Chapter 3 ISM - TopCatMathtopcatmath.com/docs/Math030/Chapter3.ISM.pdfChapter 3 Systems of Linear Equations and Inequalities Section 3.1 Practice Exercises 1. a. system 2. The lines

208

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


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


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


18. 2 5 2h x x g x x



211

19. 1 25 2

3 3k x x f x x


20. 1 52 2

2 2f x x g x x


21. 4 2 3x y x


22. 3 2 6 3

2 3 6

33

2

x y yy x

y x


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


Inconsistent system.


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


1, 2

2x y y x

. Dependent

equations.

27. 12 4 1

22 2

x y

x y


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


213


, 3 6x y x y . Dependent

equations.


, 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 .


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


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



216

7. 2 3 6 3 9

3 6 9

2 3

y x x yy xy x


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


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


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


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


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


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


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


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


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


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


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


218

3 2 4 6

6 4 6

4 0

0

yyyy


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


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


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


, 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


, 2 11x y x y ; dependent equations.


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



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



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



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



equations.

30. 4 1

12 3 3

x yy x

12 3 4 1 3

12 12 3 3

0 0

y yy y



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


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


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


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


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


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


221





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


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


41. 300 125 1350

2 8 6

x yy y

300 125 6 1350

300 750 1350

300 2100

7

xx

xx


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



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




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


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



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


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


53. 2 5 7 2 12 6

4 3 1 3 3 1

x x xy y y


54. 2 4 2 6 2 3

7 5 5 7 0 0

y y yx x x


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


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



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.


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


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


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:


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


solve for x:

2 3 1 3

2 3 3

2 6

3

xx

xx


2

2 5 7 4 10 14

3 10 13 3 10 1 3

1

1

x y x yx y x y

xx


solve for y:

2 1 5 7

2 5 7

5 5

1

yyyy


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


solve for x:

3 7 8 20

3 56 20

3 36

12

xx

xx


10. 6 9 15

5 2 40

x yx y


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


solve for y:

6 10 9 15

60 9 15

9 45

5

yyyy


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



solve for y:


5 6 4 5 6 4

5 1 3 3 5 1

x y x yy x x y


second equation by 5, add the results

and solve for y:


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


solve for x:

3 10 1 13

3 10 13

3 3

1

xxxx


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


solve for x:

5 6 1 4

5 6 4

5 10

2

xxxx


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



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


solve for x:

12 6 1 30

12 6 30

12 24

2

xx

xx



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



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


solve for x:

18 8 4 14

18 32 14

18 18

1

xx

xx



3 2 4 2 3 4 0

7 3 7 3 0

x y x yx y x y


4 3 2 3 2 4 0

5 3 3 5 0

y x x yy x x y


228



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


solve for y:

3 0 4 0

0 4 0

4 0

0

yyyy

The solution is 0. 0 .



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


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



2

3 2 1 6 4 2

6 4 2 6 4 2

0 0

x y x yx y x y


, 3 2 1x y x y ; dependent

equations.

18. 3 4

6 2 8

x yx y



2

3 4 6 2 8

6 2 8 6 2 8

0 0

x y x yx y x y




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


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


229






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



equations.


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



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



24. 0.2 0.1 1.2

1 32

x y

x y



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



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


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


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



2

2 5 9 4 10 18

4 7 16 4 7 16

17 34

2

x y x yx y x y

yy


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


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


for x:

0.1 0.5 2 0.7

0.1 1.0 0.7

0.1 0.3

3

xx

xx



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


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


39. 4 6 5

2 3 7

x yx y



2

4 6 5 4 6 5

2 3 7 4 6 14

0 9

x y x yx y x y



40. 3 6 7

2 4 5

x yx y

Multiply the first equation by –2 and the


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



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



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


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



5 3 1 05 5 3 105

5 12 120 5 12 120

15 225

15

x y x yx y x y

yy


232


solve for y:

3 12 2 24

36 2 24

2 60

30

yyyy



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


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


35

, x y y x . The equations are

dependent.


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



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


solve for y:


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



5

3 10 15 5 50

4 5 50 4 5 50

19 0

0

x y x yx y x y

xx


solve for y:


233

2 0 5 20

0 5 20

5 20

4

yyyy


3 0 10

0 10

10

yyy


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



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


solve for y:

4 30 5 170

120 5 170

5 50

10

yyyy


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


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


solve for x:

6 50 320

300 320

20

20

xx

xx


51. Write in standard form: 52. Write in standard form:


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



3

9 5 17 9 5 17

3 4 0 9 12 0

17 17

1

x y x yx y x y

yy


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



1

4 5 4 5

5 4 7 5 4 7

6 12

2

x y x yx y x y

xx


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



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



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:


235

1 2 11

2 10

5

yyy


6 5 16

5 10

2

yyy


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



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



equations.


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


solve for c:


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


solve for c:

9 7 2268

9 7 189 2268

9 1323 2268

9 945

105

c hc

ccc


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


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



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



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


, 26 13

.

61. 4 10 19

5 12 41

x yx y



results and solve for x:



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


, 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


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;


238

3 2 2

6 2

4

y



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


solve for y:

3 2 2

6 2

4

yyy


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;


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


Problem Recognition Exercises: Solving Systems of Linear Equations

239

b. 5 51

2 20 1

x x


system

b. 2 3 1 4 4

1

y x xx

2 3 1 1 3 1 4

2

yy



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


system


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


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



240

7. 5 3 2

7 4 30

x yx y



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


solve for y:

5 2 3 2

10 3 2

3 12

4

yyyy



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


solve for y:

2 4 6

4 8

2

yyy



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




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


solve for y:

7 3 25

21 25

4

4

yyyy


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


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



243

Multiply the first equation by –0.18,


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


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


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



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


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


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


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


solve for y:

10 100

90

yy

The mixture contains 10 mL of 30%

acid solution and 90 mL of 10% acid

solution.


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



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


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


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


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


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


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


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



for y:


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




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


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


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 _


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.


247

25. Let p = the speed of the plane in still

air


p + w = speed of the plane with the

wind


Distance Rate Time

Tailwind 3200 p + w 4 _

Headwind 3200 p – w 5 _


4 3200

5 3200

p wp w



and solve:

4

5

4 3200 800

5 3200 640

2 1440

720

div

div

p w p wp w p w

pp


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.


water

Let c = the speed of the current



Distance Rate Time

With current 100 b + c 2.5

Against current 100 b – c 10/3


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


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


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 _



248


20 100

30 60

x yx y



and solve:

20

30

20 100 5

30 60 2

2 7

3.5

div

div

x y x yx y x y

xx


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



2

3

2 24 12

3 24 8

2 20

10

div

div

b w b wb w b w

bb


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


3 180

3 183

61

61 3 58

x x xxxy

The angles measure 61º, 61º, and 58º.

31. Let x = one angle


3 2

180

y xx y




5

180

y xx y



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.



2 6

90

y xx y


2 6 90

3 84

28

2 28 6

56 6

62

x xxxy

The two angles measure 28º and 62º.



2 15

90

y xx y


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



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




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.


water



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


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


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




Distance Rate Time

With current 31.5 b + c 7

Against current 31.5 b – c 9


7 31.5

9 31.5

b cb c



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


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.


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



2

6 2 2330 12 4 4660

4 4 2020 4 4 2020

8 2640

330

x y x yx y x y

xx


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


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.


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



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


solve for x:



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



8000 .03 .03 240

.03 .018 222 .03 .018 222

0.012 18

1500

x y x yx y x y

yy


solve for x:

1500 8000

6500

xy


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


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.



1

490

y x

x y


190

45

904

72

172 18

4

x x

x

x

y


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.


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


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


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


0 3 0 6 0 3 3 6

0 6 9 6


Shade the region below the boundary line.

19.

5 2 10

Graph the related equation 5 2 10 by using a dashed line.


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.


256

20.

3 8



0 3 0 8 0 3 4 8

0 8 12 8


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.


2 0 6 3 12 2 0 6 0 12

0 6 0 12


Shade the

x yx y

region below the boundary line.


257

22.

4 3 12



4 0 3 0 12 4 0 3 5 12

0 12 0 3


Shade the

x yx y

region above the boundary line.

23.

2 4


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


24.

6 2


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



258

25.

2

Graph the related equation 2 by using a solid line.


0 2 3 2


Shade the region above the boundar

yy

y line.

26.

5



6 5 0 5


Shade the region above the boundary line.

yy


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.


x xx

29.

24

52

Graph the related equation 4 by using a solid line.5


2 20 0 4 5 0 4

5 50 4 5 4


Shade the reg

y x

y x

ion above the boundary line.


260

30.

54

25



5 50 0 4 5 0 4

2 20 4 5 4


Shade the

y x

y x

region above the boundary line.

31.

16

31



1 17 0 6 0 0 6

3 37 6 0 6


Shade the region be

y x

y x



261

32.

12

41



1 13 0 2 0 0 2

4 43 2 0 2


Shade the regio

y x

y x

n below the boundary line.

33.

5 0



3 5 0 0 3 5 0 0

3 0 3 0


Shade the region a

y xy x



262

34.

10

21

Graph the related equation 0 by using a dashed line.2


1 13 0 0 3 0 0

2 23 0 3 0


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


0 5 0 01 1

5 4 5 45

1 0 14


Shade the region

x y

x y

below the boundary line.


263

36.

22



5 00 2 0 2

2 25

2 0 22


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.


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.


38.

0.3 0.2 0.6

Graph the related equation 0.3 0.2 0.6 by using a dashed line.


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


264



39.

2

32

Graph the related equation by using a solid line.3


2 20 3 0 3

3 30 2 0 2


Shade the region be

x y

x y


40.

5

45



5 50 3 0 3

4 415 15

0 04 4


Shade the regio

x y

x y

n above the boundary line.


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



The solution is the intersection of the graphs.

42. 3 and 2 6


Shade the region below the boundary line using a dashed line border.


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





266

43.

2 5 or 3



2 0 6 5 2 0 0 5

6 5 0 5


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



0 3 4 3 0 3 0 3

12 3 0 3


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.


267

45.

3 and 4 6



0 4 3 0 0 3

4 3 0 3


Shade the reg

x y x yx y

ion below the boundary line.



4 0 7 6 4 0 0 6

7 6 0 6


x y



46.



3 3 3 9 3 0 0 9

12 9 0 9


Shade the region below the

x y

boundary line.



268

4 and 3 9



0 5 4 0 0 4

5 4 0 4


Shade the reg

x y x yx y

ion below the boundary line.

47.

2 2 or 2 3 6



2 0 0 2 2 0 3 2

0 2 3 2


Shade t

x y x yx y

he region above the boundary line.



2 0 3 3 6 2 0 3 0 6

9 6 0 6


Shade the region above the

x y

boundary line.

The solution is the union of the graphs.


269

48.

3 2 4 or 3



3 0 2 3 4 3 0 2 0 4

6 4 0 4


Shade th

x y x yx y

e region above the boundary line.



0 0 3 0 4 3

0 3 4 3

0, 0 is a solution. 0, 4 is not a solut

x y

ion.



49. 4 and 2


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


270

50. 3 and 4


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


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



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


271

53. 0 and 6


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.


0 7 6 0 0 6

7 6 0 6



The solution is the intersection o

f the graphs.

54. 0 and 2


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.


0 3 2 0 0 2

3 2 0 2




272


55. 0 or 4


Shade the region below the boundary line using a solid line border.


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




56.

3 and 0 and 0



0 4 3 0 0 3

4 3 0 3


Shade t

x y x yx y

he region below the boundary line.



line border.

x x


273

0 represents the points above the horizontal line 0.

Shade the region above the boundary line using a solid line border.


y y

57.

2 and 0 and 0



0 0 2 0 3 2

0 2 3 2


Shad

x y x yx y

e the region above the boundary line.



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.


58. 0 and 0 and 8 and 3 5 30



line border.

0 represents the p

x y x y x yx x

y

oints above the horizontal line 0.


y


274



0 9 8 0 0 8

9 8 0 8


Shade the region below the bound

x y

ary line.



3 0 5 7 30 3 0 5 0 30

35 30 0 30


Shade the r

x y

egion below the boundary line.


59. 0 and 0 and 5 and 2 6



line border.

0 represents the poi

x y x y x yx x

y

nts above the horizontal line 0.



y

x y


0 6 5 0 0 5

6 5 0 5




275



0 2 4 6 0 2 0 6

8 6 0 6


Shade the region below the bo

x y

undary line.


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



2 0 2 27 50 2 0 2 0 50

54 50 0 50


Shade the region

x y



276

61. a. 2 2 40x y

b. 0 and 0 and 2 2 40




x y x yx x

y

the horizontal line 0. Shade the region above the


y



2 0 2 21 40 2 0 2 0 40

42 40 0 40


Shade the region

x y


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 th

x y x y x yx x

y

e points above the horizontal line 0. Shade the region above the



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




277


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.


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



x x

y y

Shade the region below t


Graph the related equation 65 by using a solid line.x y


278


0 66 65 0 0 65

66 65 0 65



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


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


b. 2 5 3

4 10 3

x yx y



2

2 5 3 4 10 6

4 10 3 4 10 3

0 9

x y x yx y x y



5. Let b = the speed of the bike in still air


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_


424

3

2 24

b w

b w

Divide the first equation by 4/3, the second


4 /3

2

424 18

3

2 24 12

2 30

15

div

div

b w b w

b w b wb

b


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.


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.


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.


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


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


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.


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 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 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 z


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


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



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 zx z


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 z


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


14. 6 5 7

5 3 2 0

2 3 11

x y zx y zx y z



6 5 7 6 5 7

2 3 11 10 5 15 55

4 14 62


x z


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 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



3

4 3 2 1 2 4 3 2 12

2 7 8 6 3 21 24

2 19 36


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 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


283

4

4 14 62 4 14 62

3 4 4 12

10 50

5

x z x zx z x z

zz


3

5 3

2

x zx

x

2 3 11

2 2 3 5 11

4 15 11

0

x y zy

yy



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


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


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

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


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


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


eliminate z:


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


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


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


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


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


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


285


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


2 5

2 2 5

4 5

1

1

x zzzzz

3 2

3 1 2

3 3

1

y zy

yy



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


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


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


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

.


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


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.




180

2

5 4

x y zy xz x



2 5 4 180

8 4 180

8 184

23

x x xx

xx


2

2 23

46

y xyy

5 4

5 23 4

115 4 111

z xzz


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


eliminate x:

55

1

2 54

27

x y zx y z

zz


eliminate x:

8

1

2 9

x yx y z

y z


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


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


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



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 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 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


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.


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

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


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



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


289

3 3 18

5 3 18

5 15

3

x x xx

xx


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


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

Multiply the third equation by –1 and add to the second equation to eliminate z:


290

1

2 2 2 170 2 2 2 1 70

3 2 180 3 2 180

10


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


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:


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


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


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



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%).


293

35. 2 3 2

2 4

2 2 4 8

x y zx y zx y z


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


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 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



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


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



2

3 2 3 6 4 2 6

6 4 2 1 6 4 2 1

0 7



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:


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


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



eliminate z:

2 2 9

3 3 2 1

4 10

x y zx y zx y

Multiply the third equation by 2 and


2

2 4 12 2 4 12

3 3 2 1 6 6 4 2

8 5 14


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


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



2

3 8 6 2 2 16

2 3 2 1 2 3 2 1

8 17


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


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


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 .


8 17

8 2 17

16 17

1

x yyyy

3 8

3 2 1 8

6 1 8

3

3

x y zzzzz


42. 2 3 3 1 5

3 6 6 23

9 3 6 8

x y zx y zx y z


eliminate z:

3 6 6 23

9 3 6 8

6 9 15

x y zx y z

x y


add to the second equation to

eliminate 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


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 z y x y z


add to the third equation to eliminate z:

44. 2 3

2 16 10

7 3 4 8

x yy z

x y z



z:


296

2

2 3 3 4 2 6 6

4 2 6 6 4 2 6 6

0 0



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


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



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


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


solution.


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


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


solution.

47. 2 4 8 0

3 0

2 5 0

x y zx y z

x y z


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


eliminate z:

2 4 0

3 0

3 7 0

x y zx y zx y


297


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


y z


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


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




2

3 0 2 6 2 0

3 2 0 3 2 0

5 7 0


x y


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


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


49. 4 2 3 0

8 0

32 0

2

x y zx y z

x y z


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


50. 5 0

4 0

5 5 0

x yy z

x y z


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


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



298


1. a. matrix; rows; columns c. coefficient; augmented

b. column; one; square d. row echelon

2. Let x = amount of pure (100%) acid sol


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



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


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



6 9 6 9

2 13 2 1 3

8 4

1

2

x y x yx y x y

y

y


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


eliminate z:

8

2 3

2 11

x y zx y zx y



5. 2 4

3 7

3 4 22

x y zx y z

x y z


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


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


2 11

2 1 11

2 12

6

x yx

xx

2 3

6 2 1 3

6 2 3

1

x y zzzz


3

2 4 6 3 3 12

3 4 22 3 4 22

7 10


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


4 11

1 4 11

4 12

3

x zzzz

3 7

1 3 3 7

1 9 7

3

x y zy

yy


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


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


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


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


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


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



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


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


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



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


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


54. 2 3 2 11

3 8 1

3 14 2

x y zx y zx y z


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


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


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


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


57.

58.

59.

60.

61.

62.


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


2. False 3. True

4. True

5.

1

2 4

f x xg x x


6. 2 7

5

y xy x


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


308



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


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


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


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


309



equations.


, 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



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


solve for y:

2 1 3 5

2 3 5

3 3

1

yyyy


18. 4 3 5

3 4 10

x yx y



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


for y:

4 2 3 5

8 3 5

3 3

1

yyyy


19. 3 4 2

2 5 1

x yx y

20. Multiply each equation by its LCD:


310



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


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




2 3 8 3 2 8

6 4 4 6 4 4

y x x yx y x y



2

3 2 8 6 4 16

6 4 4 6 4 4

0 20

x y x yx y x y


inconsistent.


3 16 3 16

3 2 2 2 2

x y x yx y y x x y


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


for y:

3 6 16

18 16

2

yyy



4 2 9 6 9

2 2 10 2 2 10

y x x x yx y x y




for y:

6 2 9

12 9

3

yyy


311

2

6 9 12 2 18

2 2 10 2 2 10

14 28

2

x y x yx y x y

xx



4 35 3 4 3 35

15 15

x y x yx y x y


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



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



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


for y:

4 0 3 18

0 3 18

6

yyy



0.02 0.01 0.11 2 11

0.01 0.04 0.26 4 26

x y x yx y x y



4

3

2 11 8 4 44

4 26 4 26

9 18

2

x y x yx y x y

xx


for y:

2 2 11

4 11

7

7

yyyy


Section 3.4 27. Let x = the amount invested at 5%


28. Let x = the number of student tickets

Let y = the number of adult tickets

Let 1x = receipts from student tickets


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


0.05 2 0.035 303.75

0.10 0.035 303.75

0.135 303.75

2250

y yy y

yy


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



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


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


p + w = speed of the plane with the wind


Distance Rate Time_

Tailwind 280 p + w 1.75_

Headwind 280 p – w 2__


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


150 160 10w w

The speed of the plane is 150 mph in still

air and the speed of the wind is 10 mph.


313

31. a. 915 275f x x 32. Let x = one angle


5 6

90

y xx y


5 6 90

6 84

14

5 14 6 70 6 76

x xxxy


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



2 0 6 5 2 0 0 5

0 1 0 5


Shade the region

x yx y

above the boundary line.

34.

2 8 3



2 0 8 3 0 2 0 8 3 3

0 8 0 1


Shade the r

x yx y

egion below the boundary line.


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.


xx

37.

1

21



1 10 2 0 2

2 20 1 0 1


x y

x y


315


38.

2

52

Graph the related equation by using a dashed line.5


2 20 5 0 5

5 50 2 0 2


Shade the region above

x y

x y

the boundary line.

39.

2 2 and 2 2



2 0 3 2 2 0 0 2

3 2 0 2


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


316


2 0 0 2 2 0 3 2

0 2 3 2




40.

3 6 or 3 6



3 0 7 6 3 0 0 6

7 6 0 6


Shade the r

x y x yx y

egion above the boundary line.



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.




317

41.

or

Graph the related equation by using a solid line.


5 0 5 0


y x y xy x


Graph the related equation by using a solid line.


5 1 5 1


Shade the region below the boundary li

y x

ne.


42. 20 and 0 and 4



line border.


x y y x

x x

y

the horizontal line 0.


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





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.


line border.


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.



Test point above

x y

0,1 : Test point below 0, 1 :

0 4 1 0 4 1

0 4 0 4





319

Section 3.6 44. 5 5 5 30

2

10 6 2 4

x y zx y zx y z



5

5 5 5 30 5 5 5 30

2 5 5 5 10

10 20

2


xx

Multiply the second equation by 2 and add


2

2 2 2 2 4

10 6 2 4 1 0 6 2 4

8 8 8


x y


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


45. 5 3 5

2 6

2 8

x y zx y zx y z


eliminate z:

2 6

2 8

0 14

x y zx y z


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




320

47. 3 4 5

2 3 2

2 5 8

x zy z

x y


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


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


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


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


eliminate x:

950

150

2 800

400

x y zx y z

zz




180

9 9

3 26

x y zx y y xz x



9 3 26 180

5 35 180

5 145

29

x x xx

xx



321


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


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


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


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


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


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


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


6.

3

32

2

f x x

g x x


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:


324

3 5 9 13

3 5 45 13

8 32

4

9

4 9

5

x xx x

xxy x


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


for y:

16 8 5

2

3 8 5

8 2

1

4

y

yy

y

The solution is 1 12 4, .


7 5 21 5 7 21

9 2 27 2 9 27

y x x yy x x y



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


solve for x:

2 9 3 27

2 27 27

2 0

0

xx

xx


10. 3 5 7

18 30 42

x yx y



6

3 5 7 18 30 42

18 30 42 18 30 42

0 0

x y x yx y x y



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



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


solve for y:

2 2 5 34

4 5 34

5 30

6

yyyy


4 5 2 2 4

4 5 4 8

0 3

x xx x

There is no solution; . The system

is inconsistent.


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



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


solve for y:

6 5 3 30

30 3 30

3 0

0

yyyy



326

14.

2 5 10



2 0 5 0 10 2 0 5 3 10

0 10 15 10


Shade the

x yx y

region below the boundary line.

15.

3 and 3 2 6



0 4 3 0 0 3

4 3 0 3


Shade the re

x y x yx y

gion below the boundary line.



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.



Chapter 3 Test

327

16. 5 5 or 0

Graph the related equation 5 5 or 1 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




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.


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



The solution is the intersection

of the graphs.


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


xx


eliminate y:

3 2 29

44

4 73

x y zx y zx 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


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


add to the first equation to eliminate x:

1

3 2 6 3 2 6

2 2 8 2 2 8

2


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


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


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


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



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


solve for x:

120 200

80

xx

The mixture contains 80 L of 20% acid

solution and 120 L of 60% acid

solution.



2 60

90 90

y xx y y x


2 90 60

180 2 60

3 240

80

90 80 10

x xx xxxy



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


eliminate x:

504

104

2 400

200

x y zx y z

zz


eliminate x:

504

20

2 524

x y zx y

y z


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


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


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


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


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



0 5 0 5 0 5 3 5

0 5 15 5


Shade the region ab

x yx y

ove the boundary line.


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


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.


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


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


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



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


y z



x:

2

4 5 7 1 4 5 7 1

2 3 2 6 4 6 4 12

11 11 11

1


y zy z


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


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


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


Documents

Chapter 3 ISM - TopCatMathtopcatmath.com/docs/Math030/Chapter3.ISM.pdfChapter 3 Systems of Linear Equations and Inequalities Section 3.1 Practice Exercises 1. a. system 2. The lines