Young Cat2e Ssm Ch9

663 663

CHAPTER 9 Section 9.1 Solutions -------------------------------------------------------------------------------- 1.

Solve the system: 1 ( )1 ( )

x yx y = + =

12

Solve (1) for x: 1x y= + (3) Substitute (3) into (2) and solve for y:

(1 ) 11 2 1

0

y yyy

+ + =+ =

=

Substitute 0y = into (1) to find x: 1x = . So, the solution is ( )1,0 .

3.


x yx y+ = =

12

Solve (1) for y: 7y x= (3) Substitute (3) into (2) and solve for x:

(7 ) 97 9

8

x xx x

x

= + =

=

Substitute 8x = into (1) to find y: 1y = . So, the solution is ( )8, 1 .

5.

Solve the system: 2 3 ( )

3 4 ( )x y

x y = =

12

Solve (2) for x: 3 4x y= + (3) Substitute (3) into (1) and solve for y:

2(3 4) 35 5

1

y yyy

+ == =

Substitute 1y = into (1) to find x: 2 ( 1) 3

2 21

xxx

===

.

So, the solution is ( )1, 1 .

7.

Solve the system: 3 5 ( )2 5 8 ( )x yx y+ = =

12

Solve (1) for y: 5 3y x= (3) Substitute (3) into (2) and solve for x:

2 5(5 3 ) 82 25 15 8

17 171

x xx x

xx

= + =

==

Substitute 1x = into (3) to find y: 2y = So, the solution is ( )1,2 .

Chapter 9

664

9.

Solve the system: 2 5 7 ( )3 5 ( )

u vu v+ = =

12

Solve (2) for v: 3 5v u= (3) Substitute (3) into (1) and solve for u:

3217

2 5(3 5) 717 32

u uuu

+ ===

Substitute 3217u = into (2) to find v: 32 96 85 1117 17 173( ) 5v v = = = .

So, the solution is 32 1117 17,u v= = .

11.

Solve the system: 2 7 ( )2 5 ( )

x yx y+ = =

12

Solve (1) for y: 7 2y x= (3) Substitute (3) into (2) and solve for x:

2 (7 2 ) 57 5

x x = =

So, the system is inconsistent. Thus, there is no solution.

13.

Solve the system: 4 1 ( )8 2 2 ( )

r sr s = =

12

Solve (1) for s: 4 1s r= (3) Substitute (3) into (2) and solve for r:

8 2(4 1) 22 2

r r ==

So, the system is consistent. There are infinitely many solutions.

15.

Solve the system: 5 3 15 ( )

10 6 30 ( )r sr s = + =

12

Solve (1) for r: 15 (3 15)r s= + (3) Substitute (3) into (2) and solve for s:

1510 (3 15) 6 30

6 30 6 3030 30

s ss s

+ + = + =

=

So, the system is consistent. There are infinitely many solutions.

17. Solve the system: 2 3 7 ( )3 7 24 ( )

x yx y = + =

12

Solve (1) for x: ( )12 3 7x y= (3) Substitute (3) into (2) and solve for y:

( )129 212 2

23 692 2

3 3 7 7 247 24

3

y yy y

y

y

+ = + =

==

Substitute this value back into (3) to obtain that 1x = .

19. Solve the system: 1 13 4

323 4

0 ( )2 ( )

x yx y = + =

12

First, clear the fractions by multiplying both equations by 12 to obtain the equivalent system:

4 3 08 9 24 ( )

x yx y = + =

(3)4

Solve (3) for x: 34x y= (5) Substitute (5) into (4) and solve for y:

( )348 9 243 24

8

y yy

y

+ ==

=

Substitute this back into (5) to obtain that 6x = .

Section 9.1

665

21. Solve the system: 7.2 4.1 7 ( )

3.5 16.5 2.4 ( )x y

x y = + =

12

Solve (1) for x: ( )17.2 7 4.1x y= + (3) Substitute (3) into (2) and solve for y:

( )( )17.23.5 7 4.1 16.5 2.40.4

y y

y

+ + ==

Substitute this value back into (3) to obtain that 1.2x = .

23.


x yx y = + =

12

Add (1) and (2) to eliminate y: 2 6

3 ( )xx== 3

Substitute (3) into (1) and solve for y: 3 2

1yy

==

So, the solution is ( )3,1 . 25.

Solve the system: 3 ( )

7 ( )x yx y = + =

12


2 ( )xx== 3


5yy

+ ==

So, the solution is ( )2,5 .

27.

Solve the system: 5 3 3 ( )3 3 21 ( )

x yx y+ = =

12


3 ( )xx= = 3

Substitute (3) into (1) and solve for y: 5( 3) 3 3

3 124

yyy

+ = ==

So, the solution is ( )3,4 . 29.

Solve the system: 2 7 4 ( )5 7 3 ( )

x yx y = + =

12


1 ( )xx== 3

Substitute (3) into (2) and solve for y:

27

5(1) 7 37 2

yyy

+ == =

So, the solution is ( )271, .

31.

Solve the system: 2 5 7 ( )3 10 5 ( )

x yx y+ = =

12

Multiply (1) by 2 : 4 10 14 ( )x y+ = 3 Add (2) and (3) to eliminate y:

197

7 19xx== (4)

Substitute (4) into (1) and solve for y: 197

1135

2( ) 5 7yy

+ ==

So, the solution is ( )19 117 35, .

Chapter 9

666

33. Solve the system: 2 5 5 ( )4 10 10 ( )

x yx y+ = =

12

Multiply (1) by 2: 4 10 10 ( )x y+ = 3 Add (2) and (3) to eliminate y: 0 0= (4) So, the system is consistent. Thus, there are infinitely many solutions. 35.

Solve the system: 3 2 12 ( )4 3 16 ( )x yx y = + =

12

Multiply (1) by 4: 12 8 48 ( )x y = 3 Multiply (2) by -3: 12 9 48x y = (4) Add (3) and (4) to eliminate x:

17 00

yy

== (5)

Substitute (5) into (1) and solve for x: 3 2(0) 12

4x

x =

= So, the solution is ( )4,0 .

37.

Solve the system: 6 3 15 ( )7 2 12 ( )

x yx y = + =

12

Multiply (1) by 2: 12 6 30 ( )x y = 3 Multiply (2) by 3: 21 6 36x y+ = (4) Add (3) and (4) to eliminate y:

33 66 2x x= = (5) Substitute (5) into (1) and solve for y:

( )6 2 3 153 3

1

yyy

= =

=

So, the solution is ( )2,1 . 39. Solve the system:

0.02 0.05 1.25 ( )0.06 0.15 3.75 ( )

x yx y+ = =

12

Multiply (1) by 3: 0.06 0.15 3.75 ( )x y+ = 3

Add (2) and (3) to eliminate y: 0 0= (4)

So, the system is consistent. Thus, there are infinitely many solutions.

41.

Solve the system: 1 13 2

715 2

1 ( )2 ( )

x yx y+ = + =

12

Multiply (1) by 7 : 7 73 2 7 ( )x y = 3 Add (2) and (3) to eliminate y:

3215

7532

5xx

= = (4)


3 32 21 212 96

716

( ) 1yyy

+ ===

So, the solution is ( )75 732 16, . 43. c Adding the equations yields 6 6, so thatx = 1x = . Thus,

3(1) 1, so that 2y y = = . 45. d Subtracting the equations yields 0 4= . Thus, the system is inconsistent and so, the lines should be parallel.

Section 9.1

667

47.

Notes on the graph: Solid curve is y x= Dashed curve is y x= So, the solution is (0,0).

49.

Notes on the graph: Solid curve is 2 3x y+ = Dashed curve is 2x y+ = So, the solution is ( 1, 1) .

51.

Notes on the graph: Solid curve is 1 22 3 4x y = Dashed curve is 14 6x y = So, the solution is (0, 6) .

53.

Notes on the graph: Solid curve is 1.6 4.8x y = Dashed curve is 0.8 0.5 1.5x y + = So, there is no solution.

55. Let x = # of Montblanc pens and y = # of Cross pens

Must solve the system: 69 ( )

72 10 1000 ( )x y

x y+ = + =

12

Solve (1) for y: 69y x= (3) Substitute (3) into (2): 72 10(69 ) 1000 5x x x+ = = Substitute this value of x into (3): 64y = So, have 5 Montblanc pens and 64 Cross pens.

Chapter 9

668

57. Let x = number of ml of 8% HCl. y = number of ml of 15% HCl. Must solve the system:

0.08 0.15 (0.12)(37) ( )37 ( )

x yx y

+ = + =12

Multiply (2) by 0.08 : 0.08 0.08 2.96 ( )x y = 3

Add (1) and (3) to eliminate x: 0.07 1.48 21.14y y= (4)

Substitute (4) into (2) and solve for x: 15.86x

So, should use approximately 15.86 ml of 8% HCl and 21.14 ml of 15% HCl.

59. Let x = total annual sales. Salary at Autocount: 15,000 0.10x+ Salary at Polk: 30,000 0.05x+ We need to find x such that

15,000 0.10 30,000 0.05x x+ > + . Solving this yields:

0.05 15,000 300,000x x> > So, he needs to make at least $300,000 of sales in order to make more money at Autocount.

61. Let x= gallons used for highway miles and y = gallons used for city miles Must solve the system:

26 19 349.5 ( )16 ( )

x yx y

+ = + =12

Multiply (2) by 19 : 19 19 304 ( )x y = 3 Add (1) and (3) to eliminate y: 7 45.5 6.5x x= = (4) Substitute (4) into (2) and solve for y: 9.5y = Thus, there are 169 highway miles and 180.5 city miles. 63. Let x = speed of the plane and y = wind speed.

Rate (mph) Time (hours) Distance Atlanta to Paris x y+ 8 4000 Paris to Atlanta x y 10 4000

So, using Distance = Rate Time, we see that we must solve the system:

8( ) 400010( ) 4000

x yx y+ = = , which is equivalent to

500400

x yx y+ = =

(1)(2)

Adding (1) and (2) yields: 2 900, so that 450x x= = . Substituting this into (1) then yields 50.y = So, the average ground speed of the plane is 450 mph, and the average wind speed is 50 mph.

Section 9.1

669

65. Let x = amount invested in 10% stock y = amount invested in 14% stock We must solve the system:

0.10 0.14 1260 ( )10,000 ( )

x yx y

+ = + =12

Multiply (2) by 0.10 : 0.10 0.10 1000 ( )x y = 3

Add (1) and (3) to eliminate x: 0.04 260

6500yy== (4)

Substitute (4) into (2) and solve for y: 3500x =

So, should invest $3500 in the 10% stock, and $6500 in the 14% stock.

67. Let x = # CD players sold Cost equation: 15 120y x= + (1) Revenue equation: 30y x= (2) We want the intersection of these two equations to find the break even point. To this end, substitute (2) into (1) to find that

8.x = So, must sell 8 CD players to break even.

69. Every term in the first equation is not multiplied by 1 correctly. The equation should be 2 3x y = , and the resulting solution should be 11, 25x y= = . 71. Did not distribute 1 correctly. In Step 3, the calculation should be

( 3 4) 3 4y y = + and the resulting answer should be (2, -2). 73. False. If the lines are coincident, then there are infinitely many solutions.

75. False. The lines could be coincident.

77. Substitute the pair (2, 3) into both equations to get the system:

2 3 292 3 13

A BA B = + =

(1)(2)

Add (1) and (2) to eliminate B: 4 16

4AA= = (3)

Substitute (3) into (1) and solve for B: 2( 4) 3 29

3 217

BBB

= =

=

Hence, the solution is 4, 7A B= = .

79. Let x = # cups of pineapple juice for 2% drink, and y = # cups of pomegranate juice for 2% drink. Must solve the system:

4(100 ) (4 )

0.02

0.04

yx y

yx y

+

+

= =(1)

(2)

This system, after simplification, is equivalent to:

0.02 0.98 00.04 0.96 0.16

x yx y = =

(3)(4)

Substitute (3) into (4): 0.020.98y x= (5) Substitute (5) into (4) and simplify:

7.84x = Substitute this value of x back into (5):

0.16y = So, she can make 8 cups of the 2% drink and 96 cups of the 4% drink.

Chapter 9

670

81.

Notes on the graph: Solid curve is 1.25 17.5y x= + Dashed curve is 2.3 14.1y x= The approximate solution is (8.9, 6.4).

83.

Notes on the graph: Solid curve is 23 15 7x y+ = Dashed curve is 46 30 14x y+ = Note that the graphs are the same line, so there are infinitely many solutions to this system.

85. The graph of the system of equations is as follows:

The solution is approximately (2.426, -0.059).

Section 9.2

671

Section 9.2 Solutions -------------------------------------------------------------------------------- 1. Solve the system:

630

x y zx y zx y z

+ = + + = =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2):

922 9z z= = (4)

Add (2) and (3): 322 3x x = = (5)

Step 2: Solve for the remaining variable using (4) and (5). Substitute (4) and (5) into (1):

3 92 2 6 3y y + = =

3. Solve the system: 2

36

x y zx y zx y z

+ = = + =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2): 122 1z z = = (4) Add (2) and (3), and sub. in (4):

( )122 2 3 2 2 32 4 2

x z x

x x

= = = = (5)


912 22 2y y + = =


39

x y zx y zx y z

+ = = + =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2): 2 2 1z z = = (4) Add (2) and (3), and sub. in (4):

( )2 2 12 2 2 1 122 10 5

x z x

x x

= = = = (5)


5 ( 1) 1 3y y + = =

Chapter 9

672

7. Solve the system: 2 3 4 3

2 15 2 3 7

x y zx y zx y z

+ = + + = =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (2) by 3:

3 3 6 3x y z + + = (4) Add (4) and (1) to eliminate y:

10 0x z + = (5) Next, multiply (2) by 2:

2 2 4 2x y z + + = (6) Add (6) and (3) to eliminate y:

3 9x z+ = (7) These steps yield the following system:

10 03 9

x zx z + = + =

(5)( )* (7)

Step 2: Solve system ( )* from Step 1. Multiply (5) by 3:

3 30 0x z + = (8) Add (8) and (7) to eliminate x and solve for z:

93131 9 so that z z= = (9)

Substitute (9) into (7) to find x: 931

9031

10( ) 0xx

+ == (10)

Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (9) and (10) into (2) to find y:

90 931 31

72 10331 31

( ) 2( ) 11 so that

yy y

+ + = = =

Thus, the solution is: 90 103 931 31 31, ,x y z= = = .

9. Solve the system: 4 3 5 22 3 2 32 4 3 1

x y zx y zx y z

+ + = = + + =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (2) and (3) to eliminate x:

2y z+ = (4) Next, multiply (2) by 2:

4 6 4 6x y z = (5) Add (5) and (1) to eliminate x:

3 4y z + = (6) These steps yield the following system:

23 4y zy z+ = + =

(4)( )* (6)

Step 2: Solve system ( )* from Step 1. Multiply (6) by 1 : 3 4y z = (7) Add (4) and (7) to eliminate z and solve for y:

124 2y y= = (8)

Substitute (8) into (4) to find z: 12

52

2zz

+ = = (9)

Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (8) and (9) into (2) to find y:

512 2

132

134

2 3( ) 2( ) 32

xxx

= = = (10)

Thus, the solution is: 13 514 2 2, ,x y z= = = .

Section 9.2

673


1

x y zy z

x y z

+ = = + + =

(1)(2)(3)

Add (1) and (3) to eliminate both x and y: 2 0

0zz== (4)

Substitute (4) into (2) to find y: 0 1y = (5)

Substitute (4) and (5) into (1) to find x: ( 1) 0 1

2x

x + =

= (6) Thus, the solution is:

2, 1, 0x y z= = = . 13. Solve the system:

3 2 3 14

2 3 5 14

x y zx y zx y z

= + = + + =

(1)(2)(3)


3 3 3 12x y z + = (4) Add (4) and (1) to eliminate z:

6 5 13x y = (5) Next, multiply (2) by 5 :

5 5 5 20x y z + = (6) Add (6) and (3) to eliminate z:

3 8 34x y + = (7) These steps yield the following system:

6 5 133 8 34x y

x y = + =

(5)( )* (7)

Step 2: Solve system ( )* from Step 1. Multiply (7) by 2:

6 16 68x y + = (8) Add (5) and (8) to eliminate x and solve for y:

11 555

yy== (9)

Substitute (9) into (5) to find x: 6 5(5) 13

6 122

xxx

= == (10)

Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (9) and (10) into (2) to find z:

2 5 41

zz

+ = = (11)

Thus, the solution is: 2, 5, 1x y z= = = .

Chapter 9

674

15. Solve the system: 3 2

2 3 42 4 6

x y zx y zx y z

= + = + =

(1)(2)(3)


3 6 9 12x y z+ = (4) Add (1) and (4) to eliminate x:

5 10 14y z = (5) Next, multiply (2) by -2:

2 4 6 8x y z + = (6) Add (3) and (6) to eliminate x:

5 10 2y z + = (7) These steps yield the following system:

5 10 145 10 2

y zy z = + =

(5)( )* (7)

Step 2: Solve system ( )* from Step 1. Adding (5) and (7) yields the false statement 0 = 12. Hence, this system has no solution.


4 3 152 3 12

x y zx y z

x y z

+ + = = + =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by -3:

9 6 3 12x y z = (4) Add (3) and (4) to eliminate z:

8 8 0 0x y x y = + = (5) Next, multiply (2) by 3:

12 9 3 45x y z = (6) Add (3) and (6) to eliminate z:

11 11 33 3x y x y = + = (7) These steps yield the following system:

03

x yx y+ = + =

(5)( )* (7)

Step 2: Solve system ( )* from Step 1. Subtracting (5) and (7) yields the false statement 0 = 3. Hence, this system has no solution.

Section 9.2

675


3 2 42 4 2 4

x y zx y zx y z

+ + = + = =

(1)(2)(3)


3 6 3 6x y z + + = (4) Add (4) and (2) to eliminate x:

4 4 2y z+ = (5) Next, multiply (1) by 2:

2 4 2 4x y z + + = (6) Add (6) and (3) to eliminate x:

0 0= (7) Hence, we know that the system has infinitely many solutions.

Let .z a= Then, substituting this value into (5), we can find the value of y:

12

2 2 1( )

y ay a= = +

Now, substitute the values of z and y into (1) to find x:

122( ) 2

2 1 21

x a ax a a

x a

+ + = + =

=

Thus, the solution is

121 , ( ),x a y a z a= = + = , where a is

any real number.


2 3 1110

x y zx y zx y z

= + = + + =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (3):

0 0= (4) Add (1) and (2):

3 4 1x y = (5)

Hence, we know that the system has infinitely many solutions.

To determine them, let x a= . Substitute this into (5) to find y:

14

3 4 1(3 1)

a yy a

= = + Now, substitute the

values of x and y into (1) to find z: 14

4 3 1 404

414

(3 1) 10a a

a

a a zzz

+ ===

Thus, the solution is:

4114 4, (3 1), ax a y a z = = + = .

Equivalently, 41 4 , 31 3 ,x a y a z a= + = + = .

Chapter 9

676


1 2 3

1 2 3

3 13

2 0

x x xx x xx x x

+ = + = + + =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2) to eliminate 2 3 and x x :

11

1 2

4 2xx= = (4)

Add (2) and (3) to eliminate 2x :

1 33 2 3x x+ = (5) These steps yield the following system:

11 2

1 33 2 3x

x x= + =

(4)( )* (5)

Step 2: Solve system ( )* from Step 1. Substitute (4) into (5):

132

33 2

33 4

3( ) 2 32

xxx

+ = = = (6)

Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (4) and (6) into (1) to find 2x :

3122 4

72 4

3( ) ( ) 1xx

+ == (7)

Thus, the solution is: 7 31

1 2 32 4 4, ,x x x= = = .


2 33 4 7 1

x yx y zx y z

+ = + = + =

(1)(2)(3)


7 14 7 21x y z+ = (4) Add (4) and (3) to eliminate z:

4 10 22x y+ = (5) These steps yield the following system:

2 5 94 10 22

x yx y+ = + =

(1)( )* (5)

Step 2: Solve system ( )* from Step 1. Multiply (1) by 2 :

4 10 18x y = (6) Add (5) and (6) to eliminate x and solve for y:

0 4= (7) Hence, we conclude from (7) that the system has no solution.

Section 9.2

677


1 2 3

1 2 3

2 32

2 2 2 4

x x xx x xx x x

+ = + = + =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by 1 :

1 2 32 3x x x + = (4) Add (2) and (4) to eliminate 2 3 and x x :

1

1

11

xx

= = (5)

Substitute (5) into (2): 2 3

2 3

1 21

x xx x

+ = + = (6)

Substitute (5) into (3): 2 3

2 3

2 2 2 41

x xx x

+ = = (7)

These steps yield the following system:

2 3

2 3

11

x xx x + = =

(6)( )* (7)

Step 2: Solve system ( )* from Step 1. Add (6) and (7): 0 = 0 (8) Hence, we conclude from (8) that there are infinitely many solutions. To determine them, let 3x a= . Substitute this value into (7) to see that 2 1x a= .


1 2 31, 1 ,x x a x a= = + = .


6x y z

x y z+ = =

(1)(2)

Since there are two equations and three unknowns, we know there are infinitely many solutions (as long as neither statement is inconsistent). To this end, let z a= . Add (1) and (2): 3 2 8x z = (3) Substitute z a= into (3) to find x: 823 3x a= + (4) Finally, substitute z a= and (4) into (2) to find y: 1013 3y a= Thus, the solution is

8 102 13 3 3 3, ,x a y a z a= + = = .

Chapter 9

678

31. Let x = number of touchdowns y = number of extra points z = number of field goals Solve the system:

186 3 66

4

x y zx y z

x z

+ + = + + = = +

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Substitute (3) into (1) and simplify:

2 14y z+ = (4) Substitute (3) into (2) and simplify:

9 42y z+ = (5)

These steps yield the following system: 2 149 42

y zy z+ = + =

(4)( )* (5)

Step 2: Solve system ( )* from Step 1. Subtract (5) (4):

7 28 4z z= = (6) Substitute (6) into (3) to solve for x:

8x = (7) Substitute (6) and (7) into (1) to find y:

6y = (8)

Thus, there were 8 touchdowns, 6 extra points, and 4 field goals.

33. Let x = # Mediterranean chicken sand. y = # Six-inch tuna sandwiches z = # Six-inch roast beef sandwiches We must solve the system:

14350 430 290 4840

18 19 5 190

x y zx y z

x y z

+ + = + + = + + =

(1)( ) (2)*

(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Solve (1) for x: 14x y z= (4) Substitute (4) into (2) and simplify:

4 3 3y z = (5) Substitute (4) into (3) and simplify:

13 62y z = (6) These steps yield the following system:

4 3 313 62

y zy z

= = (5)

( )* (6)

Step 2: Solve system ( )* from Step 1. Solve (6) for y:

13 62y z= (7) Substitute (7) into (5) and simplify:

5z = (8) Substitute (8) into (7) and simplify:

3y = (8) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (8) and (9) into (1) to find x:

6x = (10) Thus, there are: 6 Mediterranean chicken sandwiches, 3 Six-inch tuna sandwiches, 5 Six-inch roast beef sandwiches.

Section 9.2

679

35. The system that must be solved is: 21

0 0221

0 0221

0 02

36 (1) (1)

40 (2) (2)

12 (3) (3)

a v h

a v h

a v h

= + + = + + = + +

which is equivalent to 0 0

0 0

0 0

72 2 240 2 224 9 6 2

a v ha v ha v h

= + + = + + = + +

(1)(2)(3)

Step 1: Obtain a 2 2 system: Multiply (1) by 2 , and add to (2):

0 02 3 104v h = (4) Multiply (1) by 9 , and add to (3):

0 012 16 624v h = (5) These steps yield the following 2 2 system:

0 0

0 0

2 3 10412 16 624

v hv h

= = (4)(5)

Step 2: Solve the system in Step 1. Multiply (4) by 6 and add to (5):

0

0

2 0

0

h

h

== (6)

Substitute (6) into (4): 0

0

0

2 3(0) 1042 104

52

vv

v

= =

= (7)

Step 3: Find values of remaining variables. Substitute (6) and (7) into (1):

72 2(52) 2(0)

32

a

a

= + + =

Thus, the polynomial has the equation 216 52h t t= + .

37. Since we are given that the points (20,30), (40,60), and (60,40) , the system that must be solved is:

2

2

2

30 (20) (20)60 (40) (40)40 (60) (60)

a b ca b ca b c

= + + = + + = + +

which is equivalent to 30 400 2060 1600 4040 3600 60

a b ca b ca b c

= + + = + + = + +

(1)(2)(3)


40 3 60b c = (4) Multiply (1) by 9 , and add to (3):

120 8 230b c = (5) These steps yield the following 2 2 system:

40 3 60120 8 230

b cb c

= = (4)(5)

Step 2: Solve the system in Step 1. Multiply (4) by 3 and add to (5):

50c = (6) Substitute (6) into (4):

214

40 3( 50) 6040 210

5.25

bbb

= =

= = (7)


30 400 20(5.25) ( 50)25 400

0.0625

aa

a

= + + =

=

Thus, the polynomial has the equation

20.0625 5.25 50y x x= + .

Chapter 9

680

39. Let x = amount in money market y = amount in mutual fund z = amount in stock The system we must solve is:

600020,000

0.03 0.07 0.10 1180

x yx y z

x y z

= + + + = + + =

(1)(2)(3)

Step 1: Obtain a 2 2 system: Substitute (1) into both (2) and (3):

( 6000) 20,0000.03( 6000) 0.07 0.10 1180

y y zy y z

+ + + =+ + + =

This yields the following system:

2 14,0000.10 0.10 1000

y zy z

+ = + =(4)(5)

Step 2: Solve the system in Step 1. Multiply (4) by 0.10 and add to (5) to find y:

0.10 4004000

yy

= = (6)

Substitute (6) into (1) to find x: 6000 4000 10,000x = + = (7)

Substitute (6) and (7) into (2) to find z: 10,000 4000 20,000

6000zz

+ + ==

Thus, the following allocation of funds should be made: $10,000 in money market $4000 in mutual fund $6000 in stock

41. Let x = # regular models skis y = # trick skis z = # slalom skis Solve the system:

110 ( )2 3 3 297 ( )

2 5 202 ( )

x y zx y zx y z

+ + = + + = + + =

123

Step 1: Obtain a 2 2 system: Solve (1) for x: 110 ( )x y z= 4 Substitute (4) into (2), and simplify:

77 ( )y z+ = 5 Substitute (4) into (3), and simplify:

4 92 ( )y z+ = 6

This yields the following system: 77 ( )

(*)4 92 ( )

y zy z+ = + =

56

Step 2: Solve system (*) from Step 1. Subtract (6) (5):

3 15 5z z= = (7) Substitute (7) into (5):

72 ( )y = 8 Substitute (7) and (8) into (1):

72 5 110 33x x+ + = =

So, need to sell 33 regular model skis, 72 trick skis, and 5 slalom skis.

Section 9.2

681

43. Let x = # points scored in game 1 y = # points scored in game 2 z = # points scored in game 3 Solve the system:

2,591 ( )62 ( )2 ( )

x y zx yx z

+ + = = =

123

Step 1: Obtain a 2 2 system: Solve (1) for x: 2591 ( )x y z= 4 Substitute (4) into (2), and simplify:

2 2529 ( )y z+ = 5 Substitute (4) into (3), and simplify:

2 2589 ( )y z+ = 6

This yields the following system: 2 2529 ( )

(*)2 2589 ( )

y zy z+ = + =

56

Step 2: Solve system (*) from Step 1. Multiply (6) by -2:

2 4 5178 ( )y z = 7 Add (7) + (5):

3 2649 883z z = = (8) Substitute (8) into (6):

823 ( )y = 9 Substitute (9) and (8) into (1):

823 883 2591 885x x+ + = =

So, 885 points scored in game 1, 823 points scored in game 2, and 883 points scored in game 3.

45. Equation (2) and Equation (3) must be added correctly should be 2 2x y z + = . Also, should begin by eliminating one variable from Equation (1). 47. True 49. Substitute the given points into the equation to obtain the following system:

2 2

2 2

2 2

( 2) 4 ( 2) (4) 01 1 (1) (1) 0

( 2) ( 2) ( 2) ( 2) 0

a b ca b c

a b c

+ + + + = + + + + = + + + + =

which is equivalent to (after simplification) 2 4 20

22 2 8

a b ca b c

a b c

+ + = + + = + =

(1)(2)(3)

Multiply (2) by 2 and then, add to (3): 3 12 4c c= = (4)

Multiply (2) by 2 and then, add to (1): 6 3 24b c+ = (5)

Substitute (4) into (5) to find b: 6 3( 4) 24

6 122

bbb

+ = = = (6)

Finally, substitute (4) and (6) into (2) to find a:

( 2) ( 4) 24

aa

+ + = =

Thus, the equation is 2 2 4 2 4 0x y x y+ + = .

Chapter 9

682

51. We deduce from the diagram that the following points are on the curve:

( 2,46), ( 1,51), (0,44), (1,51), (2,43) Substituting these points into the given equation gives rise to the system:

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

4 3 2

4 3 2

4 3 2

4 3 2

4 3 2

46 2 2 2 2

51 1 1 1 1

44 0 0 0 0

51 1 1 1 1

43 2 2 2 2

a b c d e

a b c d e

a b c d e

a b c d e

a b c d e

= + + + + = + + + + = + + + + = + + + + = + + + +

Observe that the third equation above simplifies to

44e = . Substitute this value into the remaining four equations to obtain the following system:

16 8 4 2 277

16 8 4 2 1

a b c da b c da b c d

a b c d

+ = + = + + + = + + + =

(1)(2)(3)(4)

Now, add (2) and (3): 2 2 14a c+ = (5) Add (1) and (4): 32 8 1a c+ = (6) Solve the system:

2 2 1432 8 1

a ca c+ = + =

(5)(6)

Multiply (5) by 4 , and then add to (6): 5524

24 55a

a

= = (7)

Substitute (7) into (5) to find c: 5524

55 22324 24

2( ) 2 14

7

c

c

+ == + = (8)

At this point, we have values for three of the five unknowns. We now use (7) and (8) to obtain a 2 2 system involving these two remaining unknowns: Substitute (7) and (8) into (1):

55 22324 242 16( ) 8 4( ) 2

36 192 48b d

b d= + = (9)

Substitute (7) and (8) into (2): 55 22324 247

0b d

b d= + = + (10)

Solve the system: 36 192 480

b db d

= = +(9)(10)

Solve (10) for b: b d= (11) Substitute (11) into (9) to find d:

14

36 192( ) 4836 144

d dd

d

= =

= (12)

Substitute (12) into (10) to find b: 14b =

Hence, the equation of the polynomial is

4 3 255 2231 124 4 24 4 44y x x x x= + + + .

Section 9.2

683

53. Solve the system 2 34 3

7 3 3 22

y zx z

x y zx y z

+ = = = =

(1)(2)(3)(4)

Step 1: Obtain a 3 3 system: Solve (2) for z: 4 3z x= + (5) Substitute (5) into each of (1), (3), and (4). Simplifying yields the following system:

2 4 03 5 11

3 1

y xy xy x

+ = = =

(6)(7)(8)

Step 2: Solve the system in Step 1. Solve (6) for y: 2y x= (9) Substitute (9) into both (7) and (8) to obtain the following two equations:

3( 2 ) 5 11 so that 11x x x = = (10) ( 2 ) 3 1 so that 1x x x = = (11)

Observe that (10) and (11) yield different values of x. As such, there can be no solution to this system. 55. Solve the system

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

3 2 2 23 4 3 4

05 3 2 1

x x x xx x x x

x x x xx x x x

+ + = + + + = + + + = + + + =

(1)(2)(3)(4)

Step 1: Obtain a 3 3 system: Add (2) and (3):

2 3 44 5 4 4x x x+ + = (5) Multiply (2) by 3, and then add to (1):

2 3 47 13 11 10x x x+ + = (6) Multiply (2) by 5, and then add to (4):

2 3 418 21 17 19x x x+ + = (7) These steps yield the following system:

2 3 4

2 3 4

2 3 4

4 5 4 47 13 11 10

18 21 17 19

x x xx x xx x x

+ + = + + = + + =

(5)(6)(7)

CONTINUED ONTO NEXT PAGE!!

These steps yield the following system: 3 4

3 4

17 16 1287 79 47

x xx x

+ = + =(8)(9)

Solve this system: Multiply (8) by 87 , (9) by 17, and then add :

4

4

49 2455

xx

= = (10)

Substitute (10) into (8): 3

3

17 16(5) 124

xx

+ == (11)

Now, substitute (10) and (11) into (5): 2

2

4 5( 4) 4(5) 41

xx

+ + == (12)

Finally, substitute (10) (12) into (3): 1

1

1 4 5 02

xx

+ + == .

Chapter 9

684

Step 2: Solve the system in Step 1. Multiply (5) by 7 , (6) by 4, and then add them: 3 417 16 12x x+ = (8) Multiply (6) by 18, (7) by 7 , and then add them: 3 487 79 47x x+ = (9)


1 2 3 42, 1, 4, 5x x x x= = = = .

57. See the solution to #21. 59. Write the system in the form:

102 3 11

10

z x yz x yz x y

= = + =

A graphical solution is given to the right: Notice that there are only two planes since the first and third equations of the system are the same. Hence, the line of intersection of these two planes constitutes the infinitely many solutions of the system. This is precisely what was found to be the case in Exercises 21 and 57.

61. If your calculator has 3-dimensional graphing capabilities, then generate the graph of the system to obtain:

Its a bit difficult to tell from the graph, but the three planes intersect in a common point with coordinates ( )80 80 487 7 7, , , which can be easily verified using your calculator.

Section 9.3

685

Section 9.3 Solutions -------------------------------------------------------------------------------- 1. d ( )2 25 ( 5)( 5)x x x x x = + has three distinct linear factors 3. a ( )2 2 25x x + has one repeated linear factor and one irreducible quadratic factor. 5. b ( )22 25x x + has one linear factor and one repeated irreducible quadratic factor. 7.

( )( )29 9

20 5 4

5 4

x x x xA B

x x

= += + +

9.

( )3 2 22

2 5 2 54 4

4

x xx x x x

A B Cx x x

+ += = + +

11. Must long divide in this case.

2 3 2

3 2

2

2

2 65 2 4 7 3

(2 2 10 )6 3 3

( 6 6 30)3 33

xx x x x x

x x xx xx x

x

+ + + +

+ + +

+

So, 3 2

2 2

2 4 7 3 3 332 65 5

x x x xxx x x x + + += ++ + + +

13.

( ) ( )3

2 222 2

3 91010 10

x x Ax B Cx Dxx x

+ + += +++ +

15. The partial fraction decomposition has the form:

( )1

1 1A B

x x x x= ++ + (1)

To find the coefficients, multiply both sides of (1) by ( 1)x x + , and gather like terms: 1 ( 1)1 ( )

A x BxA B x A

= + += + + (2)

Equate corresponding coefficients in (2) to obtain the following system: 01

A BA

+ = =(3)

( )* (4)

Now, solve system ( )* : Substitute (4) into (3) to see that 1B = . Thus, the partial fraction decomposition (1) becomes:

( )1 1 1

1 1x x x x= + +

Chapter 9

686

17. Observe that simplifying the expression first yields x

x1

1( 1) xx= .

This IS the partial fraction decomposition. 19. The partial fraction decomposition has the form:

( )( )9 113 5 3 5x A B

x x x x = + + + (1)

To find the coefficients, multiply both sides of (1) by ( )3 ( 5)x x + , and gather like terms:

( )( ) ( )

9 11 ( 5) 3

9 11 5 3

x A x B x

x A B x A B

= + + = + + (2)

Equate corresponding coefficients in (2) to obtain the following system:

95 3 11

A BA B

+ = = (3)

( )* (4)

Now, solve system ( )* : Multiply (3) by 5 :

5 5 45A B = (5) Add (5) and (3) to solve for B:

8 567

BB

= = (6)

Substitute (6) into (3) to find A: 2A =

Thus, the partial fraction decomposition (1) becomes:

( )( )9 11 2 73 5 3 5x

x x x x = + + +


( ) ( )2 23 1

11 1x A B

xx x+ = + (1)

To find the coefficients, multiply both sides of (1) by ( )21x , and gather like terms:

( )3 1 ( 1)3 1

x A x Bx Ax B A+ = ++ = + (2)


31

AB A

= =(3)

( )* (4)

Substitute (3) into (4) to find B: 3 1 4B B = =


( ) ( )2 23 1 3 4

11 1x

xx x+ = +

Section 9.3

687

23. The partial fraction decomposition has the form: ( ) ( )2 24 3

33 3x A B

xx x = +++ +

(1)

To find the coefficients, multiply both sides of (1) by ( )23x + , gather like terms: ( )4 3 ( 3) 4 3 3x A x B x Ax A B = + + = + + (2)


3 3A

A B= + =

(3)( )* (4)

Substitute (3) into (4) to find B: 3(4) 3 15B B+ = = Thus, the partial fraction decomposition (1) becomes:

( ) ( )2 24 3 4 15

33 3x

xx x = +++ +


( )( ) ( )2

2 24 32 72

1 51 5 5x x A B C

x xx x x + = + ++ +

(1)

To find the coefficients, multiply both sides of (1) by ( )( )21 5x x+ , and gather like terms:

( ) ( )( ) ( )( ) ( ) ( )

( ) ( ) ( )

22

2 2

2

4 32 72 5 5 1 1

10 25 4 5 1

10 4 25 5

x x A x B x x C x

A x x B x x C x

A B x A B C x A B C

+ = + + + += + + + += + + + + + (2)


10 4 3225 5 72

A BA B CA B C

+ = + = + =

(3)( ) (4)*

(5)

Solve (3) for B: 4B A= (6) Substitute (6) into (4):

10 4(4 ) 32 which is equivalent to 6 16A A C A C + = + = (7) Substitute (6) into (5):

25 5(4 ) 72 which is equivalent to 30 92A A C A C + = + = (8) Now, solve the 2 2 system: 6 16

30 92A CA C

+ = + =(7)

(8)

Multiply (7) by 1 , and add to (8) to find A: 36 108 so that 3A A= = . Substitute this value for A into (7) to find C: 6(3) 16 so that 2C C + = = . Finally, substitute the value of A into (3) to find B: 3 4 so that 1B B+ = = . Thus, the partial fraction decomposition (1) becomes:

( )( ) ( )2

2 24 32 72 3 1 2

1 51 5 5x x

x xx x x + = + ++ +

Chapter 9

688


( )( )2

22

5 28 64 34 3

x x A Bx Cx xx x

+ += ++ ++ + (1)

To find the coefficients, multiply both sides of (1) by ( )( )24 3x x+ + , and gather like terms: ( ) ( )( )

( ) ( ) ( )

2 2

2 2

2

5 28 6 3 4

3 4 44 3 4

x x A x Bx C x

Ax A Bx Cx Bx CA B x B C x A C

+ = + + + += + + + + += + + + + + (2)


4 283 4 6

A BB C

A C

+ = + = + =

(3)( ) (4)*

(5)


4(5 ) 28 which is equivalent to 4 8A C A C + = + = (7) Now, solve the 2 2 system:

3 4 64 8A C

A C+ = + =

(5)(7)

Multiply (7) by 4 and then add to (5) to find A: 19 38 so that 2A A= = . Substitute this value for A into (7) to find C: 4( 2) 8 so that 0C C + = = . Finally, substitute the value of A into (3) to find B: 2 5 so that 7B B + = = .


( )( )2

22

5 28 6 2 74 34 3

x x xx xx x

+ = ++ ++ +

Section 9.3

689


( ) ( )2

22

Irreducible Quadratic Term

2 17 117 3 7 57 3 7 5

x x A Bx Cx x xx x x

+ += + + +

(1)

To find the coefficients, multiply both sides of (1) by ( )( )27 3 7 5x x x + , and gather like terms: ( ) ( )( )

( ) ( ) ( )

2 2

2 2

2

2 17 11 3 7 5 7

3 7 5 7 73 7 7 5 7

x x A x x Bx C x

Ax Ax A Bx Cx Bx CA B x A B C x A C

+ = + + + = + + + = + + + + (2)

Equate corresponding coefficients in (2) to obtain the following system: 3 2

7 7 175 7 11

A BA B C

A C

+ = + = =

(3)( ) (4)*

(5)

Solve (3) for B: 2 3B A= (6) Substitute (6) into (4):

7 7( 2 3 ) 17 which is equivalent to 14 31A A C A C + = + = (7) Now, solve the 2 2 system:

5 7 1114 31

A CA C = + =

(5)(7)

Multiply (7) by 7 and then add to (5) to find A: 103 206 so that 2A A= = . Substitute this value for A into (5) to find C: 5( 2) 7 11 so that 3C C = = . Finally, substitute the value of A into (3) to find B: 3( 2) 2 so that 4B B + = = .


( )( )2

22

2 17 11 2 4 37 3 7 57 3 7 5

x x xx x xx x x

+ = + + +

Chapter 9

690


( ) ( )3

2 222 299 9

x Ax B Cx Dxx x

+ += +++ + (1)

To find the coefficients, multiply both sides of (1) by ( )22 9x + , and gather like terms: ( )( ) ( )

( ) ( )

3 2

3 2

3 2

9

9 99 9

x Ax B x Cx D

Ax Bx Ax B Cx DAx Bx A C x B D

= + + + += + + + + += + + + + + (2)


9 09 0

AB

A CB D

= = + = + =

(3)(4)

( )* (5)(6)

Substitute (3) into (5) to find C: 9(1) 0 so that 9C C+ = = Substitute (4) into (6) to find D: 9(0) 0 so that 0D D+ = =


( ) ( )3

2 222 2

999 9

x x xxx x

= ++ +

Section 9.3

691


( ) ( )3 2

2 222 2

2 3 7 211 1

x x x Ax B Cx Dxx x

+ + += +++ + (1)

To find the coefficients, multiply both sides of (1) by ( )22 1x + , and gather like terms: ( )( ) ( )

( ) ( )

3 2 2

3 2

3 2

2 3 7 2 1x x x Ax B x Cx D

Ax Bx Ax B Cx DAx Bx A C x B D

+ = + + + += + + + + += + + + + + (2)


37

2

AB

A CB D

= = + = + =

(3)(4)

( )* (5)(6)

Substitute (3) into (5) to find C: 2 7 so that 5C C+ = = Substitute (4) into (6) to find D: 3 2 so that 1D D + = =


( ) ( )3 2

2 222 2

2 3 7 2 2 3 5 111 1

x x x x xxx x

+ += +++ +

Chapter 9

692


( )( ) ( )4 22 2 23 1 3 1 3 1

1 1 1 11 1 ( 1)( 1) 1x x x A B Cx D

x x x xx x x x x+ + + += = = + + + + + + + (1)

To find the coefficients, multiply both sides of (1) by ( )2( 1)( 1) 1x x x + + , and gather like terms: ( ) ( ) ( )2 2

3 2 3 2 3 2

3 2

3 1 ( 1) 1 ( 1) 1 ( 1)( 1)

( ) ( ) ( ) ( )

x A x x B x x Cx D x x

Ax Ax Ax A Bx Bx Bx B Cx Dx Cx DA B C x A B D x A B C x A B D

+ = + + + + + + += + + + + + + + = + + + + + + + (2)


A B CA B DA B CA B D

+ + = + = + = =

(3)(4)

( )* (5)(6)

To solve this system, first obtain a 3 3 system: Add (4) and (6): 2 2 1A B = (7) This enables us to consider the following 3 3 system:

03

2 2 1

A B CA B C

A B

+ + = + = =

(3)(5)(7)

Now, to solve this system, obtain a 2 2 system: Add (3) and (5): 2 2 3A B+ = (8) This enables us to consider the following 2 2 system:

2 2 12 2 3

A BA B = + =

(7)(8)

Add (7) and (8) to find A: 4 4 so that 1A A= = Substitute this value of A into (7) to find B: 122(1) 2 1 so that B B = = . Now, substitute these values of A and B into (6) to find D:

1 12 21 1 so that D D = =

Finally, substitute these values of A and B into (3) to find C: 31

2 21 0 so that C C+ + = = .


( ) ( )4 23 1 1 1 3 1

1 1 2 1 2 1x x

x x x x+ = + + + +

Section 9.3

693


( ) ( )2

22

Irreducible Quadratic Term

5 9 81 2 11 2 1

x x A Bx Cx x xx x x

+ += + + +

(1)

To find the coefficients, multiply both sides of (1) by ( )( )25 2 3 5x x x+ + , and gather like terms: ( ) ( )( )

( ) ( ) ( )

2 2

2 2

2

5 9 8 2 1 1

22

x x A x x Bx C x


+ = + + + = + + + = + + + + (2)


2 98

A BA B C

A C

+ = + = =

(3)( ) (4)*

(5)


3 14A C+ = (7) Now, solve the 2 2 system:

83 14

A CA C

= + =(5)(7)

Add (5) and (7) to find A: 2 6 so that 3A A= = . Substitute this value for A into (7) to find C: 3(3) 14 so that 5C C+ = = . Finally, substitute the value of A into (3) to find B: 2B = .


( )( )2

22

5 9 8 3 2 51 2 11 2 1

x x xx x xx x x

+ += + + +

Chapter 9

694


( )3 223 3

1 1 1( 1) 1x x A Bx C

x x x xx x x+= = + + + + + (1)

To find the coefficients, multiply both sides of (1) by ( )2( 1) 1x x x + + , and gather like terms: ( ) ( )

( ) ( )

2

2 2

2

3 1 ( 1)

( )

x A x x Bx C x


= + + + + = + + + + = + + + + (2)


A BA B C

A C

+ = + = =

(3)( ) (4)*

(5)

Solve (3) for B: B A= (6) Solve (5) for C: C A= (7) Substitute (6) and (7) into (4): ( ) ( ) 3 so that 1A A A A + = = . Substitute this value of A into (6) to find B: 1B = Finally, substitute this value of A into (7) to find C: 1C =


3 2

3 1 11 1 1

x xx x x x

= + + +

Section 9.3

695

41. The partial fraction decomposition has the form: ( )0

0 0

i

i i

f d d A Bd d d d+ = + (1)

To find the coefficients, multiply both sides of (1) by 0id d , and gather like terms: ( )0 0

0 0

i i

i i

f d d Ad Bdfd fd Ad Bd

+ = ++ = +

Equate corresponding coefficients in (2) to obtain the following system: A fB f= =

(3)( )* (4)

Thus, the partial fraction decomposition (1) becomes ( )0

0 0

( )ii i

f d d f fd d d d+ = + 5

Hence, using (5) enables us to write the lens law as 0

1i

f fd d

+ = , or as 0

1 1 1

id d f+ = .

43. The form of the decomposition is incorrect. It should be 2 1A Bx Cx x

++ + . Once this

correction is made, the correct decomposition is 21 2 3

1x

x x++ + .

45. False. The degree of the numerator must be less than or equal to the degree of the denominator in order to apply the partial fraction decomposition procedure.

Chapter 9

696

47. The first step in forming the partial fraction decomposition of 2

3 2

4 84 4

x xx x x

+ + is to

factor the denominator. To do so, we begin by applying the Rational Root Test: Factors of 4: 1, 2, 4 Factors of 1: 1 Possible Rational Zeros: 1, 2, 4 Applying synthetic division to the zeros, one can see that 1 is a rational zero:

1 1 1 4 41 0 4

1 0 4 0

So, 3 2 24 4 ( 1)( 4) ( 1)( 2)( 2)x x x x x x x x + = = + . Now, the partial fraction decomposition has the form:

2 2

3 2

4 8 4 84 4 ( 1)( 2)( 2) 1 2 2

x x x x A B Cx x x x x x x x x

+ + = = + + + + + (1) To find the coefficients, multiply both sides of (1) by ( 1)( 2)( 2)x x x + , and gather like terms:

2

2 2 2

2

4 8 ( 2)( 2) ( 1)( 2) ( 1)( 2)( 4) ( 2) ( 3 2)

( ) ( 3 ) ( 4 2 2 )

x x A x x B x x C x xA x B x x C x xA B C x B C x A B C

+ = + + + + = + + + += + + + + + (2)


3 44 2 2 8

A B CB C

A B C

+ + = = + =

(3)( ) (4)*

(5)

Now, solve system ( )* : Multiply (3) by 4 and then, add to (5): 2 6 4B C+ = (6) This leads to the following 2 2 system:

3 42 6 4B C

B C = + =

(4)(6)

Multiply (4) by 2 and then add to (6) to find C: 12 12 so that 1C C= = Substitute this value of C into (4) to find B: 3( 1) 4 so that 1B B = = . Finally, substitute these values of B and C into (3) to find A: 1 1 1 so that 1A A+ = = . Thus, the partial fraction decomposition (1) becomes:

2

3 2

4 8 1 1 14 4 1 2 2

x xx x x x x x

+ = + + +

Section 9.3

697

49. The partial fraction decomposition has the form: 3 2 3 2

4 3 3 2 3

2 1 2 1( 1) 1

x x x x x x A B C Dx x x x x x x x+ + = = + + ++ + + (1)

To find the coefficients, multiply both sides of (1) by 3( 1)x x + , and gather like terms: ( ) ( ) ( )

( ) ( ) ( )

3 2 2 3

3 2 2 3

3 2

2 1 1 1 1x x x Ax x Bx x C x Dx

Ax Ax Bx Bx Cx C DxA D x A B x B C x C

+ = + + + + + += + + + + + += + + + + + + (2)


11

A DA BB C

C

+ = + = + = =

(3)(4)

( )* (5)(6)

Substitute (6) into (5) to find B: 1 1 so that 0B B = = . Substitute this value of B into (4) to find A: 0 1 so that 1A A+ = = . Substitute this value of A into (3) to find D: 1 2 so that 1D D+ = = . Thus, the partial fraction decomposition (1) becomes:

3 2

4 3 3

2 1 1 1 11

x x xx x x x x+ = + + +

Chapter 9

698


( ) ( ) ( )5

3 2 322 2 2

211 1 1

x Ax B Cx D Ex Fxx x x

+ + + += + +++ + + (1)

To find the coefficients, multiply both sides of (1) by ( )32 1x + , and gather like terms: ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )

4 2

25 2 2

2 15 4 3 2 3 2

5 4 3 2

2 1 1

2 22 2

x x

x Ax B x Cx D x Ex F

Ax Bx Ax Bx Ax B Cx Dx Cx D Ex FAx Bx A C x B D x A C E x B D F

+ +

+ = + + + + + + +

= + + + + + + + + + + += + + + + + + + + + + + (2)


2 02 0

02

AB

A CB D

A C EB D F

= = + = + = + + = + + =

(3)(4)(5)

( )* (6)(7)(8)

Substitute (3) into (5) to find C: 2 0 so that 2C C+ = = . Substitute (4) into (6) to find D: 0 0 so that 0D D+ = = . Substitute the values of A and C into (7) to find E: 1 ( 2) 0 so that 1E E+ + = = . Substitute the values of B and D into (8) to find F: 0 0 2 so that 2F F+ + = = . Thus, the partial fraction decomposition (1) becomes:

( ) ( ) ( )5

3 2 322 2 2

2 2 211 1 1

x x x xxx x x

+ += +++ + +

Section 9.4

699

53.

Notes on the graph: Solid curve: Graph of 1y , Dashed curve: Graph of 2y In this case, since the graphs coincide, we know 2y is, in fact, the partial fraction decomposition of 1y .

55.

Notes on the graph: Solid curve: Graph of 1y , Dashed curve: Graph of 2y In this case, since the graphs do not coincide, we know 2y is not the partial fraction decomposition of 1y .

57. Since the graphs coincide, as seen below, y2 is the partial fraction decomposition of y1.

Section 9.4 Solutions -------------------------------------------------------------------------------- 1. d Above the line y x= , and do not include the line itself. 3. b Below the line y x= , and do not include the line itself.

Chapter 9

700

5.

The equation of dashed curve is 1y x= .

7.

The equation of the solid curve is y x= .

9.

The equation of the solid curve is

3 2y x= + .

11.


2 1y x= + 13. Write the inequality as 14 (2 3 )y x< .

The equation of the dashed curve is

14 (2 3 )y x= .

15.

The equation of the dashed curve is

53 5y x= + .

Section 9.4

701

17. Write the inequality as 2 3y x .


2 3y x= .

19.


32 3y x= + .

21.

Notes on the graph: C1: 1y x= C2: 1y x= +

23.

There is no common region, hence the system has no solution. Notes on the graph: C1: 2 1y x= + C2: 2 1y x=

25.

In this case, the common region is the line itself. Notes on the graph: C1 and C2: 2y x=

27.

Chapter 9

702

29.

Notes on the graph: C1: y x=

31.

Notes on the graph: C1: y x= C2: 0x = C3: 4y =

33.

Notes on the graph: C1: 2y x= + C2: 1y = C3: 0x =

35.


Section 9.4

703

37.

39.

Notes on the graph: C1: 1y x= C2: 3y x= + C3: 2y x= +

41.

Notes on the graph: C1: 4y x= C2: 2y x= + C3: 1y = C4: 1y =

43.

Notes on the graph: C1: 3y x= + C2: 1y x= +

45. There is no solution since the regions do not overlap.

Chapter 9

704

47.

49. The shaded region is a triangle, seen below. Note that the base b has length 4, and the height h is 2. Hence, the area is:

212Area = (4)(2) 4 units=

Notes on the graph: C1: y x= C2: 2y =

51.

Area of R1: 2512 2(1)(5) units= Area of R2: 21(5) 5 units= So, the area of the shaded region is 7.5 units2.

Section 9.4

705

53. Let x = number of cases of water and y = number of generators Certainly, 0, 0x y . Then, we also have the following restrictions: Cubic feet restriction: 20 2400x y+ Weight restriction: 25 150 6000x y+

So, we obtain the following system of inequalities: 0, 020 2400

25 150 6000

x yx y

x y

+ +

Notes on the graph: C1: 20 2400x y+ = C2: 25 150 6000x y+ = 55. First, find the point of intersection of

80 0.0120 0.02

P xP x= = +

Equating these and solving for x yields x = 2000. Then, substituting this into the first equation yields P = 60. Hence, the system for consumer surplus is:

80 0.01600

P xPx

57. The graph of the system in Exercise 55 is as follows:

The consumer surplus is the area of the shaded region, which is

212 (20)(2000) 20,000 units= .

59. The shading should be above the line. 61. True. The line cuts the plane into two half planes, and one must either shade above or below the line.

Chapter 9

706

63. False A dashed curve is used. 65. Given that and a b c d< < , the solution region is a shaded rectangle which includes the upper and left sides shown below:

67. For any value of b, the following system has a solution:

y ax by ax b + +

The solution region occurs in the first and fourth quadrants, and is shown graphically below:

If 0a = , then the system becomes:

y by b

The solution to this system is the horizontal line y b= .

Section 9.5

707

69.

The solid curve is the graph of 2 3y x= .

71.

Section 9.5 Solutions -------------------------------------------------------------------------------- 1.

Vertex Objective Function 2 3z x y= +

( 1,4) 2( 1) 3(4) 10z = + = (2, 4) 2(2) 3(4) 16z = + =

( 2, 1) 2( 2) 3( 1) 7z = + = (1, 1) 2(1) 3( 1) 1z = + =

So, the maximum value of z is 16, and the minimum value of z is 7 .

3. Vertex Objective Function

1.5 4.5z x y= + ( 1,4) 1.5( 1) 4.5(4) 16.5z = + = (2, 4) 1.5(2) 4.5(4) 21z = + =

( 2, 1) 1.5( 2) 4.5( 1) 7.5z = + = (1, 1) 1.5(1) 4.5( 1) 3z = + =

So, the maximum value of z is 21, and the minimum value of z is 7.5 .

5. The region in this case is:



(0,0) 7(0) 4(0) 0z = + = (0, 4) 7(0) 4(4) 16z = + =

So, the minimum value of z is 0.

Chapter 9

708


Notes on the graph: C1: 4y x= C2: y x= C3: 0x =


(0,0) 4(0) 3(0) 0z = + = (0,4) 4(0) 3(4) 12z = + =

Additional point (4,0)

4(4) 3(0) 16z = + =

Since the region is unbounded and the maximum must occur at a vertex, we conclude that the objective function does not have a maximum in this case.


Notes on the graph: C1: 2y x= + C2: 6y x= + C3: 4x =

Vertex Objective Function 2.5 3.1z x y= +

(2, 4) 2.5(2) 3.1(4) 17.4z = + = (0,0) 2.5(0) 3.1(0) 0z = + = (4, 2) 2.5(4) 3.1(2) 16.2z = + = (0,2) 2.5(0) 3.1(2) 6.2z = + =


Section 9.5

709


Notes on the graph: C1: 5y x= + C2: 7y x= + C3: 5y x= + C4: 3y x= + We need all of the intersection points since they constitute the vertices: Intersection of 5y x= + and

7y x= + : 5 7

2 21

x xxx

+ = +==

So, the intersection point is (1,6).

Intersection of 3y x= + and 7y x= + : 3 7

2 4 so that 2x x

x x+ = +

= = So, the intersection point is (2,5). Intersection of 3y x= + and 5y x= + :

3 52 2 so that 1

x xx x

+ = += =

So, the intersection point is (1,4). Intersection of 5y x= + and 5y x= + :

5 52 0 so that 0

x xx x

+ = += =

So, the intersection point is (0,5). Now, compute the objective function at the vertices:

Vertex Objective Function 1 24 5z x y= +

(1,6) 531 24 5 20(1) (6)z = + = (2,5) 51 24 5 2(2) (5)z = + = (1,4) 371 24 5 20(1) (4)z = + = (0,5) 1 24 5(0) (5) 2z = + =

So, the maximum value of z is 5320 2.65= .

13. Let x = number of Charley T-shirts y = number of Francis T-shirts Profit from Charley T-shirts:

Revenue cost = 13 7 6x x x = Profit from Francis T-shirts:

Revenue cost = 10 5 5y y y =

So, to maximize profit, we must maximize the objective function 6 5z x y= + . We have the following constraints:

7 5 10001800, 0

x yx y

x y

+ +

CONTINUED ONTO NEXT PAGE!

Notes on the graph: C1: 75 200y x= + C2: 180y x= + C3: 0x = We need all of the intersection points since they constitute the vertices: Intersection of 7 5 1000x y+ = and 180x y+ = :

75

25

180 2002050

x xxx

+ = +==

So, the intersection point is (50, 130).

Chapter 9

710

The region in this case is:

Now, compute the objective function at the vertices:


(0,180) 6(0) 5(180) 900z = + = 1000

7( ,0) 100076( ) 5(0) 857.14z = + (50, 130) 6(50) 5(130) 950z = + =

So, to attain a maximum profit of $950, she should sell 130 Francis T-shirts, and 50 Charley T-shirts.

15. Let x = # of desktops y = # of laptops

We must maximize the objective function

500 300z x y= + subject to the following constraints:

5 3 90700 400 10,000

30, 0

x yx y

y xx y

+ +


Notes on the graph: C1: 5330y x= C2: 7425y x= C3: 0x = C4: 0y = C5: 3y x=

We need all of the intersection points since they constitute the vertices: Intersection of 5 3 90x y+ = and 700 400 10,000x y+ = :

5 73 430 25

360 20 300 2160

x xx xx

= =

=

So, the intersection point is ( 60,130) . Not a vertex, however, since x must be non-negative (see the region). Intersection of C5 and C2:

74

19 1004 19

3 2525 so that

x xx x= = =

So, the intersection point is ( )100 30019 19, Now, compute the objective function at the vertices:


(0,25) 500(0) 300(25) 7500z = + = (0,0) 500(0) 300(0) 0z = + =

( )100 30019 19, 140,000100 30019 19 19500( ) 300( )z = + = In order to attain a maximum profit of $7500, he must sell 25 laptops and 0 desktops.

Section 9.5

711

17. Let x = # first class cars y = # second class cars Let p denote the profit for each second class car. Then, the profit for each first class car is 2p. Hence, we seek to maximize the objective function

2 ( 2 )z py px p y x= + = + . We have the following constraints:

302 4

8

x yx

y x

+ =


Notes on the graph: C1: 8y x= C2: 30y x= + C3: 2x =

We need all of the intersection points since they constitute the vertices: Intersection of 30x y+ = and 8y x= :

103

30 830 9 so that

x xx x

+ == =

So, the intersection point is 10 803 3( , ) . Now, compute the objective function at the vertices:

Vertex Objective Function (2 )z p x y= +

(2, 16) (2(2) 16) 20z p p= + = (2, 28) (2(2) 28) 32z p p= + =

10 803 3( , ) 10 80 1003 3 3(2( ) )z p p= + =

(3, 27) (2(3) 27) 33z p p= + = The maximum in this case would occur at

10 803 3( , ) . However, this is not tenable since

one cannot have a fraction of a car. However, very near at the vertex (3, 27) the profit is very near to this one, as are the number of cars. Hence, to maximize profit, they should use: 3 first class cars and 27 second class cars.

Chapter 9

712

19. Let x = # of regular skis y = # of slalom skis

We must maximize the objective function 25 50z x y= + subject to the following constraints:

400200, 800, 0

x yx yx y

+


Now, compute the objective function at the vertices:


(200,80) 25(200) 50(80) 9000z = + =

(200,200) 25(200) 50(200) 15,000z = + =

(320,80) 25(320) 50(80) 12,000z = + =

So, he should sell 200 of each type of ski.

21. Should compare the values of the objective function at the vertices rather than comparing the y-values of the vertices. 23. False. The region could be the entire plane (i.e., unconstrained). 25. Assume that 2a > . Then, the region looks like:

Notes on the graph: C1: y ax a= C3: y ax a= + C2: y ax a= + C4: y ax a= Now, compute the objective function at the vertices:

Vertex Objective Function 2z x y= +

(0, )a 2(0)z a a= + = ( 1,0) 2( 1) 0 2z = + = (0, )a 2(0)z a a= = (1, 0) 2(1) 0 2z = + =

Since 2a > , the maximum in this case would occur at (0, )a and is a.

Section 9.5

713

27. This is the same as #9 the minimum occurs at (0, 0) and is 0. 29. Notes on the graph:

C1: 3.7y x= + C2: 11.2y x= + C3: 4.5y =


(0.8,4.5)

17(0.8) 14(4.5) 76.6z = + =

(6.7,4.5)

17(6.7) 14(4.5) 176.9z = + =

(3.75,7.4

17(3.75) 14(7.45) 168.05z = + =

The maximum occurs at (6.7, 4.5) and is 176.9.

31.


(-2.2, 4.58) -1.656 (0.444, 12.778) 24.9984

(3.75, -3.75) 10.125 The maximum occurs at (0.444,12.778) and is approximately 25.

Chapter 9

714

Chapter 9 Review Solutions------------------------------------------------------------------------

1. Solve the system: 3 ( )3 ( )

r sr s = + =

12

Add (1) and (2): 2 6 so that 3r r= = Substitute this value of r into (1) to find s: 3 3 so that 0s s = = . So, the solution is ( )3,0 .

3. Solve the system: 4 2 3 ( )

4 5 ( )x y

x y + = =

12

Add (1) and (2) to eliminate x: 8 ( )y = 3

Substitute (3) into (2) and solve for x:

134

4 8 54 13

xxx

===

So, the solution is ( )134 ,8 . 5. Solve the system:

3 ( )1 ( )

x yx y+ = =

12

Solve (1) for y: 3y x= (3) Substitute (3) into (2) and solve for x:

(3 ) 13 1

2 4 so that 2

x xx x

x x

= + =

= =

Substitute this value of x into (3) to find y: 1y = .

So, the solution is ( )2, 1 .

7. Solve the system: 4 4 3 ( )

4 ( )c dc d = + =

12

Solve (2) for c: 4c d= (3) Substitute (3) into (1) and solve for d:

138

4(4 ) 4 316 8 3

8 13

d dddd

= =

==

Now, substitute this value of d into (3) to find c: 13 198 84c = = So, the solution is ( )19 138 8, .

9.

Notes on the graph: Solid curve: 12y x= Dashed curve: 12 2y x= + So, the solution is ( 2,1) .

Chapter 9 Review

715

11.

Notes on the graph: (Careful! The curves are very close together in a vicinity of the point of intersection.) Solid curve: 1.3 2.4 1.6x y = Dashed curve: 0.7 1.2 1.4x y = So, the solution is (12,5.83) . 13. Solve the system:

5 3 21 ( )2 7 20 ( )

x yx y = + =

12

To eliminate x, multiply (1) by 2: 10 6 42x y = (3)

Multiply (2) by 5: 10 35 100x y + = (4) Add (3) and (4): 29 58 2y y= = Substitute this into (1) to find x:

5 15 3x x= = So, the solution is ( )3, 2 .

15. Solve the system: 10 7 24 ( )7 4 1 ( )

x yx y = + =

12

To eliminate y, multiply (1) by 4: 40 28 96x y = (3)

Multiply (2) by 7: 49 28 7x y+ = (4) Add (3) and (4): 89 89 1x x= = Substitute this into (1) to find y:

7 14 2y y = = So, the solution is ( )1,2 .

17. c The intersection point is 25 211 11( , ) 19. d Multiplying the first equation by 2 reveals that the two equations are equivalent. Hence, the graphs are the same line.

Chapter 9

716

21. Let x = number of ml of 6% NaCl. y = number of ml of 18% NaCl. Must solve the system:

0.06 0.18 (0.15)(42) ( )42 ( )

x yx y

+ = + =12

First, for convenience, simplify (1) to get the equivalent system: 3 105

42x yx y+ = + =

(3)(4)

Multiply (1) by 1 , and then add to (2): 632

2 6331.5

yy

= = = (6)

Substitute (6) into (4) to find x: 31.5 42

10.5x

x+ =

= So, should use approximately 10.5 ml of 6% NaCl and 31.5 ml of 18% NaCl. 23. Solve the system:

13

3

x y zx y z

x y z

+ + = = + + =

(1)(2)(3)

Add (2) and (3): 0 0= (4) Hence, we know that the system has infinitely many solutions. Add (1) and (2):

2 2 so that 1x x= = (5)

Substitute this value of x into (1): 1 1

2y zy z

+ + =+ = (6)

Let z a= . Substitute this into (6) to find z: 2y a= +

Thus, the solutions are:

1, 2,x y a z a= = + = .


175

x y zx y z

y z

+ + = = + =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Subtract (1) - (2) and simplify: 5y z+ = (4) Solve the system:

5 ( )(*)

5 ( )y zy z+ = + =

34

Note that equating (3) and (4) yields the false statement 5 = -5. Hence, the system has no solution.

Chapter 9 Review

717

27. Since we are given that the points (16,2), (40,6), and (65,4) , the system that must be solved is:

2

2

2

2 (16) (16)6 (40) (40)4 (65) (65)

a b ca b ca b c

= + + = + + = + +

which is equivalent to 2 256 166 1600 404 4225 65

a b ca b ca b c

= + + = + + = + +

(1)(2)(3)


1344 24 4a b+ = (4) Multiply (1) by 1 , and add to (3):

3969 49 2a b+ = (5) These steps yield the following 2 2 system:

Step 2: Solve the system in Step 1. 1344 24 43969 49 2

a ba b+ = + =

(4)(5)

Multiply (4) by 49 and (5) by 24. Then, add them together:

29,400 1480.0050

aa= (6)

Substitute (6) into (4): 1344( 0.005) 24 4

0.4486bb

+ = (7)


256( 0.005) 16(0.447) 23.8884

cc

+ + =

Thus, the polynomial has the approximate equation 20.0050 0.4486 3.8884y x x= + .

29.

2 2

4( 1) ( 3)( 5) 1 ( 1) 3 5

A B C Dx x x x x x x

= + + + + +

31.

( )2212

(4 5)(2 1) 4 5 2 1 2 1A B C D

x x x x x x x= + + ++ + + + +

33.

( )( )23 3

12 3 4 3 4A B

x x x x x x= = ++ + +

35.

( ) ( )3 2

2 222 2

3 4 56 621717 17

x x x Ax B Cx Dxx x

+ + + + += +++ +

Chapter 9

718


( )9 23

( 1) 7 1 7x A B

x x x x+ = + + + (1)

To find the coefficients, multiply both sides of (1) by ( 1)( 7)x x + , and gather like terms: 9 23 ( 7) ( 1) ( ) (7 )x A x B x A B x A B+ = + + = + + (2) Equate corresponding coefficients in (2) to obtain the following system:

97 23

A BA B+ = =

(3)( )* (4)

Now, solve system ( )* : Add (3) and (4): 8 32 4A A= = Substitute this value of A into (3) to see that 5B = . Thus, the partial fraction decomposition (1) becomes:

( )9 23 4 5

( 1) 7 1 7x

x x x x+ = + + +


( )2 2

3

13 90 25 13 90 252 50 2 ( 5) 5 2 5 5x x x x A B C

x x x x x x x x+ + = = + + + + (1)

To find the coefficients, multiply both sides of (1) by ( )2 ( 5) 5x x x + , and gather like terms:

2 2

2

13 90 25 ( 25) (2 )( 5) (2 )( 5)( 2 2 ) (10 10 ) 25

x x A x B x x C x xA B C x B C x A

+ = + + + = + + + (2)

Equate corresponding coefficients in (2) to obtain the following system: 2 2 13

10 10 9025 25

A B CB C

A

+ + = = =

(3)( ) (4)*

(5)

Now, solve system ( )* : Solve (5): 1A = Substitute this value of A into (3): 6B C+ = (6) Note that (4) is equivalent to 9B C = (7). So, solve the system:

6 ( )9 ( )

B CB C+ = =

67

Add (6) and (7): 1522 15B B= = Substitute this value of B into (6): 32C = Thus, the partial fraction decomposition (1) becomes:

( ) ( )2

3

13 90 25 1 15 32 50 2 2 5 2 5x x

x x x x x+ = + +

Chapter 9 Review

719


( )2

1 1A B

x x x x= ++ + (1)

To find the coefficients, multiply both sides of (1) by ( 1)x x + , and gather like terms: 2 ( 1)2 ( )

A x BxA B x A

= + += + + (2)


A BA

+ = =(3)

( )* (4)

Now, solve system ( )* : Substitute (4) into (3) to see that 2B = . Thus, the partial fraction decomposition (1) becomes: ( )

2 2 21 1x x x x

= ++ +


( ) ( )2 25 17

22 2x A B

xx x = +++ + (1)

To find the coefficients, multiply both sides of (1) by ( )22x + , and gather like terms: ( )

5 17 ( 2)5 17 2

x A x Bx Ax A B = + + = + + (2)


2 17A

A B= + =

(3)( )* (4)

Substitute (3) into (4) to find B: 2(5) 17 so that 27B B+ = =


( ) ( )2 25 17 5 27

22 2x

xx x = ++ +

Chapter 9

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45.

The solid curve is the graph of 2 3y x= + .

47. Shade above the dashed line 14 (5 2 )y x= .

49. Shade above the solid line 3 2y x= + .

51. Shade below the solid line 38 2y x= + .

53.

Notes on the graph: C1: 2y x= + C2: 2y x= Since there is no region in common with both inequalities, the system has no solution.

55.

Notes on the graph: C1: y x= C2: 2x =

Chapter 9 Review

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57.

Notes on the graph: C1: 34 4y x= C2: 533y x= 59. The region in this case is:

Vertex Objective Function

2z x y= + (0,0) 2(0) (0) 0z = + = (3,0) 2(3) (0) 6z = + = (0,3) 2(0) (3) 3z = + =


Chapter 9

722


We need all of the intersection points since they constitute the vertices: Intersection of 8y x= + and y x= :

82 8

4

x xxx

+ ===

So, the intersection point is (4,4). Vertex Objective Function

2.5 3.2z x y= + (0,0) 2.5(0) 3.2(0) 0z = + = (0,8) 2.5(0) 3.2(8) 25.6z = + = (4,4) 2.5(4) 3.2(4) 22.8z = + =

So, the maximum value of z is 25.6.

63. The region in this case is: Vertex Objective Function

3 5z x y= ( )3,0 9 (0,6) -30

So, the minimum value of z is -30 and occurs at (0,6).

Chapter 9 Review

723

65. Let x = number of ocean watercolor coaster sets y = number of geometric shape coaster sets Profit from ocean watercolor sets:

Revenue cost = 15x Profit from geometric shape sets:

Revenue cost = 8y So, to maximize profit, we must maximize the objective function 15 8z x y= + . We have the following constraints:

4 2 1003 2 90

0, 0

x yx y

x y

+ +


Notes on the graph: C1: 50 2y x= C2: 3245y x= C3: 0y = We need all of the intersection points since they constitute the vertices: Intersection of 4 2 100x y+ = and 3 2 90x y+ = :

32

12

50 2 455

10

x xx

x

= ==

So, the intersection point is (10, 30). Now, compute the objective function at the vertices:


(0,0) 15(0) 8(0) 0z = + = (0, 45) 15(0) 8(45) 360z = + = (10, 30) 15(10) 8(30) 390z = + = (25,0) 15(25) 8(0) 375z = + =

So, to attain maximum profit, she should sell 10 ocean watercolor coaster sets and 30 geometric shape coaster sets.

Chapter 9

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67. The graph of this system of equation is as follows:

The solution is (2, -3).

69. If your calculator has 3-dimensional graphing capabilities, graph the system to obtain the following:

It is difficult to see from this graph, but zooming in will enable you to see that the solution is (3.6, 3, 0.8).

71. Since the graphs coincide, y2 is the partial fraction decomposition of y1.

73. The region is as follows.

Chapter 9 Practice Test

725

75.


(-2.7, 2.6) -12.84 (-7.5, -11.8) -64.2

(1.8, 0.6) 12.06 The maximum is 12.06 and occurs at (1.8, 0.6).

Chapter 9 Practice Test----------------------------------------------------------------------------- 1. Solve the system:

2 1 ( )3 2 ( )

x yx y = + =

12

Add (1) and (2): 3y = Substitute 3y = into (1) to find x:

2(3) 1 7x x = = . So, the solution is ( )7,3 .

3. Solve the system: 2 ( )

2 2 4 ( )x y

x y = + =

12

Multiply (1) by 2, and then add to (2): 0 = 0

So, the system is consistent. There are infinitely many solutions of the form

, 2x a y a= = . 5. Solve the system:

12 0

2 0

x y zx y z

x y z

+ + = + + = + + =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by 2 , and then add to (2) to eliminate x:

2y z = (4) Add (1) and (3) to eliminate x:

2 3 1y z+ = (5) These steps yield system:

22 3 1

y zy z = + =

(4)( )* (5)

Step 2: Solve system ( )* from Step 1. Multiply (4) by 2, and then add to (5):

3z = (6) Substitute (6) into (4) to find y:

3 2 so that 5y y = = (7) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (6) and (7) into (1) to find x:

5 3 11

xx

+ = =

Thus, the solution is: 1, 5, 3x y z= = = .

Chapter 9

726


2 4 213 5 14

x y zx y z

x y z

+ = + + = + =

(1)(2)(3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by 2, and then add to (2) to eliminate x:

3 10 31y z + = (4) Multiply (1) by 3 , and then add to (3) to eliminate x:

8 29y z = (5) These steps lead to the system:

3 10 318 29

y zy z

+ = = (4)(5)

Step 2: Solve the system in Step 1. Multiply (5) by 3: 3 24 87y z = (6) Add (4) and (6): 14 56 4z z = = Substitute this value of z into (6) to find y:

8(4) 29 3y y = = Substitute the values of y and z into (1) to find x: 1x = .


( )2 5

1 1x A B

x x x x+ = ++ + (1)

To find the coefficients, multiply both sides of (1) by ( 1)x x + , and gather like terms:

2 5 ( 1)2 5 ( )

x A x Bxx A B x A+ = + ++ = + + (2)


25

A BA

+ = =(3)

( )* (4)

Now, solve system ( )* : Substitute (4) into (3) to see that 3B = . Thus, the partial fraction decomposition (1) becomes:

( )2 5 5 3

1 1x

x x x x+ = + +


2 2

7 5( 2) 2 ( 2)

x A Bx x x

+ = ++ + + (1) To find the coefficients, multiply both sides of (1) by 2( 2)x + , and gather like terms:

7 5 ( 2) (2 )x A x B Ax A B+ = + + = + + (2) Equate corresponding coefficients in (2) to obtain the following system:

72 5

AA B

= + =(3)

( )* (4)

Now, solve system ( )* : Substitute (3) into (4) to see that 9B = . Thus, the partial fraction decomposition (1) becomes: 2 2

7 5 7 9( 2) 2 ( 2)

xx x x

+ = + + +


727


( )( )5 3

3 3 3 3x A B C

x x x x x x = + + + + (1)

To find the coefficients, multiply both sides of (1) by ( )( )3 3x x x + , and gather like terms:

( )( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )2 2 2

2

5 3 3 3 3 3

9 3 3

3 3 9

x A x x Bx x Cx x

A x B x x C x x

A B C x B C x A

= + + + + = + + + = + + + + (2)


3 3 59 3

A B CB C

A

+ + = = =

(3)( ) (4)*

(5)

Solve (5) for A: 13A = (6) Substitute (6) into (1) to eliminate A: 13B C+ = (7) Now, solve the 2 2 system:

13

3 3 5B CB C = + =

(4)(7)

Multiply (7) by 3, and then add to (4) to find B: 236 4 so that B B= = . Substitute this value for B into (7) to find C: 2 13 3 so that 1C C+ = = . Finally, substitute the values of B and C into (1) to find A: 2 13 31 0 so that A A+ = = . Thus, the partial fraction decomposition (1) becomes:

( )( ) ( )5 3 1 2 1

3 3 3 3 3 3x

x x x x x x = + + +

15.

The dashed curve is the graph of

2 6y x= + .

17.

Notes on the graph: C1: 4y x= C2: 2y x=

Chapter 9

728

19.

Notes on the graph: C1: 52 4y x= C2: 34 3y x= + 21. The region in this case is:

Notes on the graph: C1: 3y x= + C2: 1y x= + C3: 0x =

We need all of the intersection points since they constitute the vertices: Intersection of 3y x= + and 1y x= + :

3 12 2

1

x xxx

+ = +==

So, the intersection point is (1,2).

Vertex Objective Function

5 7z x y= + (0,3) 5(0) 7(3) 21z = + = (0,1) 5(0) 7(1) 7z = + = (1,2) 5(1) 7(2) 19z = + =



729

23. Let x = amount in money market y = amount in aggressive stock z = amount in conservative stock The system we must solve is:

100030,000

0.03 0.12 0.06 1890

y zx y z

x y z

= + + + = + + =

First, simplify this system by multiplying the third equation by 100:

100030,000

3 12 6 189,000

y zx y z

x y z

= + + + = + + =

(1)(2)(3)

Step 1: Obtain a 2 2 system: Substitute (1) into both (2) and (3):

( 1000) 30,0003 12( 1000) 6 189,000

x z zx z z

+ + + =+ + + =

which are equivalent to 2 29,000

3 18 177,000x z

x z+ = + =

(4)(5)

Step 2: Solve the system in Step 1. Multiply (4) by 3 and add to (5) to find z:

12 90,000 7500z z= = (6) Substitute (6) into (1) to find y:

7500 1000 8500y = + = (7) Substitute (6) and (7) into (2) to find x:

8500 7500 30,00014,000

xx

+ + ==

Thus, the following allocation of funds should be made: $14,000 in money market $8,500 in aggressive stock $7,500 in conservative stock

25. If your calculator has 3-dimensional graphing capabilities, use it to graph the system of equations to obtain:

It is difficult to see from the graph, but if you zoom in, you will see that the solution is (11,19,1).

Chapter 9

730

Chapter 9 Cumulative Review---------------------------------------------------------------------

1. 2 (6 5)6 11 530 25

xx xx

+ =( 1)

5 (6 5)x

x

1

5x =

3. The LCD is ( 1)x x + , 0, 1x . Multiply both sides by the LCD and solve for x:

5( 1) 5 5 5 5x x+ = = Since this statement is false, the equation has no solution. 5.

2 22 3 ( 1) 3 (3 )0 0 01 1 1 1

t t t t t t tt t t t

+

CPs: 0,1,3t = 0 1 3| | |

+ +

The solution set is ( ] ( ],0 1,3 .

7. The slope is 6.2 5.0 0.55.6 3.2

m = = . To find b, use the point (3.2, 5.0):

5.0 0.5(3.2) 3.4b b= + = So, the line has equation 0.5 3.4y x= + .

9. 5 54 2 5

8(4) (2)

4 2 4 2f f = =

11. 3( 1) 1 7 2f = = . So, ( ( 1)) ( 2) 1g f g = = .

13. Since (0,7) is the vertex, we know that 2( ) ( 0) 7f x a x= + . Use the point (2,-1) to find a:

21 (2 0) 7 2a a = + = Hence, 2( ) 2 7f x x= + . 15. Factors of 10: 1, 2, 5, 10 Factors of 2: 1, 2 So, possible rational zeros: 512 21, 2, 5, 10, , Observe that

1 2 7 18 13 102 5 23 10

2 2 5 23 10 04 18 10

2 9 5 0

Thus, ( )2( ) ( 1)( 2) 2 9 5 ( 1)( 2)( 5)(2 1)P x x x x x x x x x= + + = + + . So, the rationa

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