7
8 2 56⎡ ⎤⎢ ⎥⎣ ⎦
Next we use row operations to transform the augmented matrix into row echelon form.
14
2 1 2
14
1 788 2 56
1 70 0 0
R R r⎡ ⎤⎢ ⎥ = − +⎣ ⎦⎡ ⎤⎢ ⎥⎣ ⎦
The augmented matrix is now in row echelon form. Because the bottom row consists entirely of 0’s, the system actually consists of one equation in two variables. The system is dependent and therefore has an infinite number of solutions. Any ordered pair satisfying the
equation 1 74
x y+ = , or 4 28y x= − + , is a
solution to the system.
15. 2 4 32 7 15 124 7 13 10
x y zx y zx y z
+ + = −⎧⎪ + + = −⎨⎪ + + = −⎩
We start by writing the augmented matrix for the system.
1 2 4 32 7 15 124 7 13 10
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦
Next we use row operations to transform the augmented matrix into row echelon form.
( )
2 1 2
3 1 3
2 3
3 2
3 2 3
1 2 4 32 7 15 124 7 13 10
1 2 4 32
0 3 7 6 4
0 1 3 2
1 2 4 30 1 3 2 0 3 7 6
1 2 4 30 1 3 2 30 0 2 0
1 2 4 30 1 3 20 0 1 0
R r rR r r
R rR r
R r r
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦
−⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −⎣ ⎦
−⎡ ⎤= −⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ =⎝ ⎠⎢ ⎥−⎣ ⎦
−⎡ ⎤⎢ ⎥→ − = − +⎢ ⎥⎢ ⎥−⎣ ⎦
−⎡ ⎤⎢→ −⎢⎢⎣
( )13 32 R r⎥ = −⎥
⎥⎦
The matrix is now in row echelon form. The last row represents the equation 0z = . Using 0z = we back-substitute into the equation 3 2y z+ = − (from the second row) and obtain
( )3 2
3 0 22
y zy
y
+ = −
+ = −
= −
Using 2y = − and 0z = , we back-substitute into the equation 2 4 3x y z+ + = − (from the first row) and obtain
( ) ( )2 4 3
2 2 4 0 31
x y zx
x
+ + = −
+ − + = −
=
The solution is 1x = , 2y = − , 0z = or (1, 2, 0)− .
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16. 2 2 3 52 8
3 5 8 2
x y zx y z
x y z
+ − =⎧⎪ − + =⎨⎪ + − = −⎩
We start by writing the augmented matrix for the system.
2 2 3 51 1 2 83 5 8 2
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− −⎣ ⎦
Next we use row operations to transform the augmented matrix into row echelon form.
( )
1 2
2 1
2 1 2
3 1 3
12 24
7 114 4
7 114 4
2 2 3 51 1 2 83 5 8 2
1 1 2 82 2 3 5 3 5 8 2
1 1 2 82
0 4 7 11 3
0 8 14 26
1 1 2 80 1
0 8 14 26
1 1 2 80 1
R rR r
R r rR r r
R r
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− −⎣ ⎦
−⎡ ⎤=⎛ ⎞⎢ ⎥= − ⎜ ⎟⎢ ⎥ =⎝ ⎠⎢ ⎥− −⎣ ⎦
−⎡ ⎤= − +⎛ ⎞⎢ ⎥= − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −⎣ ⎦
⎡ ⎤−⎢ ⎥
= =− −⎢ ⎥⎢ ⎥
− −⎣ ⎦
−= − − ( )3 2 3 8
0 0 0 4
R r r
⎡ ⎤⎢ ⎥
= − +⎢ ⎥⎢ ⎥
−⎣ ⎦
The last row represents the equation 0 4= − which is a contradiction. Therefore, the system has no solution and is be inconsistent.
17. ( )( ) ( )( )2 5
2 7 5 3 14 15 293 7
−= − − = − − = −
18. 2 4 61 4 01 2 4
−
− −
[ ] [ ][ ]
4 0 1 0 1 42 ( 4) 6
2 4 1 4 1 2
2 4( 4) 2(0) 4 1( 4) ( 1)(0)
6 1(2) ( 1)42( 16) 4( 4) 6(6)
32 16 3612
= − − +− − − −
= − − + − − −
+ − −
= − + − += − − += −
19. 4 3 233 5 19
x yx y
+ = −⎧⎨ − =⎩
The determinant D of the coefficients of the variables is
( )( ) ( )( )4 3
4 5 3 3 20 9 293 5
D = = − − = − − = −−
Since 0D ≠ , Cramer’s Rule can be applied.
( )( ) ( )( )23 3
23 5 3 19 5819 5xD−
= = − − − =−
( )( ) ( )( )4 23
4 19 23 3 1453 19yD
−= = − − =
58 229
xDx
D= = = −
−
145 529
yDy
D= = = −
−
The solution of the system is 2x = − , 5y = − or ( 2, 5)− − .
20. 4 3 2 152 3 15
5 5 2 18
x y zx y z
x y z
− + =⎧⎪− + − = −⎨⎪ − + =⎩
The determinant D of the coefficients of the variables is
( )
( ) ( ) ( )( ) ( ) ( )
4 3 22 1 3
5 5 2
1 3 2 3 2 14 3 2
5 2 5 2 5 5
4 2 15 3 4 15 2 10 5
4 13 3 11 2 552 33 109
D−
= − −−
− − − −= − − +
− −
= − + − + + −
= − + +
= − + += −
Since 0D ≠ , Cramer’s Rule can be applied.
( )
( ) ( ) ( )( ) ( ) ( )
15 3 215 1 3
18 5 2
1 3 15 3 15 115 3 2
5 2 18 2 18 5
15 2 15 3 30 54 2 75 18
15 13 3 24 2 579
xD−
= − −−
− − − −= − − +
− −
= − + − + + −
= − + +
= −
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Chapter 8: Systems of Equations and Inequalities
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( ) ( ) ( )( ) ( ) ( )
4 15 22 15 3
5 18 2
15 3 2 3 2 154 15 2
18 2 5 2 5 18
4 30 54 15 4 15 2 36 75
4 24 15 11 2 399
yD = − − −
− − − − − −= − +
= − + − − + + − +
= − +
= −
( )
( ) ( ) ( )( ) ( ) ( )
4 3 152 1 15
5 5 18
1 15 2 15 2 14 3 15
5 18 5 18 5 5
4 18 75 3 36 75 15 10 5
4 57 3 39 15 536
zD−
= − −−
− − − −= − − +
− −
= − + − + + −
= − + +
= −
9 19
xDx
D−
= = =−
, 9 19
yDy
D= = = −
−,
36 49
zDz
D−
= = =−
The solution of the system is 1x = , 1y = − , 4z = or (1, 1, 4)− .
21. 2 2
2
3 12
9
x y
y x
⎧ + =⎪⎨
=⎪⎩
Substitute 9x for 2y into the first equation and solve for x:
( )2
2
2
3 9 12
3 9 12 0
3 4 0( 1)( 4) 0
x x
x x
x xx x
+ =
+ − =
+ − =− + =
1 or 4x x= = − Back substitute these values into the second equation to determine y:
1x = : 2 9(1) 93
yy
= == ±
4x = − : 2 9( 4) 36
36 (not real)
y
y
= − = −
= ± −
The solutions of the system are (1, 3)− and (1, 3) .
22. 2 22 3 5
1 1y x
y x y x⎧ − =⎨
− = ⇒ = +⎩
Substitute 1x + for y into the first equation and solve for x:
( )( )
2 2
2 2
2 2
2
2
2 1 3 5
2 2 1 3 5
2 4 2 3 5
4 3 0
4 3 0( 1)( 3) 0
x x
x x x
x x x
x x
x xx x
+ − =
+ + − =
+ + − =
− + − =
− + =− − =
1 or 3x x= = Back substitute these values into the second equation to determine y:
1x = : 1 1 2y = + = 3x = : 3 1 4y = + =
The solutions of the system are (1, 2) and (3, 4) .
23. 2 2 100
4 3 0x y
x y⎧ + ≤⎨
− ≥⎩
Graph the circle 2 2 100x y+ = . Use a solid curve since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 100+ ≤ is true, shade the same side of the circle as (0, 0); that is, inside the circle.
Graph the line 4 3 0x y− = . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 1). Since 4(0) 3(1) 0− ≥ is false, shade the opposite side of the line from (0, 1). The overlapping region is the solution.
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Chapter 8 Test
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24. ( )2
3 73
xx
+
+
The denominator contains the repeated linear factor 3x + . Thus, the partial fraction decomposition takes on the form
( ) ( )2 2
3 733 3
x A Bxx x
+= +
++ +
Clear the fractions by multiplying both sides by ( )23x + . The result is the identity
( )3 7 3x A x B+ = + + or
( )3 7 3x Ax A B+ = + + We equate coefficients of like powers of x to obtain the system
37 3
AA B
=⎧⎨ = +⎩
Therefore, we have 3A = . Substituting this result into the second equation gives
( )7 37 3 32
A BB
B
= +
= +
− =
Thus, the partial fraction decomposition is
( ) ( )2 2
3 7 3 233 3
xxx x
+ −= +
++ +.
25. ( )
2
22
4 3
3
x
x x
−
+
The denominator contains the linear factor x and the repeated irreducible quadratic factor 2 3x + . The partial fraction decomposition takes on the form
( ) ( )2
2 2 22 2
4 333 3
x A Bx C Dx Ex xx x x
− + += + +
++ +
We clear the fractions by multiplying both sides
by ( )22 3x x + to obtain the identity
( ) ( )( ) ( )2 2224 3 3 3x A x x x Bx C x Dx E− = + + + + + +
Collecting like terms yields ( ) ( )( ) ( )
2 4 3 24 3 6 3
3 9
x A B x Cx A B D x
C E x A
− = + + + + +
+ + +
Equating coefficients, we obtain the system
00
6 3 43 0
9 3
A BC
A B DC E
A
+ =⎧⎪ =⎪⎪ + + =⎨⎪ + =⎪
= −⎪⎩
From the last equation we get 13
A = − .
Substituting this value into the first equation
gives 13
B = . From the second equation, we
know 0C = . Substituting this value into the fourth equation yields 0E = .
Substituting 13
A = − and 13
B = into the third
equation gives us ( ) ( )1 1
3 36 3 4
2 1 45
D
DD
− + + =
− + + ==
Therefore, the partial fraction decomposition is
( ) ( ) ( )
2
2 222 2
1 13 34 3 5
33 3
xx xx xx x x
−− = + +++ +
26. 002 8
2 3 2
xyx y
x y
≥⎧⎪ ≥⎪⎨ + ≥⎪⎪ − ≥⎩
The inequalities 0x ≥ and 0y ≥ require that the graph be in quadrant I.
2 81 42
x y
y x
+ ≥
≥ − +
Test the point ( )0,0 .
( )2 8
0 2 0 8 ?0 8 false
x y+ ≥
+ ≥
≥
The point ( )0,0 is not a solution. Thus, the graph of the inequality 2 8x y+ ≥ includes the
half-plane above the line 1 42
y x= − + . Because
the inequality is non-strict, the line is also part of the graph of the solution.
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Chapter 8: Systems of Equations and Inequalities
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2 3 22 23 3
x y
y x
− ≥
≤ −
Test the point ( )0,0 .
( ) ( )2 3 22 0 3 0 2 ?0 2 false
x y− ≥
− ≥
≥
The point ( )0,0 is not a solution. Thus, the graph of the inequality 2 3 2x y− ≥ includes the
half-plane below the line 2 23 3
y x= − .
Because the inequality is non-strict, the line is also part of the graph of the solution. The overlapping shaded region (that is, the shaded region in the graph below) is the solution to the system of linear inequalities.
The graph is unbounded. The corner points are ( )4,2 and ( )8,0 .
27. The objective function is 5 8z x y= + . We seek the largest value of z that can occur if x and y are solutions of the system of linear inequalities
02 8
3 3
xx y
x y
≥⎧⎪ + ≤⎨⎪ − ≤ −⎩
2 82 8
x yy x
+ == − +
3 33 3
1 13
x yy x
y x
− = −− = − −
= +
The graph of this system (the feasible points) is shown as the shaded region in the figure below. The corner points of the feasible region are ( )0,1 , ( )3,2 , and ( )0,8 .
4
8
y
x4 8
2 8x y+ =
3 3x y− = −
(0, 1)(3, 2)
(0, 8)
( )
( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
Corner point, , Value of obj. function, 0,1 5 0 8 1 83,2 5 3 8 2 310,8 5 0 8 8 64
x y zz
zz
= + == + == + =
From the table, we can see that the maximum value of z is 64, and it occurs at the point ( )0,8 .
28. Let j = unit price for flare jeans, c = unit price for camisoles, and t = unit price for t-shirts. The given information yields a system of equations with each of the three women yielding an equation.
2 2 4 90 (Megan) 3 42.5 (Paige) 3 2 62 (Kara)
j c tj tj c t
+ + =⎧⎪ + =⎨⎪ + + =⎩
We can solve this system by using matrices.
( )11 12
2 2 4 90 1 1 2 451 0 3 42.5 1 0 3 42.5 1 3 2 62 1 3 2 62
R r⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
( )
( )
2 1 2
3 1 3
2 2
1 2 1
3 2 3
3 312
1 1 2 450 1 1 2.5 0 2 0 17
1 1 2 450 1 1 2.5 0 2 0 17
1 0 3 42.50 1 1 2.5
20 0 2 12
1 0 3 42.50 1 1 2.5 0 0 1 6
R r rR r r
R r
R r rR r r
R r
⎡ ⎤ = − +⎛ ⎞⎢ ⎥= − − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥= − = −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤ = − +⎛ ⎞⎢ ⎥= − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥= − =⎢ ⎥⎢ ⎥⎣ ⎦
The last row represents the equation 6z = . Substituting this result into 2.5y z− = (from the second row) gives
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Chapter 8 Cumulative Review
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2.56 2.5
8.5
y zy
y
− =− =
=
Substituting 6z = into 3 42.5x z+ = (from the first row) gives
( )3 42.5
3 6 42.524.5
x zx
x
+ =
+ =
=
Thus, flare jeans cost $24.50, camisoles cost $8.50, and t-shirts cost $6.00.
Chapter 8 Cumulative Review
1. 22 0x x− = ( )2 1 0
0 or 2 1 0 2 1
1 2
x xx x
x
x
− =
= − ==
=
The solution set is 10,2
⎧ ⎫⎨ ⎬⎩ ⎭
.
2. 3 1 4x + =
( )2 23 1 4
3 1 163 15
5
x
xxx
+ =
+ ===
Check: 3 5 1 4
15 1 4
16 44 4
⋅ + =
+ =
==
The solution set is { }5 .
3. 3 22 3 8 3 0x x x− − − = The graph of 3 2
1 2 3 8 3Y x x x= − − − appears to have an x-intercept at 3x = .
Using synthetic division:
3 2 3 8 3 6 9 3
2 3 1 0
− − −
Therefore,
( )( )( )( )( )
3 2
2
2 3 8 3 0
3 2 3 1 0
3 2 1 1 013 or or 12
x x x
x x x
x x x
x x x
− − − =
− + + =
− + + =
= = − = −
The solution set is 11, ,32
⎧ ⎫− −⎨ ⎬⎩ ⎭
.
4. 13 9x x+=
( ) 12
2 2
3 3
3 32 2
2
xx
x x
x xx
+
+
=
== += −
The solution set is { }2− .
5. ( ) ( )3 3log 1 log 2 1 2x x− + + =
( )( )( )( ) ( )
( )( )
3
2
2
2
log 1 2 1 2
1 2 1 3
2 1 9
2 10 02 5 2 0
5 or 22
x x
x x
x x
x xx x
x x
− + =
− + =
− − =
− − =
− + =
= = −
Since 2x = − makes the original logarithms
undefined, the solution set is 52
⎧ ⎫⎨ ⎬⎩ ⎭
.
6. 3x e=
( )ln 3 ln
ln 3 11 0.910
ln 3
x e
x
x
=
=
= ≈
The solution set is 1 0.910ln 3
⎧ ⎫≈⎨ ⎬⎩ ⎭
.
7. 3
4
2( )1
xg xx
=+
( )( )
( )3 3
4 4
2 2( )11
x xg x g xxx
− −− = = = −
+− +
Thus, g is an odd function and its graph is symmetric with respect to the origin.
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Chapter 8: Systems of Equations and Inequalities
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8. 2 2 2 4 11 0x y x y+ − + − = 2 2
2 2
2 2
2 4 11( 2 1) ( 4 4) 11 1 4
( 1) ( 2) 16
x x y yx x y y
x y
− + + =
− + + + + = + +
− + + =
Center: (1,–2); Radius: 4
9. 2( ) 3 1xf x −= +
Using the graph of 3xy = , shift the graph horizontally 2 units to the right, then shift the graph vertically upward 1 unit.
Domain: ( , )−∞ ∞ Range: (1, )∞
Horizontal Asymptote: 1y =
10. 5( )2
f xx
=+
52
5 Inverse2
( 2) 52 5
5 25 2 5 2
yx
xy
x yxy x
xy xxy
x x
=+
=+
+ =+ =
= −−
= = −
Thus, 1 5( ) 2f xx
− = −
Domain of f = { | 2}x x ≠ − Range of f = { | 0}y y ≠
Domain of 1f − = { | 0}x x ≠
Range of 1f − = { | 2}y y ≠ − .
11. a. 3 6y x= + The graph is a line. x-intercept: y-intercept:
0 3 63 6
2
xxx
= += −= −
( )3 0 66
y = +
=
b. 2 2 4x y+ = The graph is a circle with center (0, 0) and radius 2.
c. 3y x=
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Chapter 8 Cumulative Review
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d. 1yx
=
e. y x=
f. xy e=
g. lny x=
h. 2 22 5 1x y+ = The graph is an ellipse.
2 2
2 2
1 12 5
2 52 5
1
1
x y
x y⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+ =
+ =
i. 2 23 1x y− = The graph is a hyperbola
2 2
2 2
13
11 3
3
11
x y
x y
⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎜ ⎟⎜ ⎟⎝ ⎠
− =
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Chapter 8: Systems of Equations and Inequalities
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j. 2
2
2
2
2 4 1 0
2 1 4
4 ( 1)1 ( 1)4
x x y
x x y
y x
y x
− − + =
− + =
= −
= −
12. 3( ) 3 5f x x x= − +
a. Let 31 3 5Y x x= − + .
−3
−6
9
6
The zero of f is approximately 2.28− .
b.
−3
−6
9
6
−3
−6
9
6
f has a local maximum of 7 at 1x = − and a
local minimum of 3 at 1x = .
c. f is increasing on the intervals ( , 1)−∞ − and (1, )∞ .
Chapter 8 Projects
Project I 1. 80% = 0.80 18% = 0.18 2% = 0.02
40% = 0.40 50% = 0.50 10% = 0.10 20% = 0.20 60% = 0.60 20% = 0.20
2. 0.80 0.18 0.020.40 0.50 0.100.20 0.60 0.20
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
3. 0.80 0.18 0.02 1.000.40 0.50 0.10 1.000.20 0.60 0.20 1.00
+ + =+ + =+ + =
The sum of each row is 1 (or 100%). These represent the three possibilities of educational achievement for a parent of a child, unless someone does not attend school at all. Since these are rounded percents, chances are the other possibilities are negligible.
4.
2
2
0.8 0.18 0.020.4 0.5 0.10.2 0.6 0.2
0.716 0.246 0.0380.54 0.382 0.0780.44 0.456 0.104
P⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
Grandchild of a college graduate is a college graduate: entry (1, 1): 0.716. The probability is 71.6%
5. Grandchild of a high school graduate finishes college: entry (2,1): 0.54. The probability is 54%.
6. grandchildren → k = 2. (2) (0) 2
0.716 0.246 0.038[0.277 0.575 0.148] 0.54 0.382 0.078
0.44 0.456 0.104[0.573952 0.35528 0.070768]
v v P=
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
=College: 57%≈ High School: 36%≈ Elementary: 7%≈
7. The matrix totally stops changing significantly at
30
0.64885496 0.29770992 0.053435110.64885496 0.29770992 0.053435110.64885496 0.29770992 0.05343511
P⎡ ⎤⎢ ⎥≈ ⎢ ⎥⎢ ⎥⎣ ⎦
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Chapter 8 Projects
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Project II
a. 2 2 2 2 16× × × = codewords.
b. v uG= u will be the matrix representing all of the 4-digit information bit sequences.
0 0 0 00 0 0 10 0 1 00 0 1 10 1 0 00 1 0 10 1 1 00 1 1 11 0 0 01 0 0 11 0 1 01 0 1 11 1 0 01 1 0 11 1 1 01 1 1 1
u
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥
= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
(Remember, this is mod two. That means that you only write down the remainder when dividing by 2. )
0 0 0 0 0 0 00 0 0 1 1 1 00 0 1 0 1 0 10 0 1 1 0 1 10 1 0 0 0 1 10 1 0 1 1 0 10 1 1 0 1 1 00 1 1 1 0 0 01 0 0 0 1 1 11 0 0 1 0 0 11 0 1 0 0 1 01 0 1 1 1 0 01 1 0 0 1 0 01 1 0 1 0 1 01 1 1 0 0 0 11 1 1 1 1 1 1
v uG
v
=
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥
= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
c. Answers will vary, but if we choose the 6th row and the 10th row: 0101101 1001001 1102102 → 1100100 (13th row)
d. v uGVH uGH
==
0 0 00 0 00 0 00 0 0
GH
⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦
e.
1 1 10 1 11 0 1
[0 1 0 1 0 0 0] 1 1 01 0 00 1 00 0 1
[1 0 1]
rH
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
=
error code: 0010 000 r : 0101 000 0111 000 This is in the codeword list.
Project III
a. 1 1 1 1 1 1 13 4 5 6 7 8 9
TA ⎡ ⎤= ⎢ ⎥
⎣ ⎦
b. 1( )2.357
2.0357
T TB A A A Y
B
−=
−⎡ ⎤= ⎢ ⎥
⎣ ⎦
c. 2.0357 2.357y x= −
d. 2.0357 2.357y x= −
Project IV
Answers will vary.
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