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Chapter 3 Systems of Linear Equations and Inequalities
Section 3.1 Practice Exercises
1. a. system 2. The lines intersect at (1, 3).
1, 3
b. solution c. intersect d. consistent e. the empty set, f. dependent g. independent
3.
8 5
4 3
Substitute 1,1 3 :
13 8 1 5
8 5
13
Not a solution.
Substitute 1,1 :
1 8 1 5
8 5
13
Not a solution.
Substitute 2,1 1 :
11 8 2 5
16 5
11
11 4 2 3
8 3
11
2,1 1 is a
y xy x
solution.
4.
12
34
12
34
12
92
9 12 2
5
10
Substitute 4, 7 :
7 4 5
2 5
7
7 4 10
3 10
7
4, 7 is a solution.
Substitute 0, 10 :
10 0 5
0 5
5
Not a solution.
Substitute 3, :
3
y xy x
32
132
5
5
Not a solution.
Section 3.1 Solving Systems of Linear Equations by the Graphing Method
209
5.
2 7 30
3 7
Substitute 0, 30 :
2 0 7 30 0 210 210 30
Not a solution.
3Substitute , 5 :
2
32 7 5 3 35 32 30
2
Not a solution.
Substitute 1, 4 :
2 1 7 4 2 28 30 30
4 3 1 7
3 7
4
x yy x
1, 4 is a solution.
6.
2 4
12
2Substitute 2, 3 :
2 2 3 2 6 4 4
13 2 2 1 2 3
22, 3 is a solution.
Substitute 4, 0 :
4 2 0 4 0 4 4
10 4 2 2 2 0
24, 0 is a solution.
1Substitute 3, :
2
13 2 3 1 4 4
2
1 1 33 2 2
2 2 2
x y
y x
1
21
3, is a solution.2
7.
6
4 3 4
Substitute 4, 2 :
4 2 4 2 6 6
4 4 3 2 16 6 10 4
Not a solution.
Substitute 6, 0 :
6 0 6 6
4 6 3 0 24 0 24 4
Not a solution.
Substitute 2, 4 :
2 4 2 6
Not a solution.
x yx y
8.
3 3
2 9 1
Substitute 0,1 :
0 3 1 0 3 3 3
Not a solution.
Substitute 4, 1 :
4 3 1 4 3 7 3
Not a solution.
Substitute 9, 2 :
9 3 2 9 6 3 3
2 9 9 2 18 18 0 1
Not a solution.
x yx y
Chapter 3 Systems of Linear Equations and Inequalities
210
9. a. Consistent 10. a. Consistent
b. Independent b. Independent
c. One solution c. One solution
11. a. Inconsistent 12. a. Inconsistent
b. Independent b. Independent
c. Zero solutions c. Zero solutions
13. a. Consistent 14. a. Consistent
b. Dependent b. Dependent
c. Infinitely many solutions c. Infinitely many solutions
15. 2 3 3
2 3 3
x y x yy x y x
The solution is 2,1 .
16. 4 3 12 3 4 16
3 4 12 4 3 16
4 34 4
3 4
x y x yy x y x
y x y x
The solution is 0, 4 .
17. 2 3 5 4f x x g x x
The solution is 1,1 .
18. 2 5 2h x x g x x
The solution is 1, 3 .
Section 3.1 Solving Systems of Linear Equations by the Graphing Method
211
19. 1 25 2
3 3k x x f x x
The solution is 3, 4 .
20. 1 52 2
2 2f x x g x x
The solution is 2, 3 .
21. 4 2 3x y x
The solution is 4, 5 .
22. 3 2 6 3
2 3 6
33
2
x y yy x
y x
The solution is 4, 3 .
23. 2 3 2 1
2 1
y x x yy x
There is no solution; .
Inconsistent system.
24. 12 3 9
33 9
13
3
y x x y
y x
y x
There is no solution; .
Inconsistent system.
Chapter 3 Systems of Linear Equations and Inequalities
212
25. 21 2 3 3
33 2 3
21
3
y x x y
y x
y x
Infinitely many solutions of the form
2, 1
3x y y x
. Dependent
equations.
26. 14 16 8 2
28 4 16
12
2
x y y x
y x
y x
Infinitely many solutions of the form
1, 2
2x y y x
. Dependent
equations.
27. 12 4 1
22 2
x y
x y
The solution is 2, 2 .
28. 7 6 5 2
51
2
y x
y x
The solution is 5
, 12
.
29. 3 6 6 2 12
13 6 2
31
23
x y y x
y x y x
y x
30. 3 2 4 4 6 8
32 3 4 2
23
22
x y y x
y x y x
y x
Section 3.1 Solving Systems of Linear Equations by the Graphing Method
213
Infinitely many solutions of the form
, 3 6x y x y . Dependent
equations.
Infinitely many solutions of the form
, 3 2 4x y x y . Dependent
equations.
31. 2 4 4 2 2
2 4 2 4 2
2 4 2 1
x y x yy x y xy x y x
There is no solution; . Inconsistent
system.
32. 4 4 2 8 16
4 4 8 2 16
1 11 2
4 4
x y x yy x y x
y x y x
There is no solution; . Inconsistent
system.
33. False 34. False
35. True 36. True
37. For example: The system
9
2 13
x yx y
has solution 4, 5 .
38. For example: The system
4
8
x yx y
has
solution 2, 6 .
Chapter 3 Systems of Linear Equations and Inequalities
214
39.
2 11
1 2 3 11
6 11
5
Cx yC
CC
3 9
3 1 3 9
3 3 9
3 12
4
x DyD
DDD
40.
3 22
3 2 4 22
6 4 22
4 28
7
x MyM
MMM
4 6
2 4 4 6
2 16 6
2 22
11
Nx yN
NNN
41. 5.62 15.46
1.96 11.07
y xy x
3.5, 4.21
42. 2.3 5.48
4.62 26.352
y xy x
4.6, 5.1
43. 2.4 4.8 9.36
4.8 2.4 9.36
0.5 1.95
1.8 5.4 12.456
5.4 1.8 12.456
1 173
3 75
x yy xy x
x yy x
y x
44. 36 90 36
90 36 36
0.4 0.4
15.5 5 80.75
5 15.5 80.75
3.1 16.15
x yy xy x
x yy xy x
Section 3.2 Solving Systems of Linear Equations by the Substitution Method
215
2.14, 3.02
4.5, 2.2
Section 3.2 Practice Exercises
1. 8 1 2 16 3
16 2 3
1 3
8 16
y x x yy x
y x
One solution
2. 54 6 1 10 15
25
6 4 1 15 102
2 1 2 1
3 6 3 6
x y x y
y x y x
y x y x
Infinitely many solutions
3. 2 4 0 2 9
4 2 2 9
1 1 9
2 2 2
x y x yy x y x
y x y x
No solution
4. 6 3 8 8 4 1
3 6 8 4 8 1
8 12 2
3 4
x y x yy x y x
y x y x
No solution
5.
2 10 2 11
4 2 3 10 2 4 3 11
4 6 10 8 3 11
10 10 11 11
4, 3 is not a solution.
x y x y
6. 4 3 4 12
4 4 3 12
34 3
4
x y x yy x y x
y x y x
The solution is 4, 0 .
Chapter 3 Systems of Linear Equations and Inequalities
216
7. 2 3 6 3 9
3 6 9
2 3
y x x yy xy x
The solution is 0, 3 .
8. 4 12 4
5 11
x yy x
4 12 5 11 4
4 60 132 4
64 128
2
x xx x
xx
5 11 5 2 11 10 11 1y x
The solution is 2,1 .
9. 3 1
2 3 8
y xx y
2 3 3 1 8
2 9 3 8
11 11
1
x xx x
xx
3 1 3 1 1 3 1 2y x
The solution is 1, 2 .
10. 10 34
7 31
y xx y
7 10 34 31
70 238 31
69 207
3
y yy y
yy
10 34 10 3 34 30 34 4x y
The solution is 4, 3 .
11. 3 8 1
4 11
x yx y
3 8 4 11 1
3 32 88 1
29 87
3
x xx x
xx
4 11 4 3 11 12 11 1y x
The solution is 3,1 .
12. 12 2 0
7 1 7 1
x yx y y x
12 2 7 1 0
12 14 2 0
2 2
1
x xx x
xx
7 1 7 1 1 7 1 6y x
The solution is 1, 6 .
13. 3 12 36
5 12 5 12
x yx y x y
14. 3 3 3 3
2 3 6
x y x yx y
Section 3.2 Solving Systems of Linear Equations by the Substitution Method
217
3 5 12 12 36
15 36 12 36
27 0
0
y yy y
yy
5 12 5 0 12 0 12 12x y
The solution is 12, 0 .
2 3 3 3 6
6 6 3 6
9 0
00
9
y yy y
y
y
3 3 3 0 3 0 3 3x y
The solution is 3, 0 .
15. 8 8
3 2 9
x y x yx y
3 8 2 9
3 24 2 9
5 15
3
y yy y
yy
8
3 8
5
x y
The solution is 5, 3 .
16. 5 2 10
1
x yy x
5 2 1 10
5 2 2 10
3 8
8
3
x xx x
x
x
8 51 1
3 3y x .
The solution is 8 5
, 3 3
.
17. 2 1
2
x yy x
2 2 1
4 1
1
4
x xx
x
1 1
2 24 2
y x
The solution is 1 1
, 4 2
.
18. 1 3 10 3 9 3
5 2 6
y y yx y
5 2 3 6
5 6 6
5 0
0
xx
xx
The solution is 0, 3 .
19. 2 3 7 2 4 2
3 4 6
x x xx y
20. 2 3 7
2 125 2 12
5
x yyx y x
Chapter 3 Systems of Linear Equations and Inequalities
218
3 2 4 6
6 4 6
4 0
0
yyyy
The solution is 2, 0 .
2 122 3 7
5
2 2 12 5 3 5 7
4 24 15 35
11 24 35
11 11
1
y y
y yy y
yyy
2 1 12 2 12 102
5 5 5x
The solution is 2,1 .
21. 4 5 14
73 7
3
x yxy x y
74 5 14
3
3 4 5 7 3 14
12 5 35 42
7 35 42
7 7
1
61 72
3 3
xx
x xx x
xxx
y
The solution is 1, 2 .
22. 2 0 2
2 6 15
2( 2 ) 6 15
4 6 15
10 15
15 3
10 23
22
3
x y x yx y
y yy yy
y y
x
The solution is 3
3,2
.
23. 2 6 2
3 1
x yx y
2 3 1 6 2
6 2 6 2
2 2
y yy y
Infinitely many solutions of the form
, 3 1x y x y ; dependent
equations.
24. 2 4 22
2 11
x yx y
2 2 11 4 22
4 22 4 22
22 22
y yy y
Infinitely many solutions of the form
, 2 11x y x y ; dependent equations.
Section 3.2 Solving Systems of Linear Equations by the Substitution Method
219
25. 13
77 4
y x
x y
17 3 4
7
21 4
21 4
x x
x x
There is no solution; . This is an
inconsistent system.
26. 3 1
2 24 6 7
x y
x y
3 14 6 7
2 2
6 2 6 7
2 7
y y
y y
There is no solution; . This is an
inconsistent system.
27. 5 10 5 10
2 10 5
x y y xy x
2 5 10 10 5
10 20 10 5
20 5
x xx x
There is no solution; . This is an
inconsistent system.
28. 4 8 4 8
3 3 12
x y x yx y
3 4 8 3 12
12 24 3 12
24 3
y yy y
There is no solution; . This is an
inconsistent system.
29. 3 7 3 7
14 6 2
x y y xx y
14 6 2 3 7
14 6 6 14
14 14
x xx x
Infinitely many solutions of the form
, 3 7x y x y ; dependent
equations.
30. 4 1
12 3 3
x yy x
12 3 4 1 3
12 12 3 3
0 0
y yy y
Infinitely many solutions of the form
, 4 1x y x y ; dependent equations.
31. If you get an identity, such as 0 = 0 or
5 = 5 when solving a system of
equations, then the equations are
dependent.
32. If you get a contradiction, such as 0 = 6
or 1 = 7 when solving a system of
equations, then the system is inconsistent.
33. 1.3 1.5
1.2 4.6
x yy x
34. 0.8 1.8
1.1 9.6
y xx y
Chapter 3 Systems of Linear Equations and Inequalities
220
1.3 1.2 4.6 1.5
1.56 5.98 1.5
0.56 4.48
8
x xx xxx
1.2 4.6
1.2 8 4.6
9.6 4.6
5
y x
The solution is 8, 5 .
1.1 0.8 1.8 9.6
1.1 0.8 1.8 9.6
1.9 11.4
6
x xx xxx
0.8 1.8
0.8 6 1.8
4.8 1.8
3
y x
The solution is 6, 3 .
35. 2 1
3 31 17
4 4
y x
x y
1 2 1 17
4 3 3 4
1 1 17
6 12 45 1 51
6 12 125 50
6 1250 6 300
512 5 602 1 2 1 10 1 9
53 3 3 3 3 3 33
x x
x x
x
x
x
y x
The solution is 5, 3 .
36. 1 5
6 31 21
5 5
x y
y x
1 1 21 5
6 5 5 3
1 21 5
30 30 329 21 50
30 30 3029 29
30 301
1 21 1 211
5 5 5 51 21 20
45 5 5
x x
x x
x
x
x
y x
The solution is 1, 4 .
37. 2 4 2 4
1 1 1
4 8 4
x y y x
x y
1 1 12 4
4 8 41 1 1 1
4 4 2 41 1
2 4
x x
x x
38. 8 8 8 8
1 1 1
3 24 2
x y y x
x y
1 1 18 8
3 24 21 1 1 1
3 3 3 21 1
3 2
x x
x x
Section 3.2 Solving Systems of Linear Equations by the Substitution Method
221
There is no solution; . This is an
inconsistent system.
There is no solution; . This is an
inconsistent system.
39. 3 2 6
3
x yy x
3 2 3 6
3 2 6 6
5 6 6
5 0
0
0 3 3
x xx x
xxxy
The solution is 0, 3 .
40. 4 4
1
x yy x
4 1 4
4 4 4
3 4 4
3 0
0
0 1 1
x xx x
xxxy
The solution is 0, 1 .
41. 300 125 1350
2 8 6
x yy y
300 125 6 1350
300 750 1350
300 2100
7
xx
xx
The solution is 7, 6 .
42. 200 150
4 1 5
y xy y
200 5 150
1000 150
1000 20
150 3
xx
x
The solution is 20
, 53
.
43. 2 6 2 6
1 1 1
6 12 2
x y y x
x y
1 1 12 6
6 12 21 1 1 1
6 6 2 21 1
2 2
x x
x x
Infinitely many solutions of the form
, 2 6x y x y ; dependent
equations.
44. 4 8 4 8
1 1 1
16 4 2
x y x y
x y
1 1 14 8
16 4 21 1 1 1
4 2 4 21 1
2 2
y y
y y
Infinitely many solutions of the form
, 4 8x y x y ; dependent equations.
Chapter 3 Systems of Linear Equations and Inequalities
222
45. 2.7 5.1
3.1 63.1
y xy x
3.1 63.1 2.7 5.1
5.8 58
10
3.1 63.1 3.1 10 63.1
31 63.1 32.1
x xxxy x
The solution is 10, 32.1 .
46. 6.8 2.3
4.1 56.8
y xy x
6.8 2.3 4.1 56.8
10.9 54.5
5
6.8 2.3 6.8 5 2.3
34 2.3 36.3
x xxxy x
The solution is 5, 36.3 .
47. 4 4 5
5 54 4
2 2
x y
x y x y
54 4 4 5
2
16 10 4 5
20 15
15 3
20 4
y y
y yy
y
5 3 5 5 14 4 3
2 4 2 2 2x y
The solution is 1 3
, 2 4
.
48. 2 6 2 6
6 13 12
x y y xx y
6 13 2 6 12
6 26 78 12
20 90
9
2
x xx x
x
x
92 6 2 6 9 6 3
2y x
The solution is 9
, 32
.
49. 2 2 12 2 6 2 6
6 5 8
x y x y x yx y
6 2 6 5 8
12 36 5 8
7 28
284
72 4 6
8 6
2
y yy y
y
y
x
The solution is 2, 4 .
50.
5 2 25
310 3 10 10
10
x y
x y x y
35 10 2 25
10
310 2 25
23
15 2 252
110
220
y y
y y
y y
y
y
3 320 10 10 3
10 10x
The solution is 3, 20 .
Section 3.2 Solving Systems of Linear Equations by the Substitution Method
223
51. 5 3 2 4 1 5 10 4
1 5 14
4 7 3
y x y xy x
y x
4 7 15 14 3
4 105 98 3
4 105 101
101 101
1
15 1 14
15 14
1
y yy yy yyyx
The solution is 1,1 .
52. 32 3 3 3
23 4 22
x y x y
x y
33 3 4 22
2
93 4 22
29 3 8 44
9 27 8 44
17 17
1
31 3
23
426
y y
y y
y yy y
yy
x
The solution is 6,1 .
53. 2 5 7 2 12 6
4 3 1 3 3 1
x x xy y y
The solution is 6,1 .
54. 2 4 2 6 2 3
7 5 5 7 0 0
y y yx x x
The solution is 0, 3 .
55. 0.01 0.02 0.11 2 11
0.3 0.5 2 3 5 20
y x y xx y x y
3 5 2 11 20
3 10 55 20
7 55 20
7 35
5
2 5 11
10 11
1
x xx x
xxxy
The solution is 5, 1 .
56. 0.3 0.4 1.3 3 4 13
0.01 0.03 0.01 3 1
x y x yx y x y
3 3 1 4 13
9 3 4 13
5 3 13
5 10
2
3 2 1
6 1
7
y yy y
yyyx
The solution is 7, 2 .
Chapter 3 Systems of Linear Equations and Inequalities
224
57. a. points 4,1 and 5, 5
5 1 4 4
5 4 5 4 9m
1 1 1 1 , 4,1
41 4
94 16
19 94 25
9 9
y y m x x x y
y x
y x
y x
58. a. points 0, 3 and 2, 5
5 3 84
2 0 2m
1 1 1 1 , 0, 3
3 4 0
3 4
4 3
y y m x x x y
y xy x
y x
b. points 3, 5 and 4,1
1 5 4 4
4 3 7 7m
1 1 1 1 , 3, 5
45 3
74 12
57 74 23
7 7
y y m x x x y
y x
y x
y x
b. points 4, 4 and 2, 2
2 4 6
12 4 6
m
1 1 1 1 , 4, 4
4 1 4
4 4
y y m x x x y
y xy x
y x
c. 4 25 4 23
9 9 7 74 25 4 23
63 639 9 7 7
28 175 36 207
64 32
1
2
x x
x x
x xx
x
4 25
9 94 1 25 2 25 27
39 2 9 9 9 9
y x
y
The centroid is 1
, 32
.
c. 4 3
3 3
1
x xxx
4 3
4 1 3
4 3
1
y xy
The centroid is 1,1 .
Section 3.3 Solving Systems of Linear Equations by the Addition Method
225
59. a. At Glendale Lakes:
800 250y x
At the Breakers: 750 500y x
60. a. Surfside: 159.50 24y x
Tropical Winds:
165.50y x 750 500y x
b. 800 250 750 500
50 250
5
x xxx
The amount spent is the same for 5
months.
b. 159.50 24 165.50
24 6
4
x xx
x
The cost to stay is the same for 4
nights.
Section 3.3 Practice Exercises
1. a. 3 2. No solutions – parallel lines
b. 5
3. One solution – different slopes 4. 4 7 4 7
2 8 14 4 7
x y y xy x y x
Infinitely many solutions – same line
5. Add the two equations and solve for y:
3 1
3 4 14
3 15
5
x yx y
yy
Substitute into the first equation and
solve for x:
3 5 1
3 5 1
3 6
2
xx
xx
The solution is 2, 5 .
6. Add the two equations and solve for x:
5 2 15
3 2 7
8 8
1
x yx yx
x
Substitute into the first equation and solve
for y:
5 1 2 15
5 2 15
2 10
5
yyyy
The solution is 1, 5 .
7. 2 3 3
10 2 32
x yx y
Multiply the first equation by 5, add to
the second equation and solve for y:
8. 2 5 7
3 10 13
x yx y
Multiply the first equation by –2, add to
the second equation and solve for x:
Chapter 3 Systems of Linear Equations and Inequalities
226
5
2 3 3 10 15 1 5
10 2 32 10 2 32
17 17
1
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
2 3 1 3
2 3 3
2 6
3
xx
xx
The solution is 3, 1 .
2
2 5 7 4 10 14
3 10 13 3 10 1 3
1
1
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
2 1 5 7
2 5 7
5 5
1
yyyy
The solution is 1, 1 .
9. 3 7 20
5 3 84
x yx y
Multiply the first equation by 5 and the
second equation by 3, add the results and
solve for y:
5
3
3 7 20 15 35 100
5 3 84 15 9 252
44 352
8
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
3 7 8 20
3 56 20
3 36
12
xx
xx
The solution is 12, 8 .
10. 6 9 15
5 2 40
x yx y
Multiply the first equation by 2 and the
second equation by –9, add the results
and solve for x:
2
9
6 9 15 12 18 30
5 2 40 45 18 360
33 330
10
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
6 10 9 15
60 9 15
9 45
5
yyyy
The solution is 10, 5 .
11. Write in standard form:
3 10 13 3 10 1 3
7 4 11 4 7 11
x y x yy x x y
Multiply the first equation by 4 and the
second equation by 3, add the results and
solve for y:
12. Write in standard form:
5 6 4 5 6 4
5 1 3 3 5 1
x y x yy x x y
Multiply the first equation by 3 and the
second equation by 5, add the results
and solve for y:
Section 3.3 Solving Systems of Linear Equations by the Addition Method
227
4
3
3 10 1 3 12 40 52
4 7 11 12 21 33
19 1 9
1
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
3 10 1 13
3 10 13
3 3
1
xxxx
The solution is 1, 1 .
3
5
5 6 4 15 18 12
3 5 1 1 5 25 5
7 7
1
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
5 6 1 4
5 6 4
5 10
2
xxxx
The solution is 2, 1 .
13. Multiply each equation by 10:
1.2 0.6 3 1 2 6 30
0.8 1.4 3 8 14 30
x y x yx y x y
Multiply the first equation by 2 and the
second equation by –3, add the results
and solve for y: 2
3
12 6 30 24 1 2 60
8 14 30 24 42 90
30 30
1
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
12 6 1 30
12 6 30
12 24
2
xx
xx
The solution is 2, 1 .
14. Multiply each equation by 10:
1.8 0.8 1.4 1 8 8 14
1.2 0.6 1.2 1 2 6 12
x y x yx y x y
Multiply the first equation by 2 and the
second equation by –3, add the results
and solve for y: 2
3
18 8 14 36 16 28
12 6 12 36 18 36
2 8
4
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
18 8 4 14
18 32 14
18 18
1
xx
xx
The solution is 1, 4 .
15. Write in standard form:
3 2 4 2 3 4 0
7 3 7 3 0
x y x yx y x y
16. Write in standard form:
4 3 2 3 2 4 0
5 3 3 5 0
y x x yy x x y
Chapter 3 Systems of Linear Equations and Inequalities
228
Multiply the first equation by 3 and the
second equation by –4, add the results
and solve for x:
3
–4
3 4 0 9 12 0
7 3 0 28 12 0
19 0
0
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
3 0 4 0
0 4 0
4 0
0
yyyy
The solution is 0. 0 .
Multiply the first equation by 5 and the
second equation by 4, add the results and
solve for x:
5
4
2 4 0 10 20 0
3 5 0 1 2 20 0
32 0
0
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
2 0 4 0
0 4 0
4 0
0
yyyy
The solution is 0. 0 .
17. 3 2 1
6 4 2
x yx y
Multiply the first equation by 2, add to
the second equation and solve for y:
2
3 2 1 6 4 2
6 4 2 6 4 2
0 0
x y x yx y x y
Infinitely many solutions of the form
, 3 2 1x y x y ; dependent
equations.
18. 3 4
6 2 8
x yx y
Multiply the first equation by –2, add to
the second equation and solve for y:
2
3 4 6 2 8
6 2 8 6 2 8
0 0
x y x yx y x y
Infinitely many solutions of the form
, 3 4x y x y ; dependent equations.
19. Write in standard form:
6 14 4 4 6 14
2 3 7 2 3 7
y x x yx y x y
Multiply the second equation by –2, add
to the first equation and solve for y:
2
4 6 1 4 4 6 14
2 3 7 4 6 14
0 28
x y x yx y x y
20. Write in standard form:
2 4 2 4
2 2 2 2
x y x yy x x y
Add the second equation to the first
equation and solve for y:
2 4
2 2
0 2
x yx y
Section 3.3 Solving Systems of Linear Equations by the Addition Method
229
There is no solution; . This is an
inconsistent system.
There is no solution; . This is an
inconsistent system.
21. Write in standard form:
12 4 2 1 2 4 2
6 1 2 6 2 1
x y x yx y x y
Multiply the second equation by –2,
add to the first equation and solve for y:
2
12 4 2 12 4 2
6 2 1 12 4 2
0 0
x y x yx y x y
Infinitely many solutions of the form
, 12 4 2x y x y ; dependent
equations.
22. Write in standard form:
10 15 5 1 0 1 5 5
3 2 1 2 3 1
x y x yy x x y
Multiply the second equation by 5, add to
the first equation and solve for y:
5
10 1 5 5 10 15 5
2 3 1 10 15 5
0 0
x y x yx y x y
Infinitely many solutions of the form
, 10 15 5x y x y ; dependent
equations.
23. 1 7
2 62 4.5
x y
x y
Multiply the first equation by –2, add
to the second equation and solve for y:
2
1 7 7 2
2 6 32 4.5 2 4.5
130
6
x y x y
x y x y
There is no solution; . This is an
inconsistent system.
24. 0.2 0.1 1.2
1 32
x y
x y
Multiply the first equation by 10 and the
second equation by –2, add the results
and solve for y: 10
2
0.2 0.1 1.2 2 12
1 3 2 62
0 18
x y x y
x y x y
There is no solution; . This is an
inconsistent system.
25. Use the substitution method if one
equation has x or y already isolated.
26. It would be easier to eliminate the y-
variable by multiplying the first equation
by 2. To eliminate the x-variable, we
would have to multiply the first equation
by –7 and the second equation by 3.
27. False 28. False
Chapter 3 Systems of Linear Equations and Inequalities
230
29. True 30. True
31. True 32. True
33. 2 4 8
2 1
x yy x
2 4 2 1 8
2 8 4 8
6 12
2
x xx x
xx
2 1 2 2 1 4 1 3y x
The solution is 2, 3 .
34. 8 6 8
6 10
x yx y
8 6 10 6 8
48 80 6 8
54 72
72 4
54 3
y yy y
y
y
46 10 6 10 8 10 2
3x y
The solution is 4
2, 3
.
35. 2 5 9
4 7 16
x yx y
Multiply the first equation by –2, add to
the second equation and solve for y:
2
2 5 9 4 10 18
4 7 16 4 7 16
17 34
2
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
2 5 2 9
2 10 9
2 1
12
xx
x
x
The solution is 1
, 22
.
36. 0.1 0.5 0.7
0.2 0.7 0.8
x yx y
Multiply the first equation by –2, add to
the second equation and solve for y: 2
0.1 0.5 0.7 0.2 1.4
0.2 0.7 0.8 0.2 0.7 0.8
0.3 0.6
2
x y x yx y x y
yy
Substitute into the first equation and solve
for x:
0.1 0.5 2 0.7
0.1 1.0 0.7
0.1 0.3
3
xx
xx
The solution is 3, 2 .
Section 3.3 Solving Systems of Linear Equations by the Addition Method
231
37. 0.2 0.1 0.8
0.1 0.1 0.4 0.1 0.1 0.4
4
x yx y x y
x y
0.2 4 0.1 0.8
0.2 0.8 0.1 0.8
0.1 0.8 0.8
0.1 0
0
0 4
4
y yy y
y
yyx
The solution is 4, 0 .
38. 13
24 3
y xx y
14 3 3
2
93 3
29
02
0
10 3
20 3
3
x x
x
xx
y
The solution is 0, 3 .
39. 4 6 5
2 3 7
x yx y
Multiply the second equation by –2,
add to the first equation and solve for y:
2
4 6 5 4 6 5
2 3 7 4 6 14
0 9
x y x yx y x y
There is no solution; . This is an
inconsistent system.
40. 3 6 7
2 4 5
x yx y
Multiply the first equation by –2 and the
second equation by 3, add the results and
solve for y:
2
3
3 6 7 6 12 14
2 4 5 6 12 1 5
0 1
x y x yx y x y
There is no solution; . This is an
inconsistent system.
41. Multiply each equation by the LCD:
12
30
1 12 3 2 24
4 61 1
4 5 6 1 206 5
x y x y
x y x y
Multiply the first equation by 3, add to
the second equation and solve for x:
3
3 2 24 9 6 72
5 6 120 5 6 1 20
4 48
12
x y x yx y x y
xx
42. Multiply each equation by the LCD:
15
30
1 1 7 5 3 1 05
3 51 2
4 5 12 1206 5
x y x y
x y x y
Multiply the first equation by –1, add to
the second equation and solve for y: 1
5 3 1 05 5 3 105
5 12 120 5 12 120
15 225
15
x y x yx y x y
yy
Chapter 3 Systems of Linear Equations and Inequalities
232
Substitute into the first equation and
solve for y:
3 12 2 24
36 2 24
2 60
30
yyyy
The solution is 12, 30 .
Substitute into the first equation and
solve for x:
5 3 15 105
5 45 105
5 60
12
xx
xx
The solution is 12,1 5 .
43. 1 10
3 23
2
x y
x y
1 3 10
3 2 2
1 10
2 20 0
y y
y y
Infinitely many solutions of the form
32
, x y x y . The equations are
dependent.
44. 2 20
5 33
5
x y
y x
2 2 30
5 3 5
2 20
5 50 0
x x
x x
Infinitely many solutions of the form
35
, x y y x . The equations are
dependent.
45. Write in standard form:
2 2 20 2 4 20
2 5 20
7 16 3 7 7 16 3
7 4 16
x y y x y yx y
x y y x y yx y
Multiply the first equation by 4 and the
second equation by –5, add the results
and solve for x: 4
5
2 5 20 8 20 80
7 4 1 6 35 20 80
43 0
0
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
46. Write in standard form:
3 10 4 3 3 10 4
3 10
4 2 50 3 4 8 50 3
4 5 50
x y y x y yx y
x y y x y yx y
Multiply the first equation by –5, add to
the second equation and solve for x:
5
3 10 15 5 50
4 5 50 4 5 50
19 0
0
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
Section 3.3 Solving Systems of Linear Equations by the Addition Method
233
2 0 5 20
0 5 20
5 20
4
yyyy
The solution is 0, 4 .
3 0 10
0 10
10
yyy
The solution is 0,1 0 .
47. Solve each equation:
4 10
10 5
4 2
y
y
4 3 1
4 2
2 1
4 2
xx
x
The solution is 1 5
, 2 2
.
48. Solve each equation:
9 15
15 5
9 3
x
x
3 2 1
3 1
1
3
yy
y
The solution is 5 1
, 3 3
.
49. 0.04 0.05 1.7 4 5 170
4 5 170
0.03 2.4 0.07 3 240 7
7 3 240
x y x yx y
y x y xx y
Multiply the first equation by 3 and the
second equation by –5, add the results
and solve for x: 3
5
4 5 170 12 15 510
35 15 12007 3 24023 690
30
x y x yx yx y
xx
Substitute into the first equation and
solve for y:
4 30 5 170
120 5 170
5 50
10
yyyy
The solution is 30,1 0 .
50. 0.01 0.06 3.2 6 320
6 320
0.08 0.03 4.6 8 3 460
3 8 460
x y x yx y
y x y xx y
Multiply the first equation by –3, add to
the second equation, and solve for y:
3
6 320 3 18 960
3 8 460 3 8 460
10 500
50
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
6 50 320
300 320
20
20
xx
xx
The solution is 20, 50 .
51. Write in standard form: 52. Write in standard form:
Chapter 3 Systems of Linear Equations and Inequalities
234
1 11 53 2 11 5 3 2
3 3 35 17
3 9 5 173 3
x y x y
x y x y
2 42 3 2 2 2
3 34
0 3 4 03
x y x y
x y x y
Multiply the second equation by –3,
add to the first equation and solve for y:
3
9 5 17 9 5 17
3 4 0 9 12 0
17 17
1
x y x yx y x y
yy
Substitute into the first equation and
solve:
9 5 1 17
9 5 17
9 12
12 49 3
xx
x
x
The solution is 4
, 13
.
2 2 3 2 1 4 6 2 1
4 5
1 7 5 5 7
55 4 7
y x x y x xx y
x y y x y y
x y
Multiply the first equation by –1, add to
the second equation and solve for x:
1
4 5 4 5
5 4 7 5 4 7
6 12
2
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
2 4 5
4 3
3
4
yy
y
The solution is 3
2, 4
.
53. 1 1 11
4 2 42 1 7
3 3 3
x y
x y
Multiply the first equation by 4 and the
second equation by –6, add the results
and solve for x:
4
6
1 1 11 2 1 1
4 2 42 1 7
4 2 143 3 3
3 3
1
x y x y
x y x y
xx
Substitute into the first equation above
and solve for y:
54. 1 1 8
10 2 51 11
4 2
x y
x y
Multiply the first equation by 10 and the
second equation by 20, add the results
and solve for x:
10
20
1 1 8 5 16
10 2 51 11
20 5 1104 2
21 126
6
x y x y
x y x y
xx
Substitute into the first equation above
and solve for y:
Section 3.3 Solving Systems of Linear Equations by the Addition Method
235
1 2 11
2 10
5
yyy
The solution is 1, 5 .
6 5 16
5 10
2
yyy
The solution is 6, 2 .
55. 34 3
44
23
x y x y
y x
Substitute for x and solve for y:
4 32
3 4
2
0 2
y y
y y
There is no solution; . This is an
inconsistent system.
56. 4 2 6
1 3
2 2
x y
x y
1 34 2 6
2 2
2 6 2 6
6 6
y y
y y
Infinitely many solutions of the form
, 4 2 6x y x y ; dependent
equations.
57. Multiply each equation by the LCD:
48
140
1 112 3 2 576
16 241 1
14 1 0 7 196014 20
c h c h
c h c h
Multiply the first equation by 10
3 ,
add to the second equation and solve
for h: 10
3
203 2 576 10 1920
3
10 7 1960 10 7 1960
140
3120
c h c h
c h c h
h
h
Substitute into the first equation and
solve for c:
58. Multiply each equation by the LCD:
189
105
1 112 9 7 2268
21 271 1
16 7 5 168015 21
c h c h
c h c h
Multiply the first equation by 7
9 , add to
the second equation and solve for h: 7
9
499 7 2268 7 1764
9
7 5 1680 7 5 1680
484
9189
c h c h
c h c h
h
h
Substitute into the first equation and
solve for c:
9 7 2268
9 7 189 2268
9 1323 2268
9 945
105
c hc
ccc
Chapter 3 Systems of Linear Equations and Inequalities
236
3 2 576
3 2 120 576
3 240 576
3 336
112
c hc
ccc
112 mi in the city and 120 mi on the
highway.
105 mi in the city and 189 mi on the
highway.
59. 9 11 47
5 3 23
x yx y
Multiply the first equation by –3 and
the second equation by 11, add the
results and solve for x: 3
11
9 11 47 27 33 141
5 3 23 55 33 253
82 112
112 56
82 41
x y x yx y x y
x
x
Multiply the first equation by 5 and
the second equation by 9, add the
results and solve for y: 5
9
9 11 47 45 55 235
5 3 23 45 27 207
82 442
442 221
82 41
x y x yx y x y
y
y
The solution is 56 221
, 41 41
.
60. 6 7 4
4 9 31
x yx y
Multiply the first equation by 9 and the
second equation by 7, add the results and
solve for x: 9
7
6 7 4 54 63 36
4 9 31 28 63 217
26 181
181
26
x y x yx y x y
x
x
Multiply the first equation by 2 and the
second equation by 3, add the results and
solve for y: 2
3
6 7 4 12 14 8
4 9 31 1 2 27 93
13 85
85
13
x y x yx y x y
y
y
The solution is 181 85
, 26 13
.
61. 4 10 19
5 12 41
x yx y
Multiply the first equation by 6 and
the second equation by 5, add the
results and solve for x:
Multiply the first equation by –5 and the
second equation by 4, add the results and
solve for y:
Problem Recognition Exercises: Solving Systems of Linear Equations
237
6
5
4 10 1 9 24 60 1 14
5 12 41 25 60 205
49 91
91 13
49 7
x y x yx y x y
x
x
5
4
4 10 1 9 20 50 95
5 12 41 20 48 164
98 259
259 37
98 14
x y x yx y x y
y
y
The solution is 13 37
, 7 14
.
Problem Recognition Exercises: Solving Systems of Linear Equations 1. a. 3 2 4 4
3 2 4 4
4 4
x y x yy x y x
y x
The solution is 2, 4 .
2. a. 2 43 2 4
3 32 3 4 3 2 4
32 2 3 4
23
22
x y x y
y x x y
y x y x
y x
Infinitely many solutions; dependent
equations.
b. 3 2 3 2
4 4
x y y xx y
4 3 2 4
4 3 2 4
2 4
2
x xx x
xx
b. 3 2 4
2 4
3 3
x y
x y
2 43 2 4
3 3
2 4 2 4
4 4
y y
y y
Infinitely many solutions;
Chapter 3 Systems of Linear Equations and Inequalities
238
3 2 2
6 2
4
y
The solution is 2, 4 .
, 3 2 4x y x y ; dependent
equations.
c. 3 2
4 4
x yx y
Add the equations and solve for x:
3 2
4 4
2
x yx y
x
Substitute into the first equation and
solve for y:
3 2 2
6 2
4
yyy
The solution is 2, 4 .
c. Write in standard form:
2 43 2 4
3 33 2 4
3 2 4
x y x y
x yx y
Subtract the second equation from the
first:
3 2 4
3 2 4
0 0
x yx y
Infinitely many solutions;
, 3 2 4x y x y ; dependent
equations.
3. a. 55 2
25
12
x y y x
y x
No solution; ; inconsistent
system
4. a. 2 3 1 4 4
3 11
2 2
y x x
y x x
The solution is 1, 2 .
Problem Recognition Exercises: Solving Systems of Linear Equations
239
b. 5 51
2 20 1
x x
No solution; ; inconsistent
system
b. 2 3 1 4 4
1
y x xx
2 3 1 1 3 1 4
2
yy
The solution is 1, 2 .
c. Write in standard form:
55 2 1
25 2 0 2 5 2
5 2 2
x y y x
x y y xx y
Add the equations to solve for x:
5 2 0
5 2 2
0 2
x yx y
No solution; ; inconsistent
system
c. Write in standard form:
2 3 1 3 2 1
4 4 1
y x x yx x
Multiply the second equation by 3, add
to the first equation and solve for y:
3
3 2 1 3 2 1
1 3 3
2 4
2
x y x y
x x
yy
The solution is 1, 2 .
5. 4 9
8 3 29
y xx y
Substitute the first equation into the
second and solve for x:
8 3 4 9 29
8 12 27 29
4 27 29
4 2
1
2
x xx x
xx
x
14 9 4 9 2 9
2
11
y x
The solution is1
, 112
.
6. 5 2 17
5 2 5 2
x yx y x y
Substitute the second equation into the
first and solve for y:
5 5 2 2 17
25 10 2 17
27 10 17
27 27
1
y yy y
yyy
5 1 2
5 2
3
x
The solution is 3,1 .
Chapter 3 Systems of Linear Equations and Inequalities
240
7. 5 3 2
7 4 30
x yx y
Multiply the first equation by 4 and the
second equation by 3, add the results and
solve for x:
4
3
5 3 2 20 12 8
7 4 30 21 12 90
41 82
2
x y x y
x y x yxx
Substitute into the first equation and
solve for y:
5 2 3 2
10 3 2
3 12
4
yyyy
The solution is 2, 4 .
8. Multiply each equation by the LCD:
10
12
1 2 3 4 6
10 5 53 1 13
9 4 264 3 6
x y x y
x y x y
Add the first equation to the second
equation and solve for x:
4 6
9 4 26
10 20
2
x yx yx
x
Substitute into the first equation and
solve for y:
2 4 6
4 8
2
yyy
The solution is 2, 2 .
Section 3.4 Practice Exercises
1. a. 5 $12 $60
12 12x x
e. 180
b. 0.10 20 2L
0.10 0.10x x
f. 180
c. 0.04 $5000 $200
0.04 0.04y y
g. 90
d. b c ; b c h. 90 2. Systems of linear equations can be solved by the graphing method, the substitution
method, and the addition method.
3. Substitution:
9 2
3 16
y xx y
4. Addition:
7 25
2 5 1 4
x yx y
Section 3.4 Applications of Systems of Linear Equations in Two Variables
241
3 9 2 16
3 9 2 16
5 25
5
x xx x
xx
9 2
9 2 5
9 10
1
y x
The solution is 5, 1 .
Multiply the first equation by 5, add to
the second equation and solve for x:
5
7 25 35 5 125
2 5 1 4 2 5 1 4
37 111
3
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
7 3 25
21 25
4
4
yyyy
The solution is 3, 4 .
5. Let x = the number of premium tickets
sold
y = the number of regular tickets
sold
30x = receipts from premium tickets
20y = receipts from regular tickets
1 190 1190
30 20 30,180
x y y xx y
30 20 1190 30,180
30 23,800 20 30,180
10 6380
638
1190 638 552
x xx x
xxy
There were 638 tickets sold at $30 each
and 552 tickets sold at $20 each.
6. Let x = the cost of 1 notebook
y = the cost of 1 pen
4 5 10.65
3 3 7.50 2.50 2.50
x yx y x y y x
4 5 2.50 10.65
4 12.50 5 10.65
1.85
1.85
2.50 1.85
0.65
x xx x
xxy
Notebooks cost $1.85 and pens cost
$0.65.
7. Let x = the cost of 1 hamburger
y = the cost of 1 fish sandwich
3 2 24.20
4 23.60 23.60 4
x yx y y x
8. Let x = the cost of a member
y = the cost of a nonmember
3 150
2 75 75 2
x yx y y x
Chapter 3 Systems of Linear Equations and Inequalities
242
3 2 23.60 4 24.20
3 47.20 8 24.20
5 23
4.60
23.60 4 4.60
23.60 18.40
5.20
x xx x
xxy
Hamburgers cost $4.60 and fish
sandwiches cost $5.20.
3 75 2 150
225 6 150
5 75
15
75 2 15
75 30
45
x xx x
xxy
Members pay $15 and nonmembers pay
$45.
9. Let x = fat in 1 scoop of vanilla
y = fat in 1 scoop of mud pie
2 40 40 2
2 44
x y y xx y
2 40 2 44
80 4 44
3 36
12
40 2 12
40 24
16
x xx x
xxy
Vanilla has 12 g of fat per scoop and
mud pie has 16 g of fat per scoop.
10. Let x = calories in 1 cup of popcorn
y = calories in 1 ounce of soda
2 8 216
1 2 204 204 12
x yx y x y
2 204 12 8 216
408 24 8 216
16 192
12
204 12 12
204 144
60
y yy y
yyx
One cup of popcorn has 60 calories and
1 oz of soda has 12 calories.
11. Let x = the amount of 18% moisturizer
cream
y = the amount of 24% moisturizer
cream
18% Cr 24% Cr 22% Cr
oz cream x y 12__
oz moist 0.18x 0.24y 0.22(12)
12
0.18 0.24 0.22 12
x yx y
12. Let x = the amount of 18% acid solution
y = the amount of 45% acid solution
18% acid 45% acid 36% acid_
L solution x y 16___
L acid 0.18x 0.45y 0.36(16)
16
0.18 0.45 0.36 16
x yx y
Multiply the first equation by –0.18, add
to the second equation and solve for y:
Section 3.4 Applications of Systems of Linear Equations in Two Variables
243
Multiply the first equation by –0.18,
add to the second equation and solve
for y:
12 0.18 0.18 2.16
0.18 0.24 2.64 0.18 0.24 2.64
0.06 0.48
8
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
8 12
4
xx
The mixture contains 4 oz of 18%
moisturizer and 8 oz of 24%
moisturizer.
16 0.18 0.18 2.88
0.18 0.45 5.76 0.18 0.45 5.76
0.27 2.88
210
3
x y x yx y x y
y
y
Substitute into the first equation and
solve for x:
210 16
31
53
x
x
The mixture contains 13
5 L of 18% acid
solution and 23
10 L of 45% acid solution.
13. Let x = the amount of 8% nitrogen
fertilizer
y = the amount of 12% nitrogen
fertilizer
8% nit 12% nit 11% nit
oz cream x y _ 8 __
oz moist 0.08x 0.12y 0.11(8)
8
0.08 0.12 0.11 8
x yx y
Multiply the first equation by –0.08, add
to the second equation and solve for y:
8 0.08 0.08 0.64
0.08 0.12 0.88 0.08 0.12 0.88
0.04 0.24
6
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
6 8
2
xx
The mixture contains 2 L of 8% nitrogen
fertilizer and 6 L of 12% nitrogen
fertilizer.
14. Let x = the amount of 30% acid solution
y = the amount of 10% acid solution
30% acid 10% acid 12% acid
L solution x y 100__
L acid 0.30x 0.10y
0.12(100)
100
0.30 0.10 0.12 100
x yx y
Multiply the first equation by –0.10, add
to the second equation and solve for x:
100 0.1 0.1 10
0.3 0.1 12 0.3 0.1 1 2
0.2 2
10
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
10 100
90
yy
The mixture contains 10 mL of 30%
acid solution and 90 mL of 10% acid
solution.
Chapter 3 Systems of Linear Equations and Inequalities
244
15. Let x = amount of pure (100%) bleach
sol
y = the amount of 4% bleach
solution
100% bl 4% bl 12% bl
oz solution x y 12__
oz bleach 1.00x 0.04y 0.12(12)
12
1.00 0.04 0.12 12
x yx y
Multiply the first equation by –0.04,
add to the second equation and solve
for x:
12 0.04 0.04 0.48
1 0.04 1.44 1 .00 0.04 1 .44
0.96 0.96
1
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
1 12
11
yy
The mixture contains 1 oz of pure
bleach and 11 oz of 4% bleach solution.
16. Let x = the amount of 25% fruit juice
y = the amount of 100% fruit juice
25% fr j 100% fr j 75% fr j
oz punch x y 48 __
oz fr juice 0.25x 1.00y 0.75(48)
48
0.25 1.00 0.75 48
x yx y
Multiply the first equation by –0.25, add to
the second equation and solve for y:
48 0.25 0.25 12
0.25 1.00 36 0.25 1.00 36
0.75 24
32
x y x yx y x y
yy
Substitute into the first equation and solve
for y:
32 48
16
xx
The punch contains 16 oz of 25% fruit juice
and 32 oz of 100% fruit juice.
17. Let x = the amount invested in 5%
bonds
3x = the amount invested in 8%
stocks
5% Acct 8% Acct Total
Principal x 3x ________
Interest 0.05x 0.08(3x) 435
0.05 0.08 3 435
0.05 0.24 435
0.29 435
1500
3 3 1500 4500
x xx x
xxx
18. Let x = the amount invested in 3.5%
account
12
x = the amount invested in 2.5%
account
3.5% Acct 2.5% Acct Total
Principal x 0.5x ________
Interest 0.035x 0.025(0.5x) 247
1 12 2
0.035 0.025 0.5 247
0.035 0.0125 247
0.0475 247
5200
5200 2600
x xx x
xxx
Section 3.4 Applications of Systems of Linear Equations in Two Variables
245
He invested $1500 in the bond fund and
$4500 in the stock fund.
Aliya invested $2600 in the savings
account and $5200 in the money market
account.
19. Let x = the amount borrowed at 5.5%
y = the amount borrowed at 3.5%
5.5% Acct 3.5% Acct Total
Principal x y _________
Interest 0.055x 0.035y 245
200
0.055 0.035 245
x yx y
Substitute and solve for y:
0.055 200 0.035 245
0.055 11 0.035 245
0.09 234
2600
y yy y
yy
Substitute into the first equation and
solve for x:
2600 200
2800
xx
He borrowed $2800 at 5.5% and $2600
at 3.5%.
20. Let x = the amount invested at 4%
y = the amount invested at 3%
3% Acct 4% Acct Total
Principal x y _________
Interest 0.03x 0.04y 725
5000
0.03 0.04 725
x yx y
Substitute and solve for y:
0.03 5000 0.04 725
0.03 150 0.04 725
0.07 875
12500
y yy y
yy
12500 5000
7500
x
$7500 was invested at 3% and $12,500
was invested at 4%.
21. Let x = the amount borrowed at 6%
y = the amount borrowed at 7%
6% Acct 7% Acct Total__
Principal x y 15,000
Interest 0.06x 0.07y 4750/5
15,000
47500.06 0.07 950
5
x y
x y
Multiply the first equation by –0.06,
add to the second equation and solve
for y:
22. Let x = the amount invested at 5%
y = the amount invested at 6%
5% Acct 6% Acct Total_
Principal x y 15,500_
Interest 0.05x 0.06y 3500/4_
15,500
35000.05 0.06 875
4
x y
x y
Multiply the first equation by –0.06, add
to the second equation and solve for x:
Chapter 3 Systems of Linear Equations and Inequalities
246
15,000 0.06 0.06 900
0.06 0.07 950 0.06 0.07 950
0.01 50
5000
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
5000 15,000
10,000
xx
Alina borrowed $10,000 from the bank
charging 6% interest and $5000 from
the bank charging 7% interest.
15,500 0.06 0.06 930
0.05 0.06 875 0.05 0.06 875
0.01 55
5500
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
5500 15,500
10,000
yy
Didi should invest $5500 in the 5% fund
and $10,000 in the 6% fund.
23. Let b = the speed of the boat in still
water
c = the speed of the current
b + c = speed of boat with the current
b – c = speed of boat against the current
Distance Rate Time
With current 16 b + c 2
Against current 16 b – c 4
(rate)(time) = (distance)
2 16
4 16
b cb c
Divide the first equation by 2, the
second equation by 4, add the results,
and solve:
2
4
2 16 8
4 16 4
2 12
6
div
div
b c b cb c b c
bb
Substitute and solve for c:
6 8
2
cc
The speed of the boat is 6 mph and the
speed of the current is 2 mph.
24. Let p = the speed of the plane in still air
Let w = the speed of the wind
p + w = speed of the plane with the wind
p – w = speed of plane against the wind
Distance Rate Time_
Tailwind 720 p + w 3 _
Headwind 720 p – w 4 _
(rate)(time) = (distance)
3 720
4 720
p wp w
Divide the first equation by 3, the second
equation by 4, add the results, and solve:
3
4
3 720 240
4 720 180
2 420
210
div
div
p w p wp w p w
pp
Substitute and solve for w:
210 240
30
ww
The speed of the plane is 210 mph in still
air and the speed of the wind is 30 mph.
Section 3.4 Applications of Systems of Linear Equations in Two Variables
247
25. Let p = the speed of the plane in still
air
Let w = the speed of the wind
p + w = speed of the plane with the
wind
p – w = speed of plane against the wind
Distance Rate Time
Tailwind 3200 p + w 4 _
Headwind 3200 p – w 5 _
(rate)(time) = (distance)
4 3200
5 3200
p wp w
Divide the first equation by 4, the
second equation by 5, add the results,
and solve:
4
5
4 3200 800
5 3200 640
2 1440
720
div
div
p w p wp w p w
pp
Substitute and solve for w:
720 800
80
ww
The speed of the plane is 720 km/hr in
still air and the speed of the wind is
80 km/hr.
26. Let b = the speed of the boat in still
water
Let c = the speed of the current
b + c = speed of boat with the current
b – c = speed of boat against the current
Distance Rate Time
With current 100 b + c 2.5
Against current 100 b – c 10/3
(rate)(time) = (distance)
2.5 100
10100
3
b c
b c
Divide the first equation by 2.5, the
second equation by (10/3), add the
results, and solve:
2.5
10/3
2.5 100 40
10100 30
32 70
35
div
div
b c b c
b c b c
bb
Substitute and solve for c:
35 40
5
cc
The speed of the boat is 35 mph in still
water and the speed of the current is
5 mph.
27. Let x = the walking speed
Let y = the speed of the moving
sidewalk
x + y = speed of walking with sidewalk
x–y = speed of walking against
sidewalk
Distance Rate Time_
With walk 100 x + y 20 _
Against walk 60 x – y 30 _
28. Let b = the speed of the bike in still air
Let w = the speed of the wind
b + w = speed of the bike with the wind
b – w = speed of bike against the wind
Distance Rate Time_
Tailwind 24 b + w 2 _
Headwind 24 b – w 3 _
(rate)(time) = (distance)
Chapter 3 Systems of Linear Equations and Inequalities
248
(rate)(time) = (distance)
20 100
30 60
x yx y
Divide the first equation by 20, the
second equation by 30, add the results,
and solve:
20
30
20 100 5
30 60 2
2 7
3.5
div
div
x y x yx y x y
xx
Substitute and solve for y:
3.5 5
1.5
yy
Stephen’s speed on nonmoving ground
is 3.5 ft/sec. The sidewalk moves at
1.5 ft/sec.
2 24
3 24
b wb w
Divide the first equation by 2, the second
equation by 3, add the results, and solve:
2
3
2 24 12
3 24 8
2 20
10
div
div
b w b wb w b w
bb
Substitute and solve for w:
10 12
2
ww
Kim rides 10 km/hr in still air. The wind
speed is 2 km/hr.
29. Let x = one acute angle
Let y = the other acute angle
3 6
90
x yx y
Substitute and solve:
3 6 90
4 84
21
3 21 6 63 6 69
y yyyx
The two acute angles measure 69º and
21º.
30. Let x = one of two equal angles
Let y = the other angle
3
180
y xx x y
Substitute and solve:
3 180
3 183
61
61 3 58
x x xxxy
The angles measure 61º, 61º, and 58º.
31. Let x = one angle
Let y = the other angle
3 2
180
y xx y
Substitute and solve:
32. Let x = one angle
Let y = the other angle
5
180
y xx y
Substitute and solve:
Section 3.4 Applications of Systems of Linear Equations in Two Variables
249
3 2 180
4 182
45.5
3 45.5 2
136.5 2
134.5
x xxxy
The two angles measure 45.5º and
134.5º.
5 180
6 180
30
5 30
150
x xxxy
The angles measure 30º and 150º.
33. Let x = one angle
Let y = the other angle
2 6
90
y xx y
Substitute and solve:
2 6 90
3 84
28
2 28 6
56 6
62
x xxxy
The two angles measure 28º and 62º.
34. Let x = one angle
Let y = the other angle
2 15
90
y xx y
Substitute and solve:
2 15 90
3 75
25
2 25 15
50 15
65
x xxxy
The angles measure 25º and 65º.
35. Let x = the amount of pure (100%) gold
y = the amount of 60% gold
100% gold 60% gold 75% gold
g mix x y 20___
g gold 1.00x 0.60y 0.75(20)
20
1.00 0.60 0.75 20
x yx y
Multiply the first equation by –0.60,
add to the second equation and solve
for x:
36. Let x = the amount of 15% disinfectant
y = the amount of 55% disinfectant
15% 55% 17%__
oz mix x y 50__
oz gold 0.15x 0.55y 0.17(50)
50
0.15 0.55 0.17 50
x yx y
Multiply the first equation by –0.15, add
to the second equation and solve for y:
Chapter 3 Systems of Linear Equations and Inequalities
250
15,500 0.06 0.06 930
0.05 0.06 875 0.05 0.06 875
0.01 55
5500
x y x yx y x y
xx
7.5 g of pure gold must be used.
50 0.15 0.15 7.5
0.15 0.55 8.5 0.15 0.55 8.5
0.40 1
2.5
50 2.5 47.5
x y x yx y x y
yy
x
Connie should use 47.5 gal of the 15%
solution and 2.5 gal of the 55% solution.
37. Let b = the speed of the boat in still
water
Let c = the speed of the current
b + c = speed of boat with the current
b – c = speed of boat against the
current
Distance Rate Time
With current 16 b + c 2.5
Against current 10 b – c 2.5
(rate)(time) = (distance)
2.5 16 2.5 2.5 16
2.5 10 2.5 2.5 10
b c b cb c b c
Add the two equations, and solve:
2.5 2.5 16
2.5 2.5 10
5 26
5.2
b cb cb
b
Substitute and solve for c:
2.5 5.2 2.5 16
13 2.5 16
2.5 3
1.2
cccc
The speed of the boat in still water is
5.2 mph and the speed of the current is
1.2 mph.
38. Let b = the speed of the kayak in still
water
Let c = the speed of the current
b + c = speed of boat with the current
b – c = speed of boat against the current
Distance Rate Time
With current 31.5 b + c 7
Against current 31.5 b – c 9
(rate)(time) = (distance)
7 31.5
9 31.5
b cb c
Divide the first equation by 7, the second
equation by 9, add the results, and solve:
7
9
7 31.5 4.5
9 31.5 3.5
2 8
4
div
div
b c b cb c b c
bb
Substitute and solve for c:
4 4.5
0.5
cc
The speed of the kayak in still water is 4
mph and the speed of the current is 0.5
mph.
Section 3.4 Applications of Systems of Linear Equations in Two Variables
251
39. Let x = the cost of a grandstand ticket
y = the cost of a general admission
ticket
6 2 2330
4 4 2020
x yx y
Multiply the first equation by –2, add to
the second equation and solve for x:
2
6 2 2330 12 4 4660
4 4 2020 4 4 2020
8 2640
330
x y x yx y x y
xx
Substitute and solve for y:
6 330 2 2330
1980 2 2330
2 350
175
yyyy
Grandstand tickets cost $330 and
general admission tickets cost $175.
40. Let x = the number of two-point baskets
Let y = the number of three-point baskets
8 8
2 3 19
x y x yx y
Substitute and solve:
2 8 3 19
16 2 3 19
16 19
3
8 3
5
y yy y
yyx
The player made 5 two-point baskets and
3 three-point baskets.
41. Let x = the amount invested at 2%
y = the amount invested at 1.3%
2% Acct 1.3% Acct Total
Principal x y 3,000
Interest 0.02x 0.013y 51.25
3,000
0.02 0.013 51.25
x yx y
Multiply the first equation by –0.02,
add to the second equation and solve
for y:
3,000 .02 .02 60
.02 .013 51.25 .02 .013 51.25
0.007 8.75
1250
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
42. Let x = the amount invested at 3%
y = the amount invested at 1.8%
3% Acct 1.8% Acct Total_
Principal x y 8000_
Interest 0.03x 0.018y 222_
8000
0.03 0.018 222
x yx y
Multiply the first equation by –0.03, add
to the second equation and solve for y:
8000 .03 .03 240
.03 .018 222 .03 .018 222
0.012 18
1500
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
1500 8000
6500
xy
Chapter 3 Systems of Linear Equations and Inequalities
252
1250 3000
1750
xy
Svetlana invested $1750 at 2% and
$1250 at 1.3%.
$6500 was invested at 3%.
43. Let w = the width of the rectangle
Let l = the length of the rectangle
1
2 2 42
l wl w
Substitute and solve:
2 1 2 42
2 2 2 42
4 2 42
4 40
10
10 1 11
w ww w
wwwl
The width is 10 m and the length is
11 m.
44. Let x = one angle
Let y = the other angle
1
490
y x
x y
Substitute and solve:
190
45
904
72
172 18
4
x x
x
x
y
The two angles measure 72º and 18º.
45. Let d = the number of $1 coins
f = the number of 50 cent pieces
21 21
1 0.50 15.50
d f f dd f
0.50 21 15.50
10.50 0.50 15.50
0.50 5.00
10
21 10 11
d dd d
ddf
The collection contains 10 - $1 coins
and 11 – 50 cent pieces.
46. Let d = the number of dimes
n = the number of nickels
30 30
0.10 0.05 1.90
d n n dd n
0.10 0.05 30 1.90
0.10 1.50 0.05 1.90
0.05 0.40
8
30 8 22
d dd d
ddn
Jacob has 8 dimes and 22 nickels.
47. a. 60f x x 48. a. 20 0.25c x x
b. 50 100g x x b. 30 0.20m x x
c. Substitute and solve: c. Substitute and solve:
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
253
60 50 100
10 100
10
x xxx
10 months
20 0.25 30 0.20
0.05 10
200
x xxx
The rental fees are the same when
the cars are driven 200 mi.
Section 3.5 Practice Exercises
1. a. linear c. dashed; is not
b. is not; is d. solid; is
2.
5 1 and 2 6 6
4 and 2 12
4 and 6 4, 6
x xx xx x
3.
5 4 and 6 3 3
1 and 9 3
1 and 3 1, 3
x xx xx x
4.
4 3 12 or 2 3 12
8 4 or 2 6 12
4 8 or 2 18
2 or 9
, 9 2,
y y yy y
y yy y
5.
2 4 or 3 1 13
2 or 3 12
2 or 4
, 4 2,
x xx xx x
6.
2 8x y 7. 3 5y x
a. 2 3 5 6 5 11 8 Yes a. 3 7 1 21 1 20 5 No
b. 2 1 10 2 10 8 8 No b. 3 0 5 0 5 5 5 No c. 2 4 2 8 2 10 8 Yes c. 3 0 0 0 0 0 5 Yes
d. 2 0 0 0 0 0 8 No d. 3 3 2 9 2 7 5 Yes
8.
2y 9. 5x
a. 3 2 Yes a. 4 5 No
b. 2 2 Yes b. 5 5 Yes c. 0 2 No c. 8 5 Yes
d. 2 2 No d. 0 5 No
Chapter 3 Systems of Linear Equations and Inequalities
254
10. Use a dashed line when the inequality is strict ( < or > ).
11. To choose the correct inequality
symbol, three observations must be
made. First, notice the shading occurs
below the line. Second, since the
coefficient of y is negative in the given
statement, the direction of the
inequality will change. Third, the
boundary line is dashed indicating no
equality.
Thus use the symbol > for the
inequality: 2x y .
12. To choose the correct inequality symbol,
three observations must be made. First,
notice the shading occurs below the line.
Second, since the coefficient of y is
positive in the given statement, the
direction of the inequality will not change.
Third, the boundary line is solid indicating
equality.
Thus use the symbol ≤ for the inequality:
2 3y x .
13. To choose the correct inequality
symbol, three observations must be
made. First, notice the shading occurs
above the line. Second, since the
coefficient of y is positive in the given
statement, the direction of the
inequality will not change. Third, the
boundary line is solid indicating
equality. Thus use the symbol ≥ for the
inequality:
4y .
14. Since the boundary is a vertical line, to
choose the correct inequality symbol, two
observations must be made. First, notice
the shading occurs to the left of the line.
Second, the boundary line is dashed
indicating no equality.
Thus use the symbol < for the inequality:
3x .
15. The graph of 0x includes Quadrant I
and Quadrant IV. The graph of 0y
includes Quadrant III and Quadrant IV.
The intersection of the graphs occurs in
Quadrant IV.
Thus, the statements are
0 and 0.x y
16. The graph of 0x includes Quadrant I
and Quadrant IV. The graph of 0y
includes Quadrant I and Quadrant II. The
intersection of the graphs occurs in
Quadrant I.
Thus, the statements are 0 and 0.x y
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
255
17.
2 4
Graph the related equation 2 4 by using a dashed line.
Test point above 0, 0 : Test point below 0, 3 :
0 2 0 4 0 2 3 4
0 4 6 4
0, 0 is not a solution. 0, 3 is a solution.
Shade the region be
x yx y
low the boundary line.
18. 3 6
Graph the related equation 3 6 by using a solid line.
x yx y
Test point above 0, 0 : Test point below 0, 3 :
0 3 0 6 0 3 3 6
0 6 9 6
0, 0 is not a solution. 0, 3 is a solution.
Shade the region below the boundary line.
19.
5 2 10
Graph the related equation 5 2 10 by using a dashed line.
Test point above 0, 0 : Test point below 2, 3 :
5 0 2 0 10 5 2 2 3 10
0 10 16 10
0, 0 is a solution. 2, 3 is not a solution.
Shade th
x yx y
e region above the boundary line.
Chapter 3 Systems of Linear Equations and Inequalities
256
20.
3 8
Graph the related equation 3 8 by using a dashed line.
Test point above 0, 0 : Test point below 0, 4 :
0 3 0 8 0 3 4 8
0 8 12 8
0, 0 is a solution. 0, 4 is not a solution.
Shade the region a
x yx y
bove the boundary line.
21.
2 6 12
Graph the related equation 2 6 12 by using a solid line.
Test point above 0, 3 : Test point below 0, 0 :
2 0 6 3 12 2 0 6 0 12
0 6 0 12
0, 3 is not a solution. 0, 0 is a solution.
Shade the
x yx y
region below the boundary line.
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
257
22.
4 3 12
Graph the related equation 4 3 12 by using a dashed line.
Test point above 0, 0 : Test point below 0, 5 :
4 0 3 0 12 4 0 3 5 12
0 12 0 3
0, 0 is a solution. 0, 5 is not a solution.
Shade the
x yx y
region above the boundary line.
23.
2 4
Graph the related equation 2 4 by using a solid line.
Test point above 0,1 : Test point below 0, 1 :
2 1 4 0 2 1 4 0
2 0 2 0
0,1 is not a solution. 0, 1 is a solution.
y xy x
Shade the region below the boundary line.
24.
6 2
Graph the related equation 6 2 by using a dashed line.
Test point above 0,1 : Test point below 0, 1 :
6 0 2 1 6 0 2 1
0 2 0 2
0,1 is a solution. 0, 1 is not a solution.
Shade the region a
x yx y
bove the boundary line.
Chapter 3 Systems of Linear Equations and Inequalities
258
25.
2
Graph the related equation 2 by using a solid line.
Test point above 0, 0 : Test point below 0, 3 :
0 2 3 2
0, 0 is a solution. 0, 3 is not a solution.
Shade the region above the boundar
yy
y line.
26.
5
Graph the related equation 5 by using a solid line.
Test point above 0, 6 : Test point below 0, 0 :
6 5 0 5
0, 6 is a solution. 0, 0 is not a solution.
Shade the region above the boundary line.
yy
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
259
27. 54 5 or represents all the points to the left of the vertical
45
line . The boundary is a dashed line.4
Shade the region to the left of the boundary line.
x x
x
28. 6 7 or 1 represents all the points to the left of the vertical
line 1. The boundary is a dashed line.
Shade the region to the left of the boundary line.
x xx
29.
24
52
Graph the related equation 4 by using a solid line.5
Test point above 0, 0 : Test point below 0, 5 :
2 20 0 4 5 0 4
5 50 4 5 4
0, 0 is a solution. 0, 5 is not a solution.
Shade the reg
y x
y x
ion above the boundary line.
Chapter 3 Systems of Linear Equations and Inequalities
260
30.
54
25
Graph the related equation 4 by using a solid line.2
Test point above 0, 0 : Test point below 0, 5 :
5 50 0 4 5 0 4
2 20 4 5 4
0, 0 is a solution. 0, 5 is not a solution.
Shade the
y x
y x
region above the boundary line.
31.
16
31
Graph the related equation 6 by using a solid line.3
Test point above 0, 7 : Test point below 0, 0 :
1 17 0 6 0 0 6
3 37 6 0 6
0, 7 is not a solution. 0, 0 is a solution.
Shade the region be
y x
y x
low the boundary line.
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
261
32.
12
41
Graph the related equation 2 by using a solid line.4
Test point above 0, 3 : Test point below 0, 0 :
1 13 0 2 0 0 2
4 43 2 0 2
0, 3 is not a solution. 0, 0 is a solution.
Shade the regio
y x
y x
n below the boundary line.
33.
5 0
Graph the related equation 5 0 by using a dashed line.
Test point above 0, 3 : Test point below 0, 3 :
3 5 0 0 3 5 0 0
3 0 3 0
0, 3 is a solution. 0, 3 is not a solution.
Shade the region a
y xy x
bove the boundary line.
Chapter 3 Systems of Linear Equations and Inequalities
262
34.
10
21
Graph the related equation 0 by using a dashed line.2
Test point above 0, 3 : Test point below 0, 3 :
1 13 0 0 3 0 0
2 23 0 3 0
0, 3 is a solution. 0, 3 is not a solution.
Shade the regi
y x
y x
on above the boundary line.
35.
15 4
Graph the related equation 1 by using a dashed line.5 4
Test point above 0, 5 : Test point below 0, 0 :
0 5 0 01 1
5 4 5 45
1 0 14
0, 5 is not a solution. 0, 0 is a solution.
Shade the region
x y
x y
below the boundary line.
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
263
36.
22
Graph the related equation 2 by using a solid line.2
Test point above 0, 5 : Test point below 0, 0 :
5 00 2 0 2
2 25
2 0 22
0, 5 is a solution. 0, 0 is not a solution.
Shade the region above
yx
yx
the boundary line.
37.
0.1 0.2 0.6
Graph the related equation 0.1 0.2 0.6 by using a solid line.
Test point above 0, 5 : Test point below 0, 0 :
0.1 0 0.2 5 0.6 0.1 0 0.2 0 0.6
1 0.6 0 0.6
0, 5 is not a solution. 0, 0 is a
x yx y
solution.
Shade the region below the boundary line.
38.
0.3 0.2 0.6
Graph the related equation 0.3 0.2 0.6 by using a dashed line.
Test point above 0, 0 : Test point below 0, 5 :
0.3 0 0.2 0 0.6 0.3 0 0.2 5 0.6
0 0.6 1 0.6
x yx y
Chapter 3 Systems of Linear Equations and Inequalities
264
0, 0 is a solution. 0, 5 is not a solution.
Shade the region above the boundary line.
39.
2
32
Graph the related equation by using a solid line.3
Test point above 0, 3 : Test point below 0, 3 :
2 20 3 0 3
3 30 2 0 2
0, 3 is not a solution. 0, 3 is a solution.
Shade the region be
x y
x y
low the boundary line.
40.
5
45
Graph the related equation by using a solid line.4
Test point above 0, 3 : Test point below 0, 3 :
5 50 3 0 3
4 415 15
0 04 4
0, 3 is a solution. 0, 3 is not a solution.
Shade the regio
x y
x y
n above the boundary line.
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
265
41. 4 and 2
4 represents the points below the horizontal line 4.
Shade the region below the boundary line using a dashed line border.
Graph the related equation 2 by using a dashed line.
Tes
y y xy y
y x
t point above 0, 3 : Test point below 0, 0 :
3 0 2 0 0 2
3 2 0 2
0, 3 is a solution. 0, 0 is not a solution.
Shade the region above the boundary line.
The solution is the intersection of the graphs.
42. 3 and 2 6
3 represents the points below the horizontal line 3.
Shade the region below the boundary line using a dashed line border.
Graph the related equation 2 6 by using a dashed line.
Tes
y x yy y
x y
t point above 0, 4 : Test point below 0, 0 :
0 2 4 6 0 2 0 6
8 6 0 6
0, 4 is not a solution. 0, 0 is a solution.
Shade the region below the boundary line.
The solution is the intersection of the graphs.
Chapter 3 Systems of Linear Equations and Inequalities
266
43.
2 5 or 3
Graph the related equation 2 5 by using a solid line.
Test point above 0, 6 : Test point below 0, 0 :
2 0 6 5 2 0 0 5
6 5 0 5
0, 6 is not a solution. 0, 0 is a solution.
Shade the regi
x y xx y
on below the boundary line.
3 represents the points to the right of the vertical line 3.
Shade the region to the right of the boundary line using a solid
line border. The solution is the union of the
x x
graphs.
44.
3 3 or 2
Graph the related equation 3 3 by using a solid line.
Test point above 0, 4 : Test point below 0, 0 :
0 3 4 3 0 3 0 3
12 3 0 3
0, 4 is a solution. 0, 0 is not a solution.
Shade the re
x y xx y
gion above the boundary line.
2 represents the points to the left of the vertical line 2.
Shade the region to the left of the boundary line using a solid
line border. The solution is the union of t
x x
he graphs.
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
267
45.
3 and 4 6
Graph the related equation 3 by using a dashed line.
Test point above 0, 4 : Test point below 0, 0 :
0 4 3 0 0 3
4 3 0 3
0, 4 is not a solution. 0, 0 is a solution.
Shade the reg
x y x yx y
ion below the boundary line.
Graph the related equation 4 6 by using a dashed line.
Test point above 0, 7 : Test point below 0, 0 :
4 0 7 6 4 0 0 6
7 6 0 6
0, 7 is not a solution. 0, 0 is a solution.
x y
Shade the region below the boundary line.
The solution is the intersection of the graphs.
46.
Graph the related equation 3 9 by using a dashed line.
Test point above 3, 3 : Test point below 0, 0 :
3 3 3 9 3 0 0 9
12 9 0 9
3, 3 is not a solution. 0, 0 is a solution.
Shade the region below the
x y
boundary line.
The solution is the intersection of the graphs.
Chapter 3 Systems of Linear Equations and Inequalities
268
4 and 3 9
Graph the related equation 4 by using a dashed line.
Test point above 0, 5 : Test point below 0, 0 :
0 5 4 0 0 4
5 4 0 4
0, 5 is not a solution. 0, 0 is a solution.
Shade the reg
x y x yx y
ion below the boundary line.
47.
2 2 or 2 3 6
Graph the related equation 2 2 by using a solid line.
Test point above 0, 0 : Test point below 0, 3 :
2 0 0 2 2 0 3 2
0 2 3 2
0, 0 is a solution. 0, 3 is not a solution.
Shade t
x y x yx y
he region above the boundary line.
Graph the related equation 2 3 6 by using a solid line.
Test point above 0, 3 : Test point below 0, 0 :
2 0 3 3 6 2 0 3 0 6
9 6 0 6
0, 3 is a solution. 0, 0 is not a solution.
Shade the region above the
x y
boundary line.
The solution is the union of the graphs.
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
269
48.
3 2 4 or 3
Graph the related equation 3 2 4 by using a solid line.
Test point above 0, 3 : Test point below 0, 0 :
3 0 2 3 4 3 0 2 0 4
6 4 0 4
0, 3 is a solution. 0, 0 is not a solution.
Shade th
x y x yx y
e region above the boundary line.
Graph the related equation 3 by using a solid line.
Test point above 0, 0 : Test point below 0, 4 :
0 0 3 0 4 3
0 3 4 3
0, 0 is a solution. 0, 4 is not a solut
x y
ion.
Shade the region above the boundary line.
The solution is the union of the graphs.
49. 4 and 2
4 represents the points to the right of the vertical line 4.
Shade the region to the right of the boundary line using a dashed
line border.
2 represents the points below the horizontal
x yx x
y
line 2.
Shade the region below the boundary line using a dashed
line border. The solution is the intersection of the graphs.
y
Chapter 3 Systems of Linear Equations and Inequalities
270
50. 3 and 4
3 represents the points to the left of the vertical line 3.
Shade the region to the left of the boundary line using a dashed
line border.
4 represents the points above the horizontal l
x yx x
y
ine 4.
Shade the region above the boundary line using a dashed
line border. The solution is the intersection of the graphs.
y
51. 2 or 0
2 represents the points to the left of the vertical line 2.
Shade the region to the left of the boundary line using a solid
line border.
0 represents the points below the horizontal
x yx x
y
line 0.
Shade the region below the boundary line using a solid
line border. The solution is the union of the graphs.
y
52. 0 or 3
0 represents the points to the right of the vertical line 0.
Shade the region to the right of the boundary line using a solid
line border.
3 represents the points above the horizontal
x yx x
y
line 3.
Shade the region above the boundary line using a solid
line border. The solution is the union of the graphs.
y
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
271
53. 0 and 6
0 represents the points to the right of the vertical line 0.
Shade the region to the right of the boundary line using a dashed
line border.
Graph the related equation 6 by using a
x x yx x
x y
dashed line.
Test point above 0, 7 : Test point below 0, 0 :
0 7 6 0 0 6
7 6 0 6
0, 7 is not a solution. 0, 0 is a solution.
Shade the region below the boundary line.
The solution is the intersection o
f the graphs.
54. 0 and 2
0 represents the points to the left of the vertical line 0.
Shade the region to the left of the boundary line using a dashed
line border.
Graph the related equation 2 by using a da
x x yx x
x y
shed line.
Test point above 0, 3 : Test point below 0, 0 :
0 3 2 0 0 2
3 2 0 2
0, 3 is not a solution. 0, 0 is a solution.
Shade the region below the boundary line.
Chapter 3 Systems of Linear Equations and Inequalities
272
The solution is the intersection of the graphs.
55. 0 or 4
0 represents the points below the horizontal line 0.
Shade the region below the boundary line using a solid line border.
Graph the related equation 4 by using a solid line.
Test p
y x yy y
x y
oint above 0, 5 : Test point below 0, 0 :
0 5 4 0 0 4
5 4 0 4
0, 5 is a solution. 0, 0 is not a solution.
Shade the region above the boundary line.
The solution is the union of the graphs.
56.
3 and 0 and 0
Graph the related equation 3 by using a solid line.
Test point above 0, 4 : Test point below 0, 0 :
0 4 3 0 0 3
4 3 0 3
0, 4 is not a solution. 0, 0 is a solution.
Shade t
x y x yx y
he region below the boundary line.
0 represents the points to the right of the vertical line 0.
Shade the region to the right of the boundary line using a solid
line border.
x x
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
273
0 represents the points above the horizontal line 0.
Shade the region above the boundary line using a solid line border.
The solution is the intersection of the graphs.
y y
57.
2 and 0 and 0
Graph the related equation 2 by using a solid line.
Test point above 0, 0 : Test point below 0, 3 :
0 0 2 0 3 2
0 2 3 2
0, 0 is a solution. 0, 3 is not a solution.
Shad
x y x yx y
e the region above the boundary line.
0 represents the points to the right of the vertical line 0.
Shade the region to the right of the boundary line using a solid
line border.
0 represents the points above the horizontal line 0.
Shad
x x
y y
e the region above the boundary line using a solid line border.
The solution is the intersection of the graphs.
58. 0 and 0 and 8 and 3 5 30
0 represents the points to the right of the vertical line 0.
Shade the region to the right of the boundary line using a solid
line border.
0 represents the p
x y x y x yx x
y
oints above the horizontal line 0.
Shade the region above the boundary line using a solid line border.
y
Chapter 3 Systems of Linear Equations and Inequalities
274
Graph the related equation 8 by using a solid line.
Test point above 0, 9 : Test point below 0, 0 :
0 9 8 0 0 8
9 8 0 8
0, 9 is not a solution. 0, 0 is a solution.
Shade the region below the bound
x y
ary line.
Graph the related equation 3 5 30 by using a solid line.
Test point above 0, 7 : Test point below 0, 0 :
3 0 5 7 30 3 0 5 0 30
35 30 0 30
0, 7 is not a solution. 0, 0 is a solution.
Shade the r
x y
egion below the boundary line.
The solution is the intersection of the graphs.
59. 0 and 0 and 5 and 2 6
0 represents the points to the right of the vertical line 0.
Shade the region to the right of the boundary line using a solid
line border.
0 represents the poi
x y x y x yx x
y
nts above the horizontal line 0.
Shade the region above the boundary line using a solid line border.
Graph the related equation 5 by using a solid line.
y
x y
Test point above 0, 6 : Test point below 0, 0 :
0 6 5 0 0 5
6 5 0 5
0, 6 is not a solution. 0, 0 is a solution.
Shade the region below the boundary line.
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
275
Graph the related equation 2 6 by using a solid line.
Test point above 0, 4 : Test point below 0, 0 :
0 2 4 6 0 2 0 6
8 6 0 6
0, 4 is not a solution. 0, 0 is a solution.
Shade the region below the bo
x y
undary line.
The solution is the intersection of the graphs.
60. a. 2 2 50x y
b. 0 and 0 and 2 2 50
0 represents the points to the right of the vertical line 0. Shade the region to the right
of the boundary line using a solid line border.
0 represents the points above
x y x yx x
y
the horizontal line 0. Shade the region above the
boundary line using a solid line border.
y
Graph the related equation 2 2 50 by using a solid line.
Test point above 0, 27 : Test point below 0, 0 :
2 0 2 27 50 2 0 2 0 50
54 50 0 50
0, 27 is not a solution. 0, 0 is a solution.
Shade the region
x y
below the boundary line.
Chapter 3 Systems of Linear Equations and Inequalities
276
61. a. 2 2 40x y
b. 0 and 0 and 2 2 40
0 represents the points to the right of the vertical line 0. Shade the region to the right
of the boundary line using a solid line border.
0 represents the points above
x y x yx x
y
the horizontal line 0. Shade the region above the
boundary line using a solid line border.
y
Graph the related equation 2 2 40 by using a solid line.
Test point above 0, 21 : Test point below 0, 0 :
2 0 2 21 40 2 0 2 0 40
42 40 0 40
0, 21 is not a solution. 0, 0 is a solution.
Shade the region
x y
below the boundary line.
62. a. 0, 0x y
b. 4 3 24x y
c. 3 12x y
d. 0 and 0 and 4 3 24 and 3 12
0 represents the points to the right of the vertical line 0. Shade the region to the right
of the boundary line using a solid line border.
0 represents th
x y x y x yx x
y
e points above the horizontal line 0. Shade the region above the
boundary line using a solid line border.
Graph the related equation 4 3 24 by using a solid line.
Test point above 0, 9 : Te
y
x y
st point below 0, 0 :
4 0 3 9 24 4 0 3 0 24
27 24 0 24
0, 9 is not a solution. 0, 0 is a solution.
Shade the region below the boundary line.
Section 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables
277
Graph the related equation 3 12 by using a solid line.
Test point above 0,1 3 : Test point below 0, 0 :
3 0 13 12 3 0 0 12
13 12 0 12
0,1 3 is not a solution. 0, 0 is a solution.
Shade the region bel
x y
ow the boundary line.
The solution is the intersection of the graphs.
e. Yes. The point (3, 1) represents 3 Model A desks and 1 Model B desk being
produced.
f. No. The point (5, 4) represents 5 Model A desks and 4 Model B desk being
produced. Producing this combination of desks would exceed the number of
available hours for staining and finishing and for assembly.
63. a. 0, 0x y
b. 40, 40x y
c. 65x y
d. 0 and 0 and 40 and 40 and 65x y x y x y
0x represents the points to the right of the vertical line 0x . Shade the region
to the right of the boundary line using a solid line border.
0y represents the points above the horizontal line 0y . Shade the region above
the boundary line using a solid line border.
40 represents the points to the left of the vertical line 40. Shade the region to t
of the boundary line using a solid line border.
40 represents the points below the horizontal line 40.
x x
y y
Shade the region below t
boundary line using a solid line border.
Graph the related equation 65 by using a solid line.x y
Chapter 3 Systems of Linear Equations and Inequalities
278
Test point above 0, 66 : Test point below 0, 0 :
0 66 65 0 0 65
66 65 0 65
0, 66 is a solution. 0, 0 is not a solution.
Shade the region above the boundary line.
The solution is the intersection of th
e graphs.
e. Yes. The point (35, 40) means that Karen works 35 hours and Todd works 40
hours.
f. No. The point (20, 40) means that Karen works 20 hours and Todd works 40 hours.
This does not satisfy the constraint that there must be at least 65 hours total.
Section 3.6 Practice Exercises 1. a. linear 2.
5 3 1 4 2 2
5 4 3 7 1 4 4 2 7 2
20 21 1 16 14 2
1 1 2 2
Yes, 4, 7 is a solution.
x y x y
b. ordered triples
3. a. 3 4 3 4
4 5
x y y xx y
4 3 4 5
4 3 4 5
1
x xx x
x
3 4
3 1 4
3 4 1
y x
4. a. 5 32 5 3 2 5 3
2 24 10 3
x y x y x y
x y
5 34 10 3
2 2
10 6 10 3
6 3
y y
y y
Section 3.6 Systems of Linear Equations in Three Variables and Applications
279
The solution is 1,1 . There is no solution; . This is an
inconsistent system.
b. 3 4
4 5
x yx y
Multiply the first equation by –1,
add to the second equation and
solve for x:
1
3 4 3 4
4 5 4 5
1
x y x yx y x y
x
Substitute into the first equation
and solve for y:
3 1 4
1
yy
The solution is 1,1 .
b. 2 5 3
4 10 3
x yx y
Multiply the first equation by 2, add to
the second equation and solve for x:
2
2 5 3 4 10 6
4 10 3 4 10 3
0 9
x y x yx y x y
There is no solution; . This is an
inconsistent system.
5. Let b = the speed of the bike in still air
Let w = the speed of the wind
b + w = speed of the bike with the
wind
b – w = speed of bike against the
wind
Distance Rate Time
Tailwind 24 b + w 4/3
Headwind 24 b – w 2_
(rate)(time) = (distance)
424
3
2 24
b w
b w
Divide the first equation by 4/3, the second
equation by 2, add the results, and solve:
4 /3
2
424 18
3
2 24 12
2 30
15
div
div
b w b w
b w b wb
b
Substitute and solve for w:
15 18
3
ww
Marge’s speed is 15 mph in still air. The
wind speed is 3 mph.
6. The number of solutions to a system of three equations in three variables is one solution, no
solution, or infinitely many solutions.
Chapter 3 Systems of Linear Equations and Inequalities
280
7.
2 10
4 2 3 10
3 2 8
Substitute 2,1 , 7 :
2 2 1 7 4 1 7 10 10
4 2 2 1 3 7 8 2 21 11 10
Not a solution.
x y zx y z
x y z
Substitute 3, 10, 6 :
2 3 10 6 6 10 6 10 10
4 3 2 10 3 6 12 20 18
10 10
3 3 10 2 6 3 30 12 21 8
Not a solution.
Substitute 4, 0, 2 :
2 4 0 2 8 0 2 10 10
4 4 2 0 3 2 16 0 6 10 10
4 3 0 2 2 4 0 4 8 8
4, 0, 2 is a solu
tion.
8.
3 3 6 24
9 6 3 45
9 3 9 33
Substitute 1,1 , 3 :
3 1 3 1 6 3 3 3 18 24 24
9 1 6 1 3 3 9 6 9 6 45
Not a solution.
Substitute 0, 0, 4 :
3 0 3 0 6 4 0 0 24 24 24
9 0 6 0 3 4 0 0 12 12 45
Not a soluti
x y zx y z
x y z
on.
Substitute 4, 2,1 :
3 4 3 2 6 1 12 6 6 24 24
9 4 6 2 3 1 36 12 3 45 45
9 4 3 2 9 1 36 6 9 33 33
4, 2,1 is a solution.
9.
4 6
3 1
4 4
Substitute 12, 2, 2 :
12 2 4 2 12 2 8 6 6
12 3 2 2 12 6 2 4 1
Not a solution.
Substitute 4, 2,1 :
4 2 4 1 4 2 4 10 6
Not a solution.
Substitute 1,1 ,1 :
1 1 4 1 1
x y zx y zx y z
1 4 6 6
1 3 1 1 1 3 1 1 1
4 1 1 1 4 1 1 4 4
1,1 ,1 is a solution.
10.
2 5
3 5
2 4
Substitute 0, 4, 3 :
0 2 4 3 0 8 3 5 5
0 3 4 3 0 12 3 9 5
Not a solution.
Substitute 3, 6,1 0 :
3 2 6 10 3 12 10 5 5
3 3 6 10 3 18 10 5 5
2 3 6 10 6 6 10 10 4
Not a solution.
S
x y zx y zx y z
ubstitute 3, 3,1 :
3 2 3 1 3 6 1 8 5
Not a solution.
Section 3.6 Systems of Linear Equations in Three Variables and Applications
281
11. 2 3 12
3 2 3
5 2 3
x y zx y zx y z
Multiply the first equation by 2 and add
to the second equation to eliminate y: 2
2 3 12 4 2 6 24
3 2 3 3 2 3
7 7 21
3
x y z x y zx y z x y z
x zx z
Multiply the first equation by –5 and add
to the third equation to eliminate y:
2 3 12 10 5 15 60
5 2 3 5 2 3
11 17 57
x y z x y zx y z x y z
x z
Multiply the first result by 11 and add to
the second result to eliminate x: 11
3 11 11 33
11 17 57 11 17 57
6 24
4
x z x zx z x z
zz
Substitute and solve for x and y:
3
4 3
1
x zx
x
2 3 12
2 1 3 4 12
2 12 12
2
x y zy
yy
The solution is 1, 2, 4 .
12. 3 2 4 15
2 5 3 3
4 7 1 5
x y zx y zx y z
Multiply the third equation by –2 and add
to the first equation to eliminate y:
3 2 4 15 3 2 4 15
4 7 15 8 2 14 30
11 10 45
x y z x y zx y z x y z
x z
Multiply the third equation by 5 and add
to the second equation to eliminate y:
2 5 3 3 2 5 3 3
4 7 15 20 5 35 75
22 32 78
x y z x y zx y z x y z
x z
Multiply the first result by 2 and add to
the second result to eliminate x: 2
11 10 45 22 20 90
22 32 78 22 32 78
12 12
1
x z x zx z x z
zz
Substitute and solve for x and y:
11 10 45
11 10 1 45
11 10 45
11 55
5
x zx
xxx
4 7 15
4 5 7 1 15
20 7 15
2
2
x y zy
yyy
The solution is 5, 2, 1 .
Chapter 3 Systems of Linear Equations and Inequalities
282
13. 3 4 7
5 2 2 1
4 5 6
x y zx y zx y z
Multiply the third equation by –3 and
add to the first equation to eliminate y:
3 4 7 3 4 7
4 5 6 12 3 15 18
11 11 11
1
x y z x y zx y z x y z
x zx z
Multiply the third equation by 2 and add
to the second equation to eliminate y:
5 2 2 1 5 2 2 1
4 5 6 8 2 10 12
13 8 13
x y z x y zx y z x y z
x z
Multiply the first result by 8 and add to
the second result to eliminate z:
8
1 8 8 8
13 8 13 1 3 8 13
5 5
1
x z x zx z x z
xx
Substitute and solve for y and z:
1
1 1
0
x zzz
4 5 6
4 1 5 0 6
4 0 6
2
2
x y zy
yyy
The solution is 1, 2, 0 .
14. 6 5 7
5 3 2 0
2 3 11
x y zx y zx y z
Multiply the third equation by 5 and add
to the first equation to eliminate y:
6 5 7 6 5 7
2 3 11 10 5 15 55
4 14 62
x y z x y zx y z x y z
x z
Multiply the third equation by –3 and
add to the second equation to eliminate
y:
5 3 2 0 5 3 2 0
2 3 11 6 3 9 33
11 11 33
3
x y z x y zx y z x y z
x zx z
Multiply the second result by 4 and add
to the first result to eliminate x:
15. 4 2 12 3 4 3 2 1 2
2 3 3 5 3 2 3 5
2 7 8 2 7 8
x z y x y zy x z x y zy x z x y z
Multiply the third equation by –3 and add
to the first equation to eliminate y:
3
4 3 2 1 2 4 3 2 12
2 7 8 6 3 21 24
2 19 36
x y z x y zx y z x y z
x z
Multiply the third equation by 2 and add to
the second equation to eliminate y:
2
3 2 3 5 3 2 3 5
2 7 8 4 2 14 16
11 21
x y z x y zx y z x y z
x z
Multiply the second result by 2 and add to
the first result to eliminate x:
2
2 19 36 2 19 36
1 1 21 2 22 42
3 6
2
x z x zx z x z
zz
Section 3.6 Systems of Linear Equations in Three Variables and Applications
283
4
4 14 62 4 14 62
3 4 4 12
10 50
5
x z x zx z x z
zz
Substitute and solve for x and y:
3
5 3
2
x zx
x
2 3 11
2 2 3 5 11
4 15 11
0
x y zy
yy
The solution is 2, 0, 5 .
Substitute and solve for y and z:
11 21
11 2 21
22 21
1
x zx
xx
2 7 8
2 1 7 2 8
2 14 8
4
4
x y zy
yyy
The solution is 1, 4, 2 .
16. 2 1 2 1
3 1 2 2 3 2 2 1
5 3 16 3 5 3 3 16
y x z x y zx y z x y z
x z y x y z
Multiply the first equation by –3 and add
to the third equation to eliminate z:
3
2 1 6 3 3 3
5 3 3 16 5 3 3 16
6 19
x y z x y zx y z x y z
x y
Multiply the first equation by 2 and add to
the second equation to eliminate z:
2
2 1 4 2 2 2
3 2 2 1 3 2 2 1
1
x y z x y zx y z x y z
x
Substitute and solve for y and z:
6 19
1 6 19
1 6 19
6 18
3
x yyyyy
2 1
2 1 3 1
2 3 1
5 1
4
x y zzzzz
The solution is 1, 3, 4 .
17. 6
2
2 3 1 1
x y zx y zx y z
Add the first and second equations to
eliminate z:
6
2
2 4
2
x y zx y z
yy
Add the second and third equations to
eliminate z:
18. 11
1 5
2 9
x y zx y zx y z
Add the first and third equations to
eliminate z:
11
2 9
3 2 20
x y zx y z
x y
Add the second and third equations to
eliminate z:
Chapter 3 Systems of Linear Equations and Inequalities
284
2
2 3 1 1
4 9
x y zx y z
x y
Substitute y = 2 and solve for x:
4 2 9
8 9
1
xx
x
Substitute into the first equation and solve
for z:
1 2 6
3
zz
The solution is 1,2,3 .
1 5
2 9
3 6
2
x y zx y z
xx
Substitute x = 2 into the first result and
solve for y:
3 2 2 20
6 2 20
2 26
13
yyyy
Substitute into the first equation and solve
for z:
2 13 11
0
0
zzz
The solution is 2,1 3, 0 .
19. 2 3 2 1
2 4
1
x y zx yx z
Multiply the third equation by –2 and add
to the first equation to eliminate z:
2
2 3 2 1 2 3 2 1
1 2 2 2
3 3
1
x y z x y zx z x z
yy
Substitute and solve for x and z:
2 4
2 1 4
2 4
6
x yx
xx
1
6 1
7
x zzz
The solution is 6,1 , 7 .
20. 2
2 5
3 2
x y zx z
y z
Add the first and second equations to
eliminate z:
2
2 5
3 7
x y zx zx y
21. 4 9 8
8 6 1
6 6 1
x yx z
y z
Multiply the third equation by –1 and add
to the second equation to eliminate z:
1
8 6 1 8 6 1
6 6 1 6 6 1
8 6 0
x z x zy z y z
x y
Section 3.6 Systems of Linear Equations in Three Variables and Applications
285
Add the second and third equations to
eliminate z:
2 5
3 2
2 3 7
x zy z
x y
Multiply the first result by –3 and add to
the second result to eliminate y:
3
3 7 9 3 21
2 3 7 2 3 7
7 14
2
x y x yx y x y
xx
Substitute and solve for y and z:
2 5
2 2 5
4 5
1
1
x zzzzz
3 2
3 1 2
3 3
1
y zy
yy
The solution is 2,1 , 1 .
Multiply the first equation by –2 and add
to this result to eliminate x:
2
4 9 8 8 18 16
8 6 0 8 6 0
24 16
23
x y x yx y x y
y
y
Substitute and solve for x and z:
4 9 8
24 9 8
3
4 6 8
4 2
1
2
x y
x
xx
x
8 6 1
18 6 1
2
4 6 1
6 5
5
6
x z
z
zz
z
The solution is 1 2 5
, , 2 3 6
.
22. 3 2 11
7 4
6 1
x zy z
x y
Multiply the third equation by –3 and add
to the first equation to eliminate x:
3
3 2 11 3 2 11
6 1 3 18 3
18 2 8
x z x zx y x y
y z
Multiply the second equation by –18 and
add to this result to eliminate y: 18
7 4 18 126 72
18 2 8 1 8 2 8
128 64
12
y z y zy z y z
z
z
Substitute and solve for x and y:
3 2 11
13 2 11
2
3 1 11
3 12
4
x z
x
xxx
7 4
17 4
2
74
21
2
y z
y
y
y
The solution is 1 1
4, , 2 2
.
Chapter 3 Systems of Linear Equations and Inequalities
286
23. Let x = the first angle
Let y = the second angle
Let z = the third angle
180
2 5
3 11
x y zy xz x
Substitute the second and third equations
into the first and solve for x:
2 5 3 11 180
6 6 180
6 186
31
x x xx
xx
Substitute and solve for y and z:
2 5
2 31 5
62 5 67
y xyy
3 11
3 31 11
93 11 82
z xzz
The angles are 31º, 67º, and 82º.
24. Let x = the first angle
Let y = the second angle
Let z = the third angle
180
2
5 4
x y zy xz x
Substitute the second and third equations
into the first and solve for x:
2 5 4 180
8 4 180
8 184
23
x x xx
xx
Substitute and solve for y and z:
2
2 23
46
y xyy
5 4
5 23 4
115 4 111
z xzz
The angles are 23º, 46º, and 111º.
25. Let x = the shortest side
Let y = the middle side
Let z = the longest side
55 55
8 8
1 1
x y z x y zx y x yz x y x y z
Add the first and third equations to
eliminate x:
55
1
2 54
27
x y zx y z
zz
Add the second and third equations to
eliminate x:
8
1
2 9
x yx y z
y z
Substitute and solve for x and y:
26. Let x = the shortest side
Let y = the middle side
Let z = the longest side
60 60
20 20
3 3 0
x y z x y zz x x zy x x y
Multiply the second equation by –1 and
add to the first equation to eliminate z:
1
60 60
20 20
2 40
x y z x y zx z x z
x y
Multiply the third equation by –1 and add
to this result to eliminate y:
1
3 0 3 0
2 40 2 40
5 40
8
x y x y
x y x yx
x
Substitute and solve for y and z:
Section 3.6 Systems of Linear Equations in Three Variables and Applications
287
2 9
2 27 9
2 36
18
y zy
yy
8
18 8
10
x yx
x
The lengths of the sides are 10 cm,
18 cm, and 27 cm.
3
3 8
24
y xyy
20
8 20
28
z xzz
The lengths of the sides are 8 in, 24 in,
and 28 in.
27. Let x = the fiber in the supplement
Let y = the fiber in the oatmeal
Let z = the fiber in the cereal
3 4 19
2 4 2 25
5 3 2 30
x y zx y zx y z
Multiply the first equation by 4 and add to the second equation to eliminate y:
4
3 4 19 12 4 16 76
2 4 2 25 2 4 2 25
10 14 51
x y z x y zx y z x y z
x z
Multiply the first equation by 3 and add to the third equation to eliminate y:
3
3 4 19 9 3 12 57
5 3 2 30 5 3 2 30
4 10 27
x y z x y zx y z x y z
x z
Multiply the second new equation by 2.5 and add to the first new equation to eliminate x:
2.5
14 1410 51 10 51
10 254 27 10 67.5
11 16.5
1.5
z zx xz zx x
zz
Substitute and solve for x and y:
4 10 27
4 10 1.5 27
4 15 27
4 12
3
x zx
xxx
3 4 19
3 3 4 1.5 19
9 6 19
15 19
4
x y zy
yy
y
The fiber supplement has 3 g; the oatmeal has 4 g; and the cereal has 1.5 g.
Chapter 3 Systems of Linear Equations and Inequalities
288
28. Let x = the calcium in milk
Let y = the calcium in ice cream
Let z = the calcium in the supplement
1180
2 1680
2 1260
x y zx y zx y z
Multiply the first equation by 1 and add to the second equation to eliminate y and z:
1
1180 1180
2 1680 2 1680
500
x y z x y zx y z x y z
x
Substitute into the second and third equations:
2 1680
2 500 1680
1000 1680
680
x y zy zy zy z
2 1260
500 2 1260
2 760
x y zy zy z
Multiply the first new equation by 2 and add to the second new equation to eliminate y: 2680 2 2 1360
2 760 2 760
600
600
y z x zy z y z
zz
Substitute and solve for y:
1180
500 600 1180
1100 1180
80
x y zy
yy
The milk has 500 mg; the ice cream has 80 mg; and the calcium supplement has 600 mg.
29. Let x = the number of par 3 holes
Let y = the number of par 4 holes
Let z = the number of par 5 holes
18
3
3
x y zy xz x
Substitute the second and third equations
into the first and solve for x:
30. Let x = the number of ounces of
peanuts
Let y = the number of ounces of
pecans
Let z = the number of ounces of
cashews
Section 3.6 Systems of Linear Equations in Three Variables and Applications
289
3 3 18
5 3 18
5 15
3
x x xx
xx
Substitute and solve for y and z:
3
3 3
9
y xyy
3
3 3
6
z xzz
There are three par 3 holes, nine par 4
holes, and six par 5 holes.
48 48
0
2 2 0
x y z x y zx y z x y zz y y z
Add the first and second equations
to eliminate z:
48
0
2 48
24
x y zx y z
xx
Add the second and third equations
to eliminate z:
0
2 0
3 0
x y zy z
x y
Substitute and solve for y and z:
3 0
24 3 0
3 24
8
x yyyy
2
2 8
16
z yzz
There are 24 oz of peanuts, 8 oz of
pecans, and 16 oz of cashews in the
mixture.
31. Let x = the price of a hat
Let y = the price of a T-shirt
Let z = the price of a jacket
3 2 140
2 2 2 170
3 2 180
x y zx y z
x y z
Multiply the first equation by –2 and add to the second equation to eliminate z:
3 2 140 6 4 2 280
2 2 2 170 2 2 2 170
4 2 110
x y z x y zx y z x y z
x y
Multiply the third equation by –1 and add to the second equation to eliminate z:
Chapter 3 Systems of Linear Equations and Inequalities
290
1
2 2 2 170 2 2 2 1 70
3 2 180 3 2 180
10
x y z x y zx y z x y z
x y
Multiply the second result by –2 and add to the first result to eliminate y:
2
4 2 110 4 2 110
10 2 2 20
6 90
15
x y x yx y x y
xx
Substitute and solve for y and z:
10
15 10
25
25
x yyyy
3 2 140
3 15 2 25 140
45 50 140
45
x y zzzz
Hats cost $15. T-shirts cost $25, and jackets cost $45.
32. Let x = the cost per night Paris
Let y = the cost per night Stockholm
Let z = the cost per night Oslo
4 2 1040
0.08 0.11 4 0.10 2 106
80
4 2 1 040
8 44 20 10,600
80
x y zx y z
x zx y zx y zx z
Multiply the third equation by 2 and add to the first equation to eliminate z:
2
4 2 1040 4 2 1040
80 2 2 1 60
3 4 1200
x y z x y zx z x z
x y
Multiply the third equation by 20 and add to the second equation to eliminate z:
8 44 20 10600
8 44 20 10600
80
20 20 1 600
28 44 12200
x y zx y z
x zx zx y
Multiply the first result by –11 and add to the second result to eliminate y:
Section 3.6 Systems of Linear Equations in Three Variables and Applications
291
3 4 1 200 33 44 13200
28 44 12200 28 44 1 2200
5 1000
200
x y x yx y x y
xx
Substitute and solve for y and z:
80
200 80
120
120
x zzzz
3 4 1200
3 200 4 1200
600 4 1200
4 600
150
x yyyyy
Hotel costs per night are: Paris $200, Stockholm $150, and Oslo $120.
33. Let x = the amount invested in small caps
Let y = the amount invested in global markets
Let z = the amount invested in the balanced fund
25,000 4 2 1040
0.06 0.10 0.09 2106 6 10 9 216,000
2 2
x y z x y zx y z x y z
y z y z
Substitute into the first two equations to eliminate y:
25,000
2 25,000
3 25,000
x y zx z z
x z
6 10 9 216,000
6 10 2 9 216,000
6 20 9 216,000
6 29 216,000
x y zx z z
x z zx z
Multiply the first new equation by 6 and add to the second new equation to eliminate
x: 63 25,000 6 18 150,000
6 29 216,000 6 29 216,000
11 66,000
6000
x z x zx z x z
zz
Substitute and solve for x:
3 25,000
3 6000 25,000
18,000 25,000
7000
x zx
xx
Substitute and solve for y:
Chapter 3 Systems of Linear Equations and Inequalities
292
25,000
7000 6000 25,000
13,000 25,000
12,000
x y zy
yy
Walter invested $7000 in small caps, $12,000 in global markets, and $6000 in the
balanced fund.
34. Let x = the amount deposited in checking
Let y = the amount deposited in savings
Let z = the amount deposited in money market
8000 8000
0.012 0.025 0.03 202 12 25 30 202,000
3 3
x y z x y zx y z x y zx z x z
Substitute into the first two equations to eliminate z:
8000
3 8000
4 8000
x y zx y x
x y
12 25 30 202,000
12 25 30 3 202,000
12 25 90 202,000
102 25 202,000
x y zx y x
x y xx y
Multiply the first new equation by 25 and add to the second new equation to
eliminate y: 254 8000 100 25 200,000
102 25 202,000 102 25 202,000
2 2000
1000
x y x yx y x y
xx
Substitute and solve for y
4 8000
4 1000 8000
4000 8000
4000
x yyyy
Substitute and solve for z:
8000
1000 4000 8000
5000 8000
3000
x y zzzz
Raeann deposited $1000 in checking (at 1.2%), $4000 in savings (at 2.5%), and
$3000 in the money market (at 3%).
Section 3.6 Systems of Linear Equations in Three Variables and Applications
293
35. 2 3 2
2 4
2 2 4 8
x y zx y zx y z
Add the first and second equations to
eliminate y:
2 3 2
2 4
3 5 2
x y zx y z
x z
Multiply the second equation by 2 and
add to the third equation to eliminate
y:
2
2 4 2 2 4 8
2 2 4 8 2 2 4 8
0 0
x y z x y zx y z x y z
The equations are dependent.
36. 0
2 4 2 6 2 4 2 6
3 6 3 9 3 6 3 9
x y z x y zx y z x y zx y z x y z
Multiply the second equation by –3,
multiply the third equation by 2, and add
to eliminate x:
3
2
2 4 2 6 6 12 6 18
3 6 3 9 6 12 6 1 8
0 0
x y z x y zx y z x y z
The equations are dependent.
37. 6 2 2 2
4 8 2 5
2 4 2
x y zx y zx y z
Multiply the third equation by 2 and
add to the second equation to
eliminate z:
2
4 8 2 5 4 8 2 5
2 4 2 4 8 2 4
0 1
x y z x y zx y z x y z
The system is inconsistent. There is no
solution.
38. 3 2 3
3 4
6 4 2 1
x y zx y zx y z
Multiply the first equation by 2 and add
to the third equation to eliminate z:
2
3 2 3 6 4 2 6
6 4 2 1 6 4 2 1
0 7
x y z x y zx y z x y z
The system is inconsistent. There is no
solution.
39. Multiply by the LCD of each equation:
6 51 22 3 2
1031 15 2 10
1231 13 4 4
3 4 1 5
2 5 3
4 3 9
x y x yx z x z
y z y z
Multiply the first equation by –1 and add to the third equation to eliminate y:
Chapter 3 Systems of Linear Equations and Inequalities
294
1
3 4 15 3 4 15
4 3 9 4 3 9
3 3 6
x y x yy z y z
x z
Multiply this result by 2/3 and add to the second equation to eliminate x: 2/3
3 3 6 2 2 4
2 5 3 2 5 3
7 7
1
x z x zx z x z
zz
Substitute and solve for x and y:
2 5 3
2 5 1 3
2 5 3
2 2
1
x zx
xxx
4 3 9
4 3 1 9
4 3 9
4 12
3
y zy
yyy
The solution is 1, 3,1 .
40. Multiply by the LCD of each
equation:
4 1 12 4
891 1 18 4 4 8
32 13 3
3 2 4 1 2
2 2 9
3 3 2 1
x y z x y zx y z x y zx y z x y z
Add the second and third equations to
eliminate z:
2 2 9
3 3 2 1
4 10
x y zx y zx y
Multiply the third equation by 2 and
add to the first equation to eliminate z:
2
2 4 12 2 4 12
3 3 2 1 6 6 4 2
8 5 14
x y z x y zx y z x y z
x y
Multiply the first result by –2 and add
to the second result to eliminate x:
41. 3 8
4 2 3 3
2 3 2 1
x y zx y z
x y z
Multiply the first equation by 3 and add
to the second equation to eliminate z: 3
3 8 9 3 3 24
4 2 3 3 4 2 3 3
13 5 21
x y z x y zx y z x y z
x y
Multiply the first equation by –2 and add
to the third equation to eliminate z:
2
3 8 6 2 2 16
2 3 2 1 2 3 2 1
8 17
x y z x y zx y z x y z
x y
Multiply the second result by –5 and add
to the first result to eliminate y:
5
13 5 21 13 5 21
8 17 40 5 85
53 106
2
x y x yx y x y
xx
Section 3.6 Systems of Linear Equations in Three Variables and Applications
295
2
4 10 8 2 20
8 5 14 8 5 1 4
3 6
2
x y x yx y x y
yy
Substitute and solve for x and z:
4 10
4 2 10
4 12
3
x yx
xx
2 2 9
3 2 2 2 9
3 4 2 9
2 2
1
x y zzzzz
The solution is 3, 2,1 .
Substitute and solve for y and z:
8 17
8 2 17
16 17
1
x yyyy
3 8
3 2 1 8
6 1 8
3
3
x y zzzzz
The solution is 2, 1, 3 .
42. 2 3 3 1 5
3 6 6 23
9 3 6 8
x y zx y zx y z
Add the second and third equations to
eliminate z:
3 6 6 23
9 3 6 8
6 9 15
x y zx y z
x y
Multiply the first equation by 2 and
add to the second equation to
eliminate z:
Substitute and solve for y and z:
6 9 15
6 1 9 15
6 9 15
9 9
1
x yyyyy
2 3 3 15
2 1 3 1 3 15
2 3 3 15
3 10
10
3
x y zzzz
z
The solution is 10
1,1 , 3
.
2
2 3 3 1 5 4 6 6 30
3 6 6 23 3 6 6 23
7 7
1
x y z x y zx y z x y z
xx
43.
2 3 1 2 3 3
3 2 2 1 3 2 4 1
2 2 3 6 2 4 2 6 6
x y z x y zx y z x y z
x z y x y z
Multiply the first equation by –2 and
add to the third equation to eliminate z:
44. 2 3
2 16 10
7 3 4 8
x yy z
x y z
Multiply the third equation by –4 and
add to the second equation to eliminate
z:
Chapter 3 Systems of Linear Equations and Inequalities
296
2
2 3 3 4 2 6 6
4 2 6 6 4 2 6 6
0 0
x y z x y zx y z x y z
The equations are dependent.
2 16 10 2 16 10
7 3 4 8 28 12 16 32
28 14 42
y z y zx y z x y z
x y
Multiply the first equation by –14 and
add to the this result to eliminate y: 14
2 3 28 14 42
28 14 42 28 14 42
0 0
x y x yx y x y
The equations are dependent.
45. Multiply each equation by 10:
0.1 0.2 0.2 2 2
0.1 0.1 0.1 0.2 2
0.1 0.3 0.2 3 2
y z y zx y z x y z
x z x z
Add the first and second equations to
eliminate y:
2 2
2
3 4
y zx y zx z
Add this result to the third equation to
eliminate x:
3 4
3 2
0 6
x zx z
The system is inconsistent. There is no
solution.
46. Multiply each equation by 10:
0.4 0.3 0 4 3 0
0.3 0.1 0.1 3 1
0.4 0.1 1 .2 4 1 2
x y x yy z y z
x z x z
Add the second and third equations to
eliminate z:
3 1
4 12
4 3 11
y zx zx y
Add the first equation to this result to
eliminate x:
4 3 0
4 3 11
0 11
x yx y
The system is inconsistent. There is no
solution.
47. 2 4 8 0
3 0
2 5 0
x y zx y z
x y z
Add the second and third equations to
eliminate x:
3 0
2 5 0
5 6 0
x y zx y z
y z
48. 2 4 0
3 0
3 2 0
x y zx y zx y z
Add the first and second equations to
eliminate z:
2 4 0
3 0
3 7 0
x y zx y zx y
Section 3.6 Systems of Linear Equations in Three Variables and Applications
297
Multiply the second equation by 2 and
add to the first equation to eliminate x:
2
2 4 8 0 2 4 8 0
3 0 2 6 2 0
10 10 0
x y z x y zx y z x y z
y z
Multiply the first result by –2 and add to
the second result to eliminate y: 2
5 6 0 10 12 0
10 10 0 10 10 0
2 0
0
y z y zy z y z
zz
Substitute and solve for x and y:
5 6 0
5 6 0 0
5 0
0
y zy
yy
2 5 0
2 0 5 0 0
0 0 0
0
x y zx
xx
The solution is 0, 0, 0 .
Multiply the second equation by 2 and
add to the third equation to eliminate z:
2
3 0 2 6 2 0
3 2 0 3 2 0
5 7 0
x y z x y zx y z x y z
x y
Multiply the first result by –1 and add to
the second result to eliminate y:
1
3 7 0 3 7 0
5 7 0 5 7 0
2 0
0
x y x yx y x y
xx
Substitute and solve for y and z:
3 7 0
3 0 7 0
7 0
0
x yyyy
2 4 0
2 0 4 0 0
0 0 0
0
x y zzzz
The solution is 0, 0, 0 .
49. 4 2 3 0
8 0
32 0
2
x y zx y z
x y z
Multiply the third equation by –2 and
add to the first equation to eliminate y:
2
4 2 3 0 4 2 3 0
32 0 4 2 3 0
20 0
x y z x y z
x y z x y z
The equations are dependent.
50. 5 0
4 0
5 5 0
x yy z
x y z
Multiply the second equation by –1 and
add to the third equation to eliminate z: 1
4 0 4 0
5 5 0 5 5 0
5 0
y z y z
x y z x y z
x y
Multiply the first equation by –1 and
add to the this result to eliminate y:
1
5 0 5 0
5 0 5 0
0 0
x y x yx y x y
The equations are dependent.
Chapter 3 Systems of Linear Equations and Inequalities
298
Section 3.7 Practice Exercises
1. a. matrix; rows; columns c. coefficient; augmented
b. column; one; square d. row echelon
2. Let x = amount of pure (100%) acid sol
y = the amount of 50% acid solution
100% 50% 70%__
oz solution x y 20__
oz bleach 1.00x 0.50y 0.70(20)
20
1.00 0.50 0.70 20
x yx y
Multiply the first equation by –0.50, add
to the second equation and solve for x:
20 0.50 0.50 10
1 0.50 14 1.00 0.50 1 4
0.50 4
8
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
8 20 12y y
Mix 12 L of the 50% mixture with 8L of
pure acid.
3. 6 9
2 13
x yx y
Multiply the first equation by –1, add to
the second equation and solve for y: 1
6 9 6 9
2 13 2 1 3
8 4
1
2
x y x yx y x y
y
y
Substitute into the first equation and solve
for x:
6 9
16 9
2
3 9
12
x y
x
xx
The solution is 1
12, 2
.
4. 8
2 3
3 2 7
x y zx y zx y z
Add the first and second equations to
eliminate z:
8
2 3
2 11
x y zx y zx y
Multiply the second equation by –2 and
add to the third equation to eliminate z:
5. 2 4
3 7
3 4 22
x y zx y z
x y z
Add the first and second equations to
eliminate y:
2 4
3 7
4 11
x y zx y zx z
Multiply the first equation by 3 and add to
the third equation to eliminate y:
Section 3.7 Solving Systems of Linear Equations by Using Matrices
299
2
2 3 2 4 2 6
3 2 7 3 2 7
7 1
x y z x y zx y z x y z
x y
Multiply the second result by 2 and add
to the first result to eliminate x:
2
2 11 2 11
7 1 2 14 2
13 13
1
x y x yx y x y
yy
Substitute and solve for x and z:
2 11
2 1 11
2 12
6
x yx
xx
2 3
6 2 1 3
6 2 3
1
x y zzzz
The solution is 6,1 , 1 .
3
2 4 6 3 3 12
3 4 22 3 4 22
7 10
x y z x y zx y z x y z
x z
Multiply the second result by 4 and add to
the first result to eliminate z:
4
4 11 4 11
7 1 0 28 4 40
29 29
1
x z x zx z x z
xx
Substitute and solve for y and z:
4 11
1 4 11
4 12
3
x zzzz
3 7
1 3 3 7
1 9 7
3
x y zy
yy
The solution is 1, 3, 3 .
6. 4 1 , column matrix 7. 3 1 , column matrix
8. 3 3 , square matrix 9. 2 2 , square matrix
10. 1 2 , row matrix 11. 1 4 , row matrix
12. 2 4 , none of these 13. 2 3 , none of these
14. 3 2 , none of these
15. 1 2 1
2 1 7
16. 1 3 3
2 5 4
17. 1 2 1 5
2 6 3 2
3 1 2 1
18. 5 0 2 17
8 1 6 26
8 3 12 24
19. 4 3 6
12 5 6
x yx y
20. 2 5 15
7 15 45
x yx y
Chapter 3 Systems of Linear Equations and Inequalities
300
21. 4
1
7
xyz
22. 0.5
6.1
3.9
xyz
23. a. 7 24. a. –13
b. –2 b. 0
25. 11 12
1 112 2
2 1 11 1
2 1 1 2 1 1
R RZ
26. 1
2 23
1 1 7 1 1 7
0 3 6 0 1 2
R RJ
27. 1 2
5 2 1 1 4 3
1 4 3 5 2 1R RK
28. 1 2
9 6 13 7 2 19
7 2 19 9 6 13R RL
29. 1 2 23
1 5 2 1 5 2
3 4 1 0 11 5R R RM
30. 1 2 22
1 3 5 1 3 5
2 2 12 0 8 2R R RN
31. a.
1 2 24
1 3 0 1
4 1 5 6
2 0 3 10
1 3 0 1
0 11 5 10
2 0 3 10
R R R
32. a.
1 2 25
1 2 0 10
5 1 4 3
3 4 5 2
1 2 0 1 0
0 9 4 47
3 4 5 2
R R R
b. 1 3 32
1 3 0 1
0 11 5 10
0 6 3 8
R R R
b. 1 3 33
1 2 0 1 0
0 9 4 47
0 1 0 5 32
R R R
33. True 34. True
35. True 36. True
Section 3.7 Solving Systems of Linear Equations by Using Matrices
301
37. Interchange rows 1 and 2. 38. Multiply row 3 by 2. Replace row 3
with the result.
39. Multiply row 1 by –3 and add to row 2.
Replace row 2 with the result.
40. Multiply row 2 by 4 and add to row 3.
Replace row 3 with the result.
41.
12 251 2 2 2 1 12 2
2 1
2 7
1 2 1 1 2 1 1 2 1 1 0 3
2 1 7 0 5 5 0 1 1 0 1 1
R RR R R R R R
x yx y
The solution is 3, 1 .
42.
1 2 2 2 1 12 3
3 3
2 5 4
1 3 3 1 3 3 1 0 3
2 5 4 0 1 2 0 1 2R R R R R R
x yx y
The solution is 3, 2 .
43.
12 231 2 2 2 1 14 3
3 6
4 9 3
1 3 6 1 3 6 1 3 6 1 0 21
4 9 3 0 3 27 0 1 9 0 1 9
R RR R R R R R
x yx y
The solution is 21, 9 .
44.
12 271 2 1 2 22
2 3 2
2 13
2 3 2 1 2 13 1 2 1 3 1 2 13
1 2 13 2 3 2 0 7 28 0 1 4
R RR R R R R
x yx y
2 1 121 0 5
0 1 4R R R
The solution is 5, 4 .
Chapter 3 Systems of Linear Equations and Inequalities
302
45.
1 2 24
3 3
4 12 12
1 3 3 1 3 3
4 12 12 0 0 0R R R
x yx y
Infinitely many solutions of the form , 3 3x y x y . The equations are dependent.
46.
1 2 22
2 5 1
4 10 2
2 5 1 2 5 1
4 10 2 0 0 0R R R
x yx y
Infinitely many solutions of the form , 2 5 1x y x y . The equations are dependent.
47.
12 231 2 2 2 1 12
4
2 5
1 1 4 1 1 4 1 1 4 1 0 3
2 1 5 0 3 3 0 1 1 0 1 1
R RR R R R R R
x yx y
The solution is 3, 1 .
48.
12 231 2 1 2 2
2 1 1
2
1
2 0
3
2 1 0 1 1 3 1 1 3 1 1 3
1 1 3 2 1 0 0 3 6 0 1 2
1 0 1
0 1 2
R RR R R R R
R R R
x yx y
The solution is 1, 2 .
49.
12 231 2 2 2 1 13 3
3 1
3 6 12
1 3 1 1 3 1 1 3 1 1 0 10
3 6 12 0 3 9 0 1 3 0 1 3
R RR R R R R R
x yx y
The solution is 10, 3 .
Section 3.7 Solving Systems of Linear Equations by Using Matrices
303
50.
12 261 2 2 2 1 12
4
2 4 4
1 1 4 1 1 4 1 1 4 1 0 2
2 4 4 0 6 12 0 1 2 0 1 2
R RR R R R R R
x yx y
The solution is 2, 2 .
51.
1 2 22
3 4
6 2 3
3 1 4 3 1 4
6 2 3 0 0 5R R R
x yx y
There is no solution; . The system is inconsistent.
52.
1 2 23
2 4
6 3 1
2 1 4 2 1 4
6 3 1 0 0 13R R R
x yx y
There is no solution; . The system is inconsistent.
53.
12 21 2 2 2
1 3 3
2 1 1
6
2
0
1 1 1 6 1 1 1 6 1 1 1 6
1 1 1 2 0 2 0 4 0 1 0 2
1 1 1 0 0 0 2 6 0 0 2 6
1 0 1 4
0 1 0 2
0 0 2 6
R RR R RR R R
R R R
x y zx y zx y z
13 3 3 1 12
1 0 1 4 1 0 0 1
0 1 0 2 0 1 0 2
0 0 1 3 0 0 1 3
R R R R R
The solution is 1, 2, 3 .
54. 2 3 2 11
3 8 1
3 14 2
x y zx y zx y z
Chapter 3 Systems of Linear Equations and Inequalities
304
1 2 1 2 2
1 3 3
12 29 2 1
2
3
3
2 3 2 11 1 3 8 1 1 3 8 1
1 3 8 1 2 3 2 11 0 9 18 9
3 1 14 2 3 1 14 2 0 10 10 5
1 3 8 1
0 1 2 1
0 10 10 5
R R R R RR R R
R R R R
1
2 3 3
13 310 3 1 1
3 2 2
10
2
2
3 32 2
1 0 2 4
0 1 2 1
0 0 10 15
1 0 2 4 1 0 0 7
0 1 2 1 0 1 0 2
0 0 1 0 0 1
RR R R
R R R R RR R R
The solution is 32
7, 2, .
55.
1 2 2
1 3 3
2
3
2 5 2 5
2 6 3 10 2 6 3 10
3 2 5 3 2 5
1 2 1 5 1 2 1 5
2 6 3 10 0 10 1 20
3 1 2 5 0 5 5 10
R R RR R R
x y z x y zx y z x y zx y z x y z
12 210
623 1 13 3 52 1 1 11
12 3 3 3 2 210
110
6 65 5
2 1 110 105
112
1 2 1 5
0 1 2
0 5 5 10
1 0 1 1 0 1 1 0 0 1
0 1 2 0 1 2 0 1 0
0 0 0 0 0 1 0
R R
R R RR RR R RR R R R R R
2
0 0 1 0
The solution is 1, 2, 0 .
56. 5 10 15 5 1 0 15
6 23 6 23
3 12 13 3 12 13
x z x zx y z x y z
x y z x y z
Section 3.7 Solving Systems of Linear Equations by Using Matrices
305
1 2 1 2 2
1 3 3
12 25
5
5 0 10 15 1 1 6 23 1 1 6 23
1 1 6 23 5 0 10 15 0 5 40 100
1 3 12 13 1 3 12 13 0 4 18 10
1 1 6 23
0 1 8 20
R R R R RR R R
R R
2 1 1
2 3 3
13 3 3 1 114
3 2 2
4
2
8
1 0 2 3
0 1 8 20
0 4 18 10 0 0 14 70
1 0 2 3 1 0 0 13
0 1 8 20 0 1 0 20
0 0 1 5 0 0 1 5
R R RR R R
R R R R RR R R
The solution is 13, 20, 5 .
57.
58.
59.
60.
61.
62.
Chapter 3 Systems of Linear Equations and Inequalities
306
Chapter 3 Group Activity
1.
2. Answers will vary. 3. Answers will vary.
4. Answers will vary based on the
choices for the points 1 1,t y ,
2 2,t y , and 3 3,t y . However, a,
b, and c should be close to 16a ,
32.5b , and 5c .
5. Answers will vary, but should be close to
216 32.5 5y t t .The equation
represents the height of the ball t seconds
after being thrown.
6. 2
2
16 32.5 5
1 116 32.5 5
2 2
4 16.25 5
17.25
y t t
y
Answers should be close to 17.25 ft.
7. 216 1.4 32.5 1.4 5
31.36 45.5 5
19.14
19 ft
y
Answers will vary but should be close to the
observed value of 19 ft.
Chapter 3 Review Exercises
307
Chapter 3 Review Exercises Section 3.1 1. a.
5 7 4
11
2Substitute 2, 2 :
5 2 7 2 10 14 24 4
2, 2 is not a solution.
x y
y x
b.
Substitute 2, 2 :
5 2 7 2 10 14 4 4
12 2 1 1 1 2
22, 2 is a solution.
2. False 3. True
4. True
5.
1
2 4
f x xg x x
The solution is 3, 2 .
6. 2 7
5
y xy x
The solution is 4, 1 .
7. 6 2 4 3 2
2 6 4 3 2
3 2
x y x yy x y xy x
8. 12 4 8 8
28 4 8
11
2
y x x y
y x
y x
Chapter 3 Systems of Linear Equations and Inequalities
308
Infinitely many solutions of the form
, 6 2 4x y x y ; dependent
equations.
There is no solution; . The system is
inconsistent.
Section 3.2
9. 34
42 6
y x
x y
32 4 6
4
38 6
21
22
4
x x
x x
xx
3 34 4 4 3 4 1
4 4y x
The solution is 4, 1 .
10. 6 5
5 3 5 3
x yx y y x
6 5
6 5 3 5
3 5
2
x yx x
xx
5 3 5 2 3 10 3 7y x
The solution is 2, 7 .
11. 2 10 3 2 2 10 3
2 5 10
0.4 1.2 1.2 0.4
x y y x y yx y
x y y x
2 5 1.2 0.4 10
2 6 2 10
6 10
x xx x
There is no solution; . The system
is inconsistent.
12. 3 11 9
3 6
11 11
x y
y x
3 63 11 9
11 11
3 3 6 9
0 3
x x
x x
There is no solution; . The system is
inconsistent.
13. 60 5 90 300 60 90
10 2 3 2 10 3
35
2
x y x yx y y x
y x
3300 60 5 90
2
300 300 90 90
90 90
x x
x x
14. 4 7 4 7
1 74 4
x y y x
x y
1 74 7
4 47 74 47 74 4
x x
x x
Chapter 3 Review Exercises
309
Infinitely many solutions of the form
, 10 2 3x y x y ; dependent
equations.
Infinitely many solutions of the form
, 4 7x y x y ; dependent system.
15. 105 45
48.50
y xy x
105 45 48.50
105 3.5
30
x xx
x
The cost would be the same for 30
months.
16. 44 81.50
87
y xy x
44 81.50 87
44 5.50
8
x xx
x
The cost would be the same for 8 days.
Section 3.3 17. Multiply each equation by its LCD:
5
3
2 31 2 3 5
5 52 1
3 2 13 3
x y x y
x y x y
Multiply the first equation by 2 and the
second equation by 3, add the results and
solve for x: 2
3
2 3 5 4 6 10
3 2 1 9 6 3
13 13
1
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
2 1 3 5
2 3 5
3 3
1
yyyy
The solution is 1,1 .
18. 4 3 5
3 4 10
x yx y
Multiply the first equation by 4 and the
second equation by 3, add the results and
solve for x:
4
3
4 3 5 16 12 20
3 4 10 9 12 30
25 50
2
x y x yx y x y
xx
Substitute into the first equation and solve
for y:
4 2 3 5
8 3 5
3 3
1
yyyy
The solution is 2, 1 .
19. 3 4 2
2 5 1
x yx y
20. Multiply each equation by its LCD:
Chapter 3 Systems of Linear Equations and Inequalities
310
Multiply the first equation by 5 and the
second equation by –4, add the results
and solve for x: 5
4
3 4 2 15 20 10
2 5 1 8 20 4
7 14
2
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
3 2 4 2
6 4 2
4 4
1
yyyy
The solution is {(2, –1)}.
3
3 1 3 1
1 1 3 1
3 3
x y x y
x y x y
Add the two equations and solve for x:
3 1
3 1
0 0
x yx y
Infinitely many solutions of the form
, 3 1x y x y ; dependent equations.
21. Write in standard form:
2 3 8 3 2 8
6 4 4 6 4 4
y x x yx y x y
Multiply the first equation by –2, add to
the second equation and solve for x:
2
3 2 8 6 4 16
6 4 4 6 4 4
0 20
x y x yx y x y
There is no solution; . The system is
inconsistent.
22. Write in standard form:
3 16 3 16
3 2 2 2 2
x y x yx y y x x y
Multiply the first equation by –2, add to
the second equation and solve for x: 2
3 16 6 2 32
2 2 2 2
5 30
6
x y x yx y x y
xx
Substitute into the first equation and solve
for y:
3 6 16
18 16
2
yyy
The solution is 6, 2 .
23. Write in standard form:
4 2 9 6 9
2 2 10 2 2 10
y x x x yx y x y
Multiply the first equation by 2, add to
the second equation and solve for x:
Substitute into the first equation and solve
for y:
6 2 9
12 9
3
yyy
Chapter 3 Review Exercises
311
2
6 9 12 2 18
2 2 10 2 2 10
14 28
2
x y x yx y x y
xx
The solution is 2, 3 .
24. Write in standard form:
4 35 3 4 3 35
15 15
x y x yx y x y
Multiply the second equation by –3, add
to the first equation and solve for x:
3
4 3 35 4 3 35
15 3 3 45
10
10
x y x yx y x y
xx
Substitute into the second equation and
solve for y:
10 15
10 15
25
25
yyyy
The solution is 10, 25 .
25. Multiply each equation by 10:
0.4 0.3 1 .8 4 3 1 8
0.6 0.2 1.2 6 2 12
x y x yx y x y
Multiply the first equation by 2 and the
second equation by 3, add the results and
solve for x: 2
3
4 3 1 8 8 6 36
6 2 12 1 8 6 36
10 0
0
x y x yx y x y
xx
Substitute into the first equation and solve
for y:
4 0 3 18
0 3 18
6
yyy
The solution is 0, 6 .
26. Multiply each equation by 100:
0.02 0.01 0.11 2 11
0.01 0.04 0.26 4 26
x y x yx y x y
Multiply the first equation by 4, add to
the second equation and solve for x:
4
3
2 11 8 4 44
4 26 4 26
9 18
2
x y x yx y x y
xx
Substitute into the first equation and solve
for y:
2 2 11
4 11
7
7
yyyy
The solution is 2, 7 .
Section 3.4 27. Let x = the amount invested at 5%
y = the amount invested at 3.5%
28. Let x = the number of student tickets
Let y = the number of adult tickets
Let 1x = receipts from student tickets
Chapter 3 Systems of Linear Equations and Inequalities
312
5% Acct 3.5% Acct Total
Principal x y ____________
Interest 0.05x 0.035y 303.75
2
0.05 0.035 303.75
x yx y
Substitute and solve for y:
0.05 2 0.035 303.75
0.10 0.035 303.75
0.135 303.75
2250
y yy y
yy
Substitute into the first equation and
solve for x: 2 2 2250 4500x y
$4500 was invested at 5%.
1.50y = receipts from adult tickets
54 54
1 1.50 70.50
x y y xx y
1 1.50 54 70.50
81 1.50 70.50
0.50 10.50
21
54 21
33
x xx x
xxy
There were 21 student tickets and 33 adult
tickets sold.
29. Let x = the amount of 20% saline
solution
y = the amount of 50% saline solution
20% sal 50% sal 31.25% sal
L solution x y 16 _
L saline 0.20x 0.50y 0.3125(16)
16
0.20 0.50 0.3125 16
x yx y
Multiply the first equation by –0.20, add
to the second equation and solve for y:
16 0.20 0.20 3.2
0.20 0.50 5 0.20 0.50 5.0
0.30 1.8
6
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
6 16
10
xx
The mixture contains 10 L of 20% saline
solution and 6 L of 50% saline solution.
30. Let p = the speed of the plane in still air
Let w = the speed of the wind
p + w = speed of the plane with the wind
p – w = speed of plane against the wind
Distance Rate Time_
Tailwind 280 p + w 1.75_
Headwind 280 p – w 2__
(rate)(time) = (distance)
1.75 280
2 280
p wp w
Divide the first equation by 1.75, the
second equation by 2, add the results, and
solve:
1.75
2
1.75 280 160
2 280 140
2 300
150
div
div
p w p wp w p w
pp
Substitute and solve for w:
150 160 10w w
The speed of the plane is 150 mph in still
air and the speed of the wind is 10 mph.
Chapter 3 Review Exercises
313
31. a. 915 275f x x 32. Let x = one angle
Let y = the other angle
5 6
90
y xx y
Substitute and solve:
5 6 90
6 84
14
5 14 6 70 6 76
x xxxy
The two angles measure 14º and 76º.
b. 12.95 0.08g x x
c. Substitute and solve:
915 275 965
50 275
5.5
x xxx
5.5 months.
Section 3.5
33.
2 5
Graph the related equation 2 5 by using a dashed line.
Test point above 0, 6 : Test point below 0, 0 :
2 0 6 5 2 0 0 5
0 1 0 5
0, 6 is a solution. 0, 0 is not a solution.
Shade the region
x yx y
above the boundary line.
34.
2 8 3
Graph the related equation 2 8 3 by using a solid line.
Test point above 0, 0 : Test point below 0, 3 :
2 0 8 3 0 2 0 8 3 3
0 8 0 1
0, 0 is not a solution. 0, 3 is a solution.
Shade the r
x yx y
egion below the boundary line.
Chapter 3 Systems of Linear Equations and Inequalities
314
35. 3 represents all the points to the right of the vertical
line 3. The boundary is a dashed line.
Shade the region to the right of the boundary line.
xx
36. 2 represents all the points to the left of the vertical
line 2. The boundary is a solid line.
Shade the region to the left of the boundary line.
xx
37.
1
21
Graph the related equation by using a solid line.2
Test point above 0, 2 : Test point below 0, 2 :
1 10 2 0 2
2 20 1 0 1
0, 2 is not a solution. 0, 2 is a solution.
x y
x y
Chapter 3 Review Exercises
315
Shade the region below the boundary line.
38.
2
52
Graph the related equation by using a dashed line.5
Test point above 0, 5 : Test point below 0, 5 :
2 20 5 0 5
5 50 2 0 2
0, 5 is a solution. 0, 5 is not a solution.
Shade the region above
x y
x y
the boundary line.
39.
2 2 and 2 2
Graph the related equation 2 2 by using a dashed line.
Test point above 0, 3 : Test point below 0, 0 :
2 0 3 2 2 0 0 2
3 2 0 2
0, 3 is not a solution. 0, 0 is a solution.
Sha
x y x yx y
de the region below the boundary line.
Graph the related equation 2 2 by using a solid line.x y
Chapter 3 Systems of Linear Equations and Inequalities
316
Test point above 0, 0 : Test point below 0, 3 :
2 0 0 2 2 0 3 2
0 2 3 2
0, 0 is a solution. 0, 3 is not a solution.
Shade the region above the boundary line.
The solution is the intersection of the graphs.
40.
3 6 or 3 6
Graph the related equation 3 6 by using a solid line.
Test point above 0, 7 : Test point below 0, 0 :
3 0 7 6 3 0 0 6
7 6 0 6
0, 7 is a solution. 0, 0 is not a solution.
Shade the r
x y x yx y
egion above the boundary line.
Graph the related equation 3 6 by using a dashed line.
Test point above 0, 0 : Test point below 0, 7 :
3 0 0 6 3 0 7 6
0 6 7 6
0, 0 is not a solution. 0, 7 is
x y
a solution.
Shade the region below the boundary line.
The solution is the union of the graphs.
Chapter 3 Review Exercises
317
41.
or
Graph the related equation by using a solid line.
Test point above 0, 5 : Test point below 0, 5 :
5 0 5 0
0, 5 is a solution. 0, 5 is not a solution.
y x y xy x
Shade the region above the boundary line.
Graph the related equation by using a solid line.
Test point above 1, 5 : Test point below 1, 5 :
5 1 5 1
1, 5 is not a solution. 1, 5 is a solution.
Shade the region below the boundary li
y x
ne.
The solution is the union of the graphs.
42. 20 and 0 and 4
30 represents the points to the right of the vertical line 0.
Shade the region to the right of the boundary line using a solid
line border.
0 represents the points above
x y y x
x x
y
the horizontal line 0.
Shade the region above the boundary line using a solid line border.
2Graph the related equation 4 by using a solid line.
3Test point above 0, 5 : Test point below 0
y
y x
, 0 :
2 25 0 4 0 0 4
3 35 4 0 4
0, 5 is not a solution. 0, 0 is a solution.
Shade the region below the boundary line.
The solution is the intersection of the graphs.
Chapter 3 Systems of Linear Equations and Inequalities
318
43. a. 0, 0x y
b. 100x y
c. 4x y
d. 0 and 0 and 100 and 4
0 represents the points to the right of the vertical line
0. Shade the region to the right of the boundary line
using a solid line border.
0 represents the p
x y x y x yxx
y
oints above the horizontal line 0.
Shade the region above the boundary line using a solid
line border.
Graph the related equation 100 by using a solid line.
Test point above 0,1 01 : Test
y
x y
point below 0, 0 :
0 101 100 0 0 100
101 100 0 100
0,1 01 is not a solution. 0, 0 is a solution.
Shade the region below the boundary line.
Graph the related equation 4 by using a solid line.
Test point above
x y
0,1 : Test point below 0, 1 :
0 4 1 0 4 1
0 4 0 4
0,1 is not a solution. 0, 1 is a solution.
Shade the region below the boundary line.
The solution is the intersection of the graphs.
Chapter 3 Review Exercises
319
Section 3.6 44. 5 5 5 30
2
10 6 2 4
x y zx y zx y z
Multiply the second equation by –5 and
add to the first equation to eliminate z:
5
5 5 5 30 5 5 5 30
2 5 5 5 10
10 20
2
x y z x y zx y z x y z
xx
Multiply the second equation by 2 and add
to the third equation to eliminate z:
2
2 2 2 2 4
10 6 2 4 1 0 6 2 4
8 8 8
x y z x y zx y z x y z
x y
Substitute and solve for y and z:
8 8 8
8 2 8 8
16 8 8
8 8
1
x yyyyy
2
2 1 2
3 2
5
x y zzzz
The solution is 2, 1, 5 .
45. 5 3 5
2 6
2 8
x y zx y zx y z
Add the second and third equations to
eliminate z:
2 6
2 8
0 14
x y zx y z
The system is inconsistent. There is no
solution.
46. 4
2 3 6
2 4 6 1 2
x y zx y zx y z
Multiply the second equation by 2 and add
to the third equations to eliminate z:
2
2 3 6 2 4 6 12
2 4 6 1 2 2 4 6 1 2
0 0
x y z x y zx y z x y z
The equations are dependent.
Chapter 3 Systems of Linear Equations and Inequalities
320
47. 3 4 5
2 3 2
2 5 8
x zy z
x y
Multiply the first equation by 3 and the
second equation by –4, and add the
results to eliminate z:
3
4
3 4 5 9 1 2 15
2 3 2 8 12 8
9 8 7
x z x zy z y z
x y
Multiply the third equation by –9 and this
result by 2, and add to eliminate x:
9
2
2 5 8 18 45 72
9 8 7 1 8 16 1 4
29 58
2
x y x yx y x y
yy
Substitute and solve for x and z:
2 5 8
2 5 2 8
2 10 8
2 2
1
x yx
xxx
2 3 2
2 2 3 2
4 3 2
3 6
2
y zzzzz
The solution is 1, 2, 2 .
48. Let x = the shortest leg
Let y = the longer leg
Let z = the hypotenuse
30
2 2
3 2
x y zy xz x
Substitute the second and third equations
into the first equation and solve:
2 2 3 2 30
6 30
5
x x xxx
2 2
2 5 2
10 2
12
3 2
3 5 2
15 2
13
y x
y x
The lengths of the sides are 5 ft,
12 ft, and 13 ft.
49. Let x = the rate of the slowest pump
Let y = the rate of the middle rate pump
Let z = the rate of the fastest pump
950 950
150 150
150 150
x y z x y zx z x zz x y x y z
Add the first and third equations to
eliminate x:
950
150
2 800
400
x y zx y z
zz
50. Let x = the first angle
Let y = the second angle
Let z = the third angle
180
9 9
3 26
x y zx y y xz x
Substitute the second and third equations
into the first and solve for x:
9 3 26 180
5 35 180
5 145
29
x x xx
xx
Substitute and solve for y and z:
Chapter 3 Review Exercises
321
Substitute and solve for x and y:
150
400 150
250
x zxx
950
250 400 950
650 950
300
x y zyy
y
The pumps can drain 250, 300, and
400 gal/hr.
9
29 9
38
y xyy
3 26
3 29 26
87 26
113
z xzzz
The angles are 29º, 38º, and 113º.
Section 3.7 51. 3 3 52. 3 2
53. 1 4 54. 3 1
55. 1 1 3
1 1 1
56. 1 1 1 4
2 1 3 8
2 2 1 9
57. 9
3
xy
58. 5
2
8
xyz
59. a.
4
60. a.
1 2 24
1 2 0 3
4 1 1 0
3 2 2 5
1 2 0 3
0 9 1 12
3 2 2 5
R R R
b. 1 2 24
1 3 1 1 3 1
4 1 6 0 13 2R R R
b. 1 3 33
1 2 0 3
0 9 1 12
0 8 2 4
R R R
Chapter 3 Systems of Linear Equations and Inequalities
322
61.
12 21 2 2 2 2 1 11 1
3
1
1 1 3 1 1 3 1 1 3 1 0 1
1 1 1 0 2 4 0 1 2 0 1 2
R RR R R R R R
x yx y
The solution is 1, 2 .
62.
12 21 2 2 4 2 1 1
11 14
3 3
4 3 6
12 5 6
4 3 6 4 3 6 4 3 6 4 0 12
12 5 6 0 4 24 0 1 6 0 1 6
1 0 3
0 1 6
R RR R R R R R
R R
x yx y
The solution is 3, 6 .
63.
2 31 2 2
1 3 3
12 23
2
4
2 2 9
2 5
1 1 1 4 1 1 1 4 1 1 1 4
2 1 2 9 0 3 4 17 0 3 0 9
1 2 1 5 0 3 0 9 0 3 4 17
1 1 1 4
0 1
R RR R RR R R
R R
x y zx y zx y z
2 1 1
2 3 3
13 3 3 1 14
3
1 0 1 1
0 3 0 1 0 3
0 3 4 17 0 0 4 8
1 0 1 1 1 0 0 1
0 1 0 3 0 1 0 3
0 0 1 2 0 0 1 2
R R RR R R
R R R R R
The solution is 1, 3, 2 .
64. 4
2 3 8
2 2 9
x y zx y zx y z
Chapter 3 Test
323
1 2 2 2 1 1
1 3 3
3 1 1
3 2 2
2
2
2
1 1 1 4 1 1 1 4 1 0 2 4
2 1 3 8 0 1 1 0 0 1 1 0
2 2 1 9 0 0 1 1 0 0 1 1
1 0 0 6
0 1 0 1
0 0 1 1
R R R R R RR R R
R R RR R R
The solution is 6,1 , 1 .
Chapter 3 Test 1.
14
4 3 5
12 2 7
Substitute , 2 :
x yx y
1
4
14 3 2 1 6 5 5
4
112 2 2 3 4 7 7
4
, 2 is a solution.
2. b. The system is consistent and
independent. There is one solution.
3. c. The system is inconsistent and
independent. There are no solutions.
4. a. The system is consistent and
dependent. There are infinitely many
solutions.
5. 4 2 4
3 7
x yx y
The solution is 1, 4 .
6.
3
32
2
f x x
g x x
The solution is 2,1 .
7. 3 5 13
9
x yy x
8. 6 8 5
3 2 1
x yx y
Multiply the second equation by 4, add to
the first equation and solve for x:
Chapter 3 Systems of Linear Equations and Inequalities
324
3 5 9 13
3 5 45 13
8 32
4
9
4 9
5
x xx x
xxy x
The solution is 4, 5 .
4
6 8 5 6 8 5
3 2 1 1 2 8 4
18 9
1
2
x y x yx y x y
x
x
Substitute into the first equation and solve
for y:
16 8 5
2
3 8 5
8 2
1
4
y
yy
y
The solution is 1 12 4, .
9. Write in standard form:
7 5 21 5 7 21
9 2 27 2 9 27
y x x yy x x y
Multiply the first equation by 2 and the
second equation by 5, add the results
and solve for y:
2
5
5 7 21 10 14 42
2 9 27 1 0 45 135
59 177
3
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
2 9 3 27
2 27 27
2 0
0
xx
xx
The solution is 0, 3 .
10. 3 5 7
18 30 42
x yx y
Multiply the first equation by 6, add to
the second equation and solve for x:
6
3 5 7 18 30 42
18 30 42 18 30 42
0 0
x y x yx y x y
Infinitely many solutions of the form
, 3 5 7x y x y ; dependent
equations.
11. Multiply each equation by the LCD: 12. 4 5 2
2 4
x yy x
Substitute and solve for x:
Chapter 3 Test
325
1 1 17 2 5 34
5 2 5 2 5 34
1 12 3 6 2
4 6 3 2 6
x y x y
x y
x y x y
x y
Multiply the first equation by 2 and
the second equation by 5, add the
results and solve for x:
2
5
2 5 34 4 10 68
3 2 6 1 5 10 30
19 38
2
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
2 2 5 34
4 5 34
5 30
6
yyyy
The solution is 2, 6 .
4 5 2 2 4
4 5 4 8
0 3
x xx x
There is no solution; . The system
is inconsistent.
13. Multiply each equation by the LCD:
0.03 0.06 0.3 3 6 30
6 3 30
0.4 2 0.5 4 20 5
4 5 20
y x y xx y
x y x yx y
Multiply the first equation by 5 and the
second equation by 3, add the results and
solve for x:
5
3
6 3 30 30 15 150
4 5 20 1 2 15 60
42 210
5
x y x yx y x y
xx
Substitute into the first equation and
solve for y:
6 5 3 30
30 3 30
3 0
0
yyyy
The solution is 5, 0 .
Chapter 3 Systems of Linear Equations and Inequalities
326
14.
2 5 10
Graph the related equation 2 5 10 by using a solid line.
Test point above 0, 0 : Test point below 0, 3 :
2 0 5 0 10 2 0 5 3 10
0 10 15 10
0, 0 is not a solution. 0, 3 is a solution.
Shade the
x yx y
region below the boundary line.
15.
3 and 3 2 6
Graph the related equation 3 by using a dashed line.
Test point above 0, 4 : Test point below 0, 0 :
0 4 3 0 0 3
4 3 0 3
0, 3 is not a solution. 0, 0 is a solution.
Shade the re
x y x yx y
gion below the boundary line.
Graph the related equation 3 2 6 by using a dashed line.
Test point above 0, 4 : Test point below 0, 0 :
3 0 2 4 6 3 0 2 0 6
8 6 0 6
0, 4 is not a solution. 0, 0 is a
x y
solution.
Shade the region below the boundary line.
The solution is the intersection of the graphs.
Chapter 3 Test
327
16. 5 5 or 0
Graph the related equation 5 5 or 1 by using a solid line.
Shade the region to the left of the boundary line.
Graph the related equation 0 by using a solid line.
Test point above 0,
x x yx x
x y
1 : Test point below 0, 1 :
0 1 0 0 1 0
1 0 1 0
0,1 is not a solution. 0, 1 is a solution.
Shade the region below the boundary line.
The solution is the union of the graphs.
17. a. 0, 0x y
b. 300 400 1200x y
0 and 0 and 300 400 1200
0 represents the points to the right of the vertical line
0. Shade the region to the right of the boundary line
using a solid line border.
x y x yxx
c. 0 represents the points above the horizontal line 0.
Shade the region above the boundary line using a solid
line border.
Graph the related equation 300 400 1200 by using
a solid line.
Test point ab
y y
x y
ove 0, 4 : Test point below 0, 0 :
300 0 400 4 1200 300 0 400 0 1200
1600 1200 0 1200
0, 4 is a solution. 0, 0 is not a solution.
Shade the region above the boundary line.
The solution is the intersection
of the graphs.
Chapter 3 Systems of Linear Equations and Inequalities
328
18. 2 2 4 6
3 2 29
44
x y zx y zx y z
Multiply the second equation by –2
and add to the first equation to
eliminate z:
2
2 2 4 6 2 2 4 6
3 2 29 6 2 4 58
4 64
16
x y z x y zx y z x y z
xx
Add the second and third equations to
eliminate y:
3 2 29
44
4 73
x y zx y zx z
Substitute and solve for y and z:
4 73
4 16 73
64 73
9
x zzzz
3 2 29
3 16 2 9 29
48 18 29
37
x y zy
yy
The solution is 16, 37, 9 .
19. Write each equation in standard form:
2 6 3 3 2 6
2 11 2 11
2 8 2 2 8
x z x y x y zx y z x y z
x y z x y z
Multiply the third equation by –1 and
add to the first equation to eliminate x:
1
3 2 6 3 2 6
2 2 8 2 2 8
2
x y z x y zx y z x y z
y
Multiply the second equation by –2
and add to the first equation to
eliminate z:
2
3 2 6 3 2 6
2 11 4 2 2 22
3 5 16
x y z x y zx y z x y z
x y
Substitute and solve for x and z:
3 5 16
3 5 2 16
3 10 16
3 6
2
x yx
xxx
2 11
2 2 2 11
4 2 11
5
x y zzzz
The solution is 2, 2, 5 .
Chapter 3 Test
329
20. Let x = the amount borrowed at 6.5%
y = the amount borrowed at 5%
5000 5000
0.065 0.05 268
x y y xx y
Substitute and solve for x.
0.065 0.05 5000 268
0.065 250 0.05 268
0.015 18
1200
x xx x
xx
Substitute into the first equation and
solve for y.
1200 5000
3800
yy
She borrowed $1200 at 6.5% and
$3800 at 5%.
21. Let x = the amount of 20% acid
solution
Let y = the amount of 60% acid
solution
20% acid 60% acid 44% acid
L solution x y 200__
L acid 0.20x 0.60y 0.44(200)
200
0.20 0.60 0.44 200
x yx y
Multiply the first equation by –0.20,
add to the second equation and solve
for y:
200 0.20 0.20 40
0.20 0.60 88 0.20 0.60 88
0.40 48
120
x y x yx y x y
yy
Substitute into the first equation and
solve for x:
120 200
80
xx
The mixture contains 80 L of 20% acid
solution and 120 L of 60% acid
solution.
22. Let x = one angle
Let y = the other angle
2 60
90 90
y xx y y x
Substitute and solve:
2 90 60
180 2 60
3 240
80
90 80 10
x xx xxxy
The two angles measure 80º and 10º.
Chapter 3 Systems of Linear Equations and Inequalities
330
23. Let x = number of orders Joanne can
process
Let y = number of orders Kent can
process
Let z = number of orders Geoff can
process
504 504
20 20
104 104
x y z x y zy x x yz x y x y z
Add the first and third equations to
eliminate x:
504
104
2 400
200
x y zx y z
zz
Add the first and second equations to
eliminate x:
504
20
2 524
x y zx y
y z
Substitute and solve for x and y:
2 524
2 200 524
2 324
162
y zy
yy
20
162 20
142
142
x yx
xx
Joanne processes 142 orders, Kent
processes 162 orders, and Geoff
processes 200 orders.
24. For example:
2 1
0 4
2.6 7
25. a.
1 2 24
1 2 1 3
4 0 1 2
5 6 3 0
1 2 1 3
0 8 3 10
5 6 3 0
R R R
b. 1 3 35
1 2 1 3
0 8 3 1 0
0 4 8 15
R R R
26.
12 261 2 1 2 25
5 4 34
2 8
5 4 34 1 2 8 1 2 8 1 2 8
1 2 8 5 4 34 0 6 6 0 1 1
R RR R R R R
x yx y
Chapters 1 – 3 Cumulative Review Exercises
331
2 1 12
1 0 6
0 1 1R R R
The solution is 6, 1 .
27.
1 2 2 2 2
2 1 1
2 3 3
2
2
1
2 0
2 5
1 1 1 1 1 1 1 1 1 1 1 1
2 1 0 0 0 1 2 2 0 1 2 2
0 2 1 5 0 2 1 5 0 2 1 5
1 0 1 1
0 1 2 2
0 0 3 9
R R R R R
R R RR R R
x y zx y
y z
13 33 3 1 1
3 2 22
1 0 1 1 1 0 0 2
0 1 2 2 0 1 0 4
0 0 1 3 0 0 1 3
R R R R RR R R
The solution is 2, 4, 3 .
Chapters 1 – 3 Cumulative Review Exercises 1.
2 22 3 8 7 5 2 3 8 2
2 9 8 2 2 9 16
2 7 14
2. 7 4 2 5 3 2 1 20
7 4 2 10 6 3 20
7 8 11 20
56 77 20 56 97
w ww w
ww w
3. 5 2 1 2 3 1 7 2 8 1
10 5 6 2 7 16 2
16 3 16 5
3 5
x x xx x x
x x
There is no solution; .
4.
1 3 12 2 1
2 4 61 3 1
12 2 2 1 122 4 6
6 2 9 2 1 2
6 12 18 9 2
12 21 2
12 19
19 19
12 12
a a
a a
a aa a
aa
a
Chapter 3 Systems of Linear Equations and Inequalities
332
5.
4 2 3 9
5 2 3
2 3 5 or 2 3 5
2 8 or 2 2
4 or 1 4, 1
x
xx x
x xx x
6.
3 2 1 8
3 2 2 8
5 2 8
5 10
2 2,
y yy y
yyy
7.
4 16 or 6 3 9
4 or 6 12
4 or 2 , 2 4,
x xx x
x x
8.
4 16 and 6 3 9
4 and 6 12
4 and 2
x xx xx x
9. 3 90 5
60 3 9 30
9 3 39
3 13 3,1 3
x
xx
x
10.
4 1 11
4 10
10 4 10
6 14 6,1 4
x
xx
x
11.
4 2 4
2 4 4 or 2 4 4
2 0 or 2 8
0 or 4
, 4 0,
xx x
x xx x
12.
5 5
Graph the related equation 5 5 by using a solid line.
Test point above 0, 0 : Test point below 0, 3 :
0 5 0 5 0 5 3 5
0 5 15 5
0, 0 is a solution. 0, 3 is not a solution.
Shade the region ab
x yx y
ove the boundary line.
Chapters 1 – 3 Cumulative Review Exercises
333
13.
5 2 15
2 5 15
5 15
2 25 15
Slope: -intercept: 0, 2 2
5 2 0 15
5 15
3 -intercept: 3, 0
x yy x
y x
y
xxx x
14. 14
3y x
15. 2x
16. 10 10
6 410 10
20
02
m
17.
4 8
2 34
1
4
8 4 3
8 4 12
4 4
m
y xy x
y x
18. Multiply the second equation by the
LCD: 2 3 6x y
1 31 2 3 4
2 4x y x y
Multiply the second equation by –1, add
to the first equation and solve for x:
1
2 3 6 2 3 6
2 3 4 2 3 4
0 2
x y x yx y x y
There is no solution; {}. The system is
inconsistent.
Chapter 3 Systems of Linear Equations and Inequalities
334
19. 2 4
3 1
x yy x
2 3 1 4
5 1 4
5 5
1
3 1 3 1 1 3 1 2
x xx
xxy x
The solution is 1, 2 .
20. Let x = the length of the longest side
y = the length of the middle side
z = the length of the shortest side
7 7
2 2
1 1
x y z x y zx z x zy z x x y z
Substitute the second equation into the
other equations to eliminate x:
7
2 7
3 7
x y zz y z
z y
1
2 1
2 1
1
1
x y zz y zz y z
y zy z
Substitute the second new equation into
the first new equation to eliminate y:
3 7
3 1 7
4 6
3 11
2 2
z yz z
z
z
Substitute and solve for y in the second
new equation:
1
31
25 1
22 2
y z
y
y
Substitute and solve for x in the second
original equation:
2
12 1
2
3
x z
x
x
The lengths of the sides are 1
12
m,
12
2m, and 3m.
21. a. 12 225f x x c. 10 300 12 225
2 75
x xx
37.5x
30 months
b. 10 300g x x
Chapters 1 – 3 Cumulative Review Exercises
335
22. 3 3 2 3 3 2 3 3
4 5 7 1 4 5 7 1
2 3 2 6 2 3 2 6
x y z x y zx y z x y zx y z x y z
Multiply the first equation by –2 and the
third equation by 3, and add the results
to eliminate x:
2
3
3 2 3 3 6 4 6 6
2 3 2 6 6 9 6 1 8
5 12 12
x y z x y zx y z x y z
y z
Multiply the third equation by –2 and
add to the second equation to eliminate
x:
2
4 5 7 1 4 5 7 1
2 3 2 6 4 6 4 12
11 11 11
1
x y z x y zx y z x y z
y zy z
Multiply the second result by 5 and add
to the first result to eliminate y:
5
5 12 12 5 12 12
1 5 5 5
7 7
1
y z y zy z y z
zz
Substitute and solve for y and z:
1
1 1
0
0
y zy
yy
2 3 2 6
2 3 0 2 1 6
2 0 2 6
2 4
2
x y zx
xxx
The solution is 2, 0, 1 .
23. 2 3 24. For example:
2 3 1 7
1 4 2 6
25.
112 21 1 92 1 2 2
2 1 1
4
2
2 4 2
4 5
2 4 2 1 2 1 1 2 1 1 2 1
4 1 5 4 1 5 0 9 9 0 1 1
1 0 1
0 1 1
R RR R R R R
R R R
x yx y
The solution is 1,1 .
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