testbankcollege.eutestbankcollege.eu/sample/Solution-Manual... · 63 Chapter 2 Equations, Problem Solving, and Inequa lities 2.1 The Addition and Subtraction Properties of Equa lity

63

Chapter 2 Equations, Problem Solving, and Inequalities

2.1 The Addition and Subtraction Properties of Equality

Problems 2.1

1. a. If x is 7, then x − 5 = 3 becomes

7 − 5 = 3, which is a false statement.

Hence, 7 is not a solution of the equation.

b. If y is 4, then 1 = 5 − y becomes

1 = 5 − 4, which is a true statement.

Thus, 4 is a solution of the equation.

c. If z is 6, 1

2 03

z − = becomes

1(6) 2 0,

3− = which is a true statement.


2. a. 5 7

5 5 7 5

12

x

x

x

− =− + = +

=

The solution is 12.

CHECK x − 5 0 7

12 − 5 7

7

b. 1 3

5 51 1 3 1

5 5 5 54

5

x

x

x

− =

− + = +

=

The solution is 4

.5

CHECK 315 5

x − 0

4 15 5

− 35

35

3. a. 5 2 4 3 17

5 17

5 5 17 5

12

y y

y

y

y

+ − + =+ =

+ − = −=

Thus 12 is the solution of the equation.

CHECK 5y + 2 − 4y + 3 0 17

5(12) + 2 − 4(12) + 3 17

60 + 2 − 48 + 3

17

b. 3 1 5

5 68 8 8

2 5

8 82 2 5 2

8 8 8 83

8

z z

z

z

z

− + + − =

+ =

+ − = −

=

Thus, 3

8 is the solution of the equation.

CHECK 3 18 8

5 6z z− + + − 0 58

( ) ( )3 3 3 18 8 8 8

5 6− + + − 58

15 3 18 18 8 8 8

− + + −

58

4. a. 5 7

5 7 7 7

2

2

x

x

x

x

= +− = − +− =

= −

The solution is −2.

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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities

64

CHECK 5 0 7 + x

5 7 + (−2)

5

b. 5 2 4 3

5 2 2 4 3 2

5 4 5

5 4 4 4 5

5

x x

x x

x x

x x x x

x

− = +− + = + +

= +− = − +

=

The solution is 5.

CHECK 5x − 2 0 4x + 3

5(5) − 2 4(5) + 3

25 − 2 20 + 3

23 23

c. 0 3( 3) 7 2

0 3 9 7 2

0 2

0 2 2 2

2

y y

y y

y

y

y

= − + −= − + −= −

+ = − +=

The solution is 2.

CHECK 0 0 3(y − 3) + 7 − 2y

0 3(2 − 3) + 7 − 2(2)

3(−1) + 7 − 4

−3 + 7 − 4

0

d. 3( 1) 4 8

3 3 4 8

3 3 3 4 8 3

3 4 5

3 4 4 4 5

5

5

z z

z z

z z

z z

z z z z

z

z

+ = ++ = +

+ − = + −= +

− = − +− =

= −


CHECK 3(z + 1) 0 4z + 8

3(−5 + 1) 4(−5) + 8

3(−4) −20 + 8

−12 −12

5. 6 5 7 2

6 5 5 7 2 5

6 7 3

6 7 7 7 3

3

3

y y

y y

y y

y y y y

y

y

+ = ++ − = + −

= −− = − −

− = −=

The solution is 3.

CHECK 6y + 5 0 7y + 2

6(3) + 5 7(3) + 2

18 + 5 21 + 2

23 23

6. 5 3( 1) 4 3

5 3 3 4 3

2 3 4 3

2 4 3 4 4 3

2 3 3

2 3 3 3 3

2 0

z z

z z

z z

z z

z z

z z z z

+ − = ++ − = +

+ = +− + = − +− + =

− + − = −− =

Since this statement is false, the equation has no solution.

7. 3 4( 2) 11 4

3 4 8 11 4

11 4 11 4

y y

y y

y y

+ + = ++ + = +

+ = +

Since both sides are identical, this equation is an identity. The solution is all real numbers.

8. a. Let x = 2003 − 1998 = 5

400 7400 400(5) 7400

2000 7400

9400

x + = += +=

The estimated tuition and fees in 2003 are $9400.


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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM

65

b. 400 7400 9000

400 9000 7400

400 1600

4

x

x

x

x

+ == −==

This means 4 years after 1998 or in 1998 + 4 = 2002, tuition and fees are estimated to be $9000.

Exercises 2.1

1. If x is 3, then x − 1 = 2 becomes 3 − 1 = 2,

which is a true statement. Thus, 3 is a solution of the equation.

2. If x is 4, then 6 = x − 10 becomes

6 = 4 − 10, which is a false statement.


3. If y is −2, then 3y + 6 = 0 becomes

3(−2) + 6 = 0, which is a true statement.

Thus, −2 is a solution of the equation.

4. If z is −3, then −3z + 9 = 0 becomes

−3(−3) + 9 = 0, which is a false statement.

Hence, −3 is not a solution of the equation.

5. If n is 2, then 12 − 3n = 6 becomes

12 − 3(2) = 6, which is a true statement.


6. If m is 1

3 ,2

then 1

3 72

m+ = becomes

1 13 3 7,

2 2+ = which is a true statement.

Thus 1

32

is a solution of the equation.

7. If d is 10, then 2

1 35

d + = becomes

2(10) 1 3,

5+ = which is a false statement.


8. If c is 2.3, then 3.4 = 2c − 1.4 becomes

3.4 = 2(2.3) − 1.4, which is a false

statement. Hence 2.3 is not a solution of the equation.

9. If a is 2.1, then 4.6 = 11.9 − 3a becomes

4.6 = 11.9 − 3(2.1), which is a false

statement. Hence 2.1 is not a solution of the equation.

10. If x is 1

,10

then 7

0.2 510

x= − becomes

7 10.2 5 ,

10 10

= −

which is a true

statement. Thus 1

10 is a solution of the

equation.

11. 5 9

5 5 9 5

14

x

x

x

− =− + = +

=

The solution is 14.

CHECK x − 5 0 9

14 − 5 9

9

12. 3 6

3 3 6 3

9

y

y

y

− =− + = +

=

The solution is 9.

CHECK y − 3 0 6

9 − 3 6

6

13. 11 8

11 8 8 8

19

m

m

m

= −+ = − +

=

The solution is 19.

CHECK 11 0 m − 8

11 19 − 8

11




66

14. 6 2

6 2 2 2

8

n

n

n

= −+ = − +

=

The solution is 8.

CHECK 6 0 n − 2

6 8 − 2

6

15. 2 8

3 32 2 8 2

3 3 3 310

3

y

y

y

− =

− + = +

=

The solution is 10

.3

CHECK x − 23

0 83

10 23 3

− 83

83

16. 4 35

3 34 4 35 4

3 3 3 339

313

R

R

R

R

− =

− + = +

=

=

The solution is 13.

CHECK y − 43

0 353

43

13 − 353

353

17. 2 6 10 5

16 5

16 16 5 16

21

k k

k

k

k

− − − =− =

− + = +=

The solution is 21.

CHECK 2k − 6 − k − 10 0 5

2(21) − 6 − 21 − 10 5

42 − 6 − 21 − 10

5

18. 3 4 2 6 7

2 7

2 2 7 2

9

n n

n

n

n

+ − − =− =

− + = +=

The solution is 9.

CHECK 3n + 4 − 2n − 6 0 7

3(9) + 4 − 2(9) − 6 7

27 + 4 − 18 − 6

7

19. 1 2

24 31 2

4 31 2 2 2

4 3 3 33 8

12 1211

12

z z

z

z

z

z

= − −

= −

+ = − +

+ =

=

The solution is 11

.12

CHECK 14

0 23

2z z− −

14

( )11 2 1112 3 12

2 − −

822 1112 12 12

− −

312

14




67

20. 7 1

3 22 57 1

2 57 1 1 1

2 5 5 535 2

10 1037

10

v v

v

v

v

v

= − −

= −

+ = − +

+ =

=

The solution is 37

.10

CHECK 72

0 15

3 2v v− −

72

( ) ( )37 37110 5 10

3 2− −

74111 210 10 10

− −

3510

72

21. 3

0 2 22

3 40

2 27

02

7 7 70

2 2 27

2

x x

x

x

x

x

= − − −

= − −

= −

+ = − +

=

The solution is 7

.2

CHECK 0 0 32

2 2x x− − −

0 ( )7 3 72 2 2

2 2− − −

3 72 2

7 2− − −

102

5 −

5 − 5

0

22. 1 1

0 3 24 2

1 20

4 43

04

3 3 30

4 4 43

4

y y

y

y

y

y

= − − −

= − −

= −

+ = − +

=

The solution is 3

.4

CHECK 0 0 1 14 2

3 2y y− − −

0 ( ) ( )3 31 14 4 4 2

3 2− − −

9 61 24 4 4 4

− − −

04

0

23. 1 1

4 35 51 1

5 51 1 1 1

5 5 5 50

c c

c

c

c

= + −

= +

− = + −

=

The solution is 0.




68

CHECK 15

0 15

4 3c c+ −

15

15

4(0) 3(0)+ −

15

0 0+ −

15

24. 19 1

0 6 52 2

180

20 9

0 9 9 9

9

b b

b

b

b

b

= + − −

= +

= +− = + −− =


CHECK 0 0 19 12 2

6 5b b+ − −

0 19 12 2

6( 9) 5( 9)− + − − −

19 12 2

54 45− + + −

182

9− +

−9 + 9

0

25. 3 3 4 0

3 0

3 3 0 3

3

x x

x

x

x

− + + =+ =

− + = −= −


CHECK −3x + 3 + 4x 0 0

−3(−3) + 3 + 4(−3) 0

9 + 3 − 12

0

26. 5 4 6 0

4 0

4 4 0 4

4

y y

y

y

y

− + + =+ =

− + = −= −


CHECK −5y + 4 + 6y 0 0

−5(−4) + 4 + 6(−4) 0

20 + 4 − 24

0

27. 3 1 1 1

4 4 4 41 1

4 41 1 1 1

4 4 4 40

y y

y

y

y

+ + =

+ =

+ − = −

=

The solution is 0.

CHECK 3 1 14 4 4

y y+ + 0 14

3 1 14 4 4

(0) (0)+ + 14

14

0 0+ +

14

28. 1 2 1 1

2 3 2 32 1

3 32 2 1 2

3 3 3 31

3

y y

y

y

y

+ + =

+ =

+ − = −

= −

The solution is 1

.3

−




69

CHECK 1 2 12 3 2

y y+ + 0 13

( ) ( )1 1 2 1 12 3 3 2 3

− + + − 13

1 2 16 3 6

− + −

2 23 6

−

2 13 3

−

13

29. 3.4 3 0.8 2 0.1

3.4 0.9

3.4 0.9 0.9 0.9

2.5

2.5

c c

c

c

c

c

= − + + += − +

− = − + −= −

− =

The solution is −2.5.

CHECK

3.4 0 −3c + 0.8 + 2c + 0.1

3.4 −3(−2.5) + 0.8 + 2(−2.5) + 0.1

7.5 + 0.8 − 5 + 0.1

3.4

30. 1.7 3 0.3 4 0.4

1.7 0.7

1.7 0.7 0.7 0.7

1

c c

c

c

c

= − + + += +

− = + −=

The solution is 1.

CHECK 1.7 0 −3c + 0.3 + 4c + 0.4

1.7 −3(1) + 0.3 + 4(1) + 0.4

−3 + 0.3 + 4 + 0.4

1.7

31. 6 9 5

6 6 9 5 6

9

9

p p

p p p p

p

p

+ =− + = −

= −− =


CHECK 6p + 9 0 5p

6(−9) + 9 5(−9)

−54 + 9 −45

−45

32. 7 4 6

7 7 4 6 7

4

4

q q

q q q q

q

q

+ =− + = −

= −− =


CHECK 7q + 4 0 6q

7(−4) + 4 6(−4)

−28 + 4 −24

−24

33. 3 3 2 4

5 3 4

5 5 3 4 5

3

3

x x x

x x

x x x x

x

x

+ + =+ =

− + = −= −

− =


CHECK 3x + 3 + 2x 0 4x

3(−3) + 3 + 2(−3) 4(−3)

−9 + 3 − 6 −12

−12

34. 2 4 6 7

8 4 7

8 8 4 7 8

4

4

y y y

y y

y y y y

y

y

+ + =+ =

− + = −= −

− =





70

CHECK 2y + 4 + 6y 0 7y

2(−4) + 4 + 6(−4) 7(−4)

−8 + 4 + (−24) −28

−28

35. 4( 2) 2 3 0

4 8 2 3 0

6 0

6 6 0 6

6

m m

m m

m

m

m

− + − =− + − =

− =− + = +

=

The solution is 6.

CHECK 4(m − 2) + 2 − 3m 0 0

4(6 − 2) + 2 − 3(6) 0

4(4) + 2 − 18

16 + 2 − 18

0

36. 3( 4) 2 2

3 12 2 2

3 14 2

3 3 14 2 3

14

14

n n

n n

n n

n n n n

n

n

+ + =+ + =

+ =− + = −

= −− =


CHECK 3(n + 4) + 2 0 2n

3(−14 + 4) + 2 2(−14)

3(−10) + 2 −28

−30 + 2

−28

37. 5( 2) 4 8

5 10 4 8

5 10 10 4 8 10

5 4 18

5 4 4 4 18

18

y y

y y

y y

y y

y y y y

y

− = +− = +

− + = + += +

− = − +=

The solution is 18.

CHECK 5(y − 2) 0 4y + 8

5(18 − 2) 4(18) + 8

5(16) 72 + 8

80 80

38. 3( 1) 4 1

3 3 4 1

3 3 3 4 1 3

3 4 4

3 4 4 4 4

4

4

z z

z z

z z

z z

z z z z

z

z

− = +− = +

− + = + += +

− = − +− =

= −


CHECK 3(z − 1) 0 4z + 1

3(−4 − 1) 4(−4) + 1

3(−5) −16 + 1

−15 −15

39. 3 1 2( 4)

3 1 2 8

3 1 1 2 8 1

3 2 7

3 2 2 2 7

7

a a

a a

a a

a a

a a a a

a

− = −− = −

− + = − += −

− = − −= −


CHECK 3a − 1 0 2(a − 4)

3(−7) − 1 2(−7 − 4)

−21 − 1 2(−11)

−22 −22




71

40. 4( 1) 5 3

4 4 5 3

4 4 4 5 3 4

4 5 7

4 5 5 7 5

7

7

b b

b b

b b

b b

b b b b

b

b

+ = −+ = −

+ − = − −= −

− = − −− = −

=

The solution is 7.

CHECK 4(b + 1) 0 5b − 3

4(7 + 1) 5(7) − 3

4(8) 35 − 3

32 32

41. 5( 2) 6 2

5 10 6 2

5 10 10 6 2 10

5 6 8

5 6 6 6 8

8

8

c c

c c

c c

c c

c c c c

c

c

− = −− = −

− + = − += +

− = − +− =

= −


CHECK 5(c − 2) 0 6c − 2

5(−8 − 2) 6(−8) − 2

5(−10) −48 − 2

−50 −50

42. 4 6 5 3 8

4 6 3 13

4 6 13 3 13 13

4 7 3

4 4 7 3 4

7

R R

R R

R R

R R

R R R R

R

− + = − +− + = − +

− + − = − + −− − = −

− + − = − +− =


CHECK −4R + 6 0 5 − 3R + 8

−4(−7) + 6 5 − 3(−7) + 8

28 + 6 5 + 21 + 8

34 34

43. 3 5 2 1 6 4 6

6 4

6 6 4 6

2

x x x x

x

x

x

+ − + = + −+ =

+ − = −= −


CHECK

3x + 5 − 2x + 1 0 6x + 4 − 6x

3(−2) + 5 − 2(−2) + 1 6(−2) + 4 − 6(−2)

−6 + 5 + 4 + 1 −12 + 4 + 12

4 4

44. 6 2 4 2 5 3

2 2 5

2 2 2 5 2

2 7

2 7

7

f f f f

f f

f f

f f

f f f f

f

− − = − + +− = +

− + = + += +

− = − +=

The solution is 7. CHECK

6f − 2 − 4f 0 −2f + 5 + 3f

6(7) − 2 − 4(7) −2(7) + 5 + 3(7)

42 − 2 − 28 −14 + 5 + 21

12 12

45. 2 4 5 6 1 14

7 4 8 1

7 4 4 8 1 4

7 8 3

7 8 8 8 3

3

g g g g

g g

g g

g g

g g g g

g

− + − = + −− + = − +

− + − = − + −− = − −

− + = − + −= −


CHECK

−2g + 4 − 5g 0 6g + 1 − 14g

−2(−3) + 4 − 5(−3) 6(−3) + 1 − 14(−3)

6 + 4 + 15 −18 + 1 + 42

25 25




72

46. 2 3 9 6 1

7 3 6 1

7 3 3 6 1 3

7 6 4

7 6 6 6 4

4

x x x

x x

x x

x x

x x x x

x

− + + = −+ = −

+ − = − −= −

− = − −= −


CHECK

−2x + 3 + 9x 0 6x − 1

−2(−4) + 3 + 9(−4) 6(−4) − 1

8 + 3 − 36 −24 − 1

−25 −25

47. 6( 4) 4 2 4

6 24 4 2 4

4 28 4

4 4 28 4 4

28 0

x x x

x x x

x x

x x x x

+ + − =+ + − =

+ =− + = −

=


48. 6( 1) 2 2 8 4

6 6 2 2 8 4

8 8 8 4

8 8 8 8 8 4

8 4

y y y

y y y

y y

y y y y

− − + = +− − + = +

− = +− − = − +

− =


49. 10( 2) 10 2 8( 1) 18

10 20 10 2 8 8 18

8 10 8 10

z z z

z z z

z z

− + − = + −− + − = + −

− = −


50. 7( 1) 1 6( 1)

7 7 1 6 6

6 6 6 6

a a a

a a a

a a

+ − − = ++ − − = +

+ = +


51. 3 6 2 2( 2) 4

6 2 4 4

6 2

6 2

6

b b b

b b

b b

b b b b

b

+ − = − ++ = − ++ =

− + = −=

The solution is 6.

CHECK 3b + 6 − 2b 0 2(b − 2) + 4

3(6) + 6 − 2(6) 2(6 − 2) + 4

18 + 6 − 12 2(4) + 4

12 8 + 4

12

52. 3 2 3( 2) 5

2 2 3 6 5

2 2 3 1

2 3 3

3

b b b

b b

b b

b b

b

+ − = − +

+ = − ++ = −

+ ==

The solution is 3.

CHECK 3b + 2 − b 0 3(b − 2) + 5

3(3) + 2 − 3 3(3 − 2) + 5

9 + 2 − 3 3(1) + 5

8 8

53.

2 12 5 4 7

3 32 1

3 4 73 3

2 2 1 23 4 7

3 3 3 322 2

3 43 3

203 4

320

3 4 4 43

20

3

p p p

p p

p p

p p

p p

p p p p

p

+ − = − +

− = − +

− − = − + −

− = − + −

− = − +

− + = − + +

=

The solution is 20

.3




73

CHECK

23

2 5p p+ − 0 13

4 7p− +

( ) ( )20 2023 3 3

2 5+ − ( )20 13 3

4 7− +

40 10023 3 3

+ − 80 223 3

− +

583

− 583

−

54. 2 2

4 6 3 27 7

2 162 3

7 72 3 2

2

q q q

q q

q q

q

+ − =− +

− + =− +

− =− +=

The solution is 2.

55.

3 15 9 5 1

8 23 3

4 58 2

3 3 3 34 5

8 8 2 89

4 58

94 5 5 5

89

8

r r r

r r

r r

r r

r r r r

r

+ − = − +

− + = − +

− + − = − + −

− = − +

− + = − + +

=

The solution is 9

.8

CHECK

38

5 9r r+ − 0 12

5 1r− +

( ) ( )9 3 98 8 8

5 9+ − ( )9 18 2

5 1− +

45 3 818 8 8

+ − 45 128 8

− +

338

− 338

−

56. Let p = old price. 7 23

16

p

p

+ =

=

The old price was $16.

57. Let h = average hourly earnings the previous year.

9.81 0.40

9.81 0.40 0.40 0.40

9.41

h

h

h

= +− = + −

=

The average hourly earnings the previous year were $9.41.

58. Let x = Consumer Price Index for previous year. 169.6 5.7

163.9

x

x

= +

=

The Consumer Price Index for housing was 163.9 the previous year.

59. Let y = cost 6 years ago. 142.2 326.9

142.2 142.2 326.9 142.2

184.7

y

y

y

+ =+ − = −

=

The cost of medical care was 184.7 points 6 years ago.

60. Let s = score 10 years ago. 476 16

492

s

s

= −

=

Ten years ago, the mathematics score was 492.

61. Let w = waste generated in 1960. 236.2 148.1

236.2 148.1 148.1 148.1

88.1

w

w

w

= +

− = + −=

In 1960 there was 88.1 million tons of waste generated. Let i = increase from 1960.

88.1 234

88.1 88.1 234 88.1

145.9

i

i

i

+ =

− + = −=

The increase from 1960 was 145.9 million tons.




74

62. Let a = waste recovered in 1960. 66.7 72.3

5.6

a

a

+ =

=

In 1960 there was 5.9 million tons of waste recovered. Let b = waste recovered in 2000.

3.4 72.3

68.9

b

b

+ ==

In 2000 there was 68.9 million tons of waste recovered.

63. Let m = percent of males. 38 17

38 17 17 17

21

m

m

m

= +− = − +

=

21% of males engage in exercise-walking.

64. Let f = percent of females 26 1

25

f

f

= +=

25% of females participate in swimming.

65. Let p = percent that attended art museums. 35 8

35 8 8 8

27

p

p

p

= +

− = + −=

27% attended art museums.

66. Let x = consumption in 1970. 16 113.5

129.5

x

x

− =

=

The per capita consumption in 1970 was 129.5 pounds. Let y = consumption in 1999.

113.5 1.1

112.4

y

y

= +

=

The per capita consumption in 1999 was 112.4 pounds.

67. a. 100 5( 60)

120 100 5(62 60)

120 100 5(2)

120 110 is false

W h= + −

= + −

= +

=

No, her weight does not satisfy the equation.

b. 100 5( 60)

100 5(62 60)

100 5(2)

110

W h= + −

= + −

= +

=

Her weight should be 110 lb.

68. a. 110 5( 60)

160 110 5(70 60)

160 160 is true

W h= + −

= + −

=

Yes, his weight does satisfy the equation.

b. 110 5(70 60)

110 5(10)

160 lb

W = + −

= +

=

69. a. 50 2.3( 60)

75 50 2.3(70 60)

75 50 2.3(10)

75 73 is false

W h= + −

= + −

= +

=

No, his weight doesn’t satisfy the equation.

b. He is 75 – 73 = 2 kg overweight.

70. a. 45.5 2.3( 60)

68.5 45.5 2.3(70 60)

68.5 68.5 is true

W h= + −

= + −

=

Yes, her weight does satisfy the equation.

b. 45.5 2.3( 60)

45.5 2.3(70 60) 68.5 kg

W h= + −

= + − =

71. a. 52 1.9( 60)

70 52 1.9(70 60)

70 52 1.9(10)

70 71 is false

W h= + −

= + −

= +

=

No, his weight does not satisfy the equation.

b. He is 71 – 70 = 1 kg underweight.




75

72. a. 49 1.7( 60)

65 49 1.7(70 60)

65 66 is false

W h= + −

= + −

=

No, her weight doesn’t satisfy the equation.

b. She is 66 – 65 = 1 kg underweight.

73. a. 56.2 1.41( 60)

70.3 56.2 1.41(70 60)

70.3 56.2 14.1

70.3 70.3 is true

W h= + −

= + −

= +

=

Yes, his weight does satisfy the equation.

b. 56.2 1.41(70 60)

56.2 1.41(10)

56.2 14.1

70.3 lb

W = + −

= +

= +

=

74. a. 12 500

940 12(120) 500

940 940 is true

C W= −

= −

=

Yes, this caloric intake does satisfy the equation.

b. 12(120) 500 1440 500 940C = − = − =

75. a. 14 500

1700 12(150) 500

1700 1600 is false

C W= −

= −

=

No, this caloric intake does satisfy the equation.

b. He is 1700 – 1600 = 100 calories over.

76. a. 17 500

2900 17(200) 500

2900 2900 is true

C W= −

= −

=

Yes, this caloric intake does satisfy the equation.

b. 17(200) 500

3400 500

2900

C = −

= −

=

77. If f = 40 and H = 120, then H = 1.95f + 72.85 becomes 120 = 1.95(40) + 72.85 which is a false statement. Therefore, the bone cannot belong to the missing female.

78. If f = 40 and H = 150.85, then H = 1.95f + 72.85 becomes 150.85 = 1.95(40) + 72.85 which is a true statement. Therefore, the bone can belong to the missing female.

79. If t 36,L = a t144, 30L V= = and a 50,V =

then 2

2 a ta

t

L VV

L= becomes

22 144(30 )

50 ,36

=

which is a false statement. Therefore, you cannot believe him.

80. If t a t36, 144, 30L L V= = = and a 90,V =

then 2

2 a ta

t

L VV

L= becomes

22 144(30 )

90 ,36

=

which is a false statement. Therefore, you cannot believe him.

81. Answers may vary.



84. If −x = −5, then x = 5. Answers may vary.

85. a c b c+ = +

86. a c b c− = −

87. 5 4( 1) 3 4

5 4 4 3 4

9 4 3 4

9 4 4 3 4 4

9 3

x x

x x

x x

x x x x

+ + = ++ + = +

+ = ++ − = + −

=

Since this statement is false, there is no solution.




76

88. 2 4( 1) 5 3

2 4 4 5 3

4 6 5 3

4 6 6 5 3 6

4 5 3

4 5 5 5 3

3

3

x x

x x

x x

x x

x x

x x x x

x

x

+ + = ++ + = +

+ = ++ − = + −

= −− = − −

− = −=

The solution is 3.

89. 5 4

5 5 4 5

9

x

x

x

− =− + = +

=

The solution is 9.

90. 1 3

5 51 1 3 1

5 5 5 54

5

x

x

x

− =

− + = +

=

The solution is 4

.5

91. 2.3 3.4

2.3 2.3 3.4 2.3

5.7

x

x

x

− =− + = +

=

The solution is 5.7.

92. 1 1

7 41 1 1 1

7 7 4 77 4

28 2811

28

x

x

x

x

− =

− + = +

= +

=

The solution is 11

.28

93. 2 6 2 12

8 12

8 8 12 8

4

x x

x

x

x

+ − + =+ =

+ − = −=

The solution is 4.

94. 3 5 2 3 7

8 7

8 8 7 8

1

x x

x

x

x

+ − + =+ =

+ − = −= −


95. 2 4 5

5 69 9 9

2 5

9 92 2 5 2

9 9 9 97

9

x x

x

x

x

− + + − =

− =

− + = +

=

The solution is 7

.9

96. 5 2 4 1

5 2 2 4 1 2

5 4 3

5 4 4 4 3

3

y y

y y

y y

y y y y

y

− = +− + = + +

= +− = − +

=

The solution is 3.

97. 0 4( 3) 5 3

0 4 12 5 3

0 7

0 7 7 7

7

z z

z x

z

z

z

= − + −= − + −= −

+ = − +=

The solution is 7.

98. 2 (4 1) 1 4

2 4 1 1 4

1 4 4

x x

x x

x x

− + = −− − = −

− = −

Since both sides are identical, this is an identity. The solution is all real numbers.

99. 3( 2) 3 2 (1 3 )

3 6 3 2 1 3

3 9 1 3

3 3 9 1 3 3

9 1

x x

x x

x x

x x x x

+ + = − −+ + = − +

+ = +− + = + −

=

This statement is false, so there is no solution.




77

100. 3( 1) 3 2 5

3 3 3 2 5

3 2 5

3 2 2 2 5

5

x x

x x

x x

x x x x

x

+ − = −+ − = −

= −− = − −

= −


101. If z = −7, then 1

1 07

z − = becomes

1( 7) 1 0,

7− − = which is a false statement.

Hence, −7 is not a solution of the equation.

102. If 3

,5

x = − then 5

1 03

x + = becomes

5 31 0,

3 5

− + =

which is a true statement.

Thus, 3

5− is a solution of the equation.

103. 4(−5) = −20

104. 6(−3) = −18

105. 2 3 6 1

3 4 12 2

− = − = −

106. 5 7 35 1

7 10 70 2

− = − = −

107. The reciprocal of 3

2 is

2.

3

108. The reciprocal of 2

5 is

5.

2

109. 6 = 2 ⋅ 3

16 = 2 ⋅ 2 ⋅ 2 ⋅ 2

LCM = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 = 48

110. 9 = 3 ⋅ 3

12 = 2 ⋅ 2 ⋅ 3

LCM = 2 ⋅ 2 ⋅ 3 ⋅ 3 = 36

111. 10 = 2 ⋅ 5

8 = 2 ⋅ 2 ⋅ 2

LCM = 2 ⋅ 2 ⋅ 2 ⋅ 5 = 40

112. 30 = 2 ⋅ 3 ⋅ 5

18 = 2 ⋅ 3 ⋅ 3

LCM = 2 ⋅ 3 ⋅ 3 ⋅ 5 = 90

2.2 The Multiplication and Division Properties of Equality

Problems 2.2

1. a. 35

5 5 35

15

x

x

x

=

⋅ = ⋅

=

The solution is 15.

CHECK 5x 0 3

155

3

3

b. 54

4 4( 5)4

20

y

y

y

=−

⋅ = −

=−


CHECK

4

y 0 −5

204

− −5

−5




78

2. a. 3 12

3 12

3 34

x

x

x

=

=

=

The solution is 4.

CHECK 3x 0 12

3 ⋅ 4 12

12

b. 7 21

7 21

7 73

x

x

x

=−

−=

=−


CHECK 7x 0 −21

7(−3) −21

−21

c. 5 20

5

x− =

−

5

x

−

20

5

4x

=−=−


CHECK −5x 0 20

−5(−4) 20

20

3. a. 3

125

5 3 5(12)

3 5 3

5 121

3

x

x

x

=

=

⋅ = ⋅5 4

1 120x

⋅=

=

The solution is 20.

CHECK 35

x 0 12

35

(20) 12

12

b. 2

65

5 2 5(6)

2 5 2

5 6 5 31

2 1 115

x

x

x

x

− =

− − =−

⋅⋅ =− ⋅ =−

=−


CHECK 25

x− 0 6

25

( 15)− − 6

6

c. 4

85

5 4 5( 8)

4 5 4

5 ( 8)1

x

x

x

− =−

− − =− −

− −⋅ =

5( 2)

4 110x

− −=

=

The solution is 10.

CHECK 45

x− 0 −8

45

(10)− −8

−8




79

4. a. 2 10 6

5 3

LCM of 10 and 6 is 2 ⋅ 5 ⋅ 3 = 30.

810 6

30 30 30 810 6

3 5 30 8

8 30 8

30

x x

x x

x x

x

x

+ =

⋅ + ⋅ = ⋅

+ = ⋅= ⋅=

The solution is 30.

CHECK 10 6x x+ 0 8

30 3010 6

+ 8

3 + 5

8

b. The LCM of 4 and 5 is 4 ⋅ 5 = 20.

14 5

20 20 20 14 5

5 4 20

20

x x

x x

x x

x

− =

⋅ − ⋅ = ⋅

− ==

The solution is 20.

CHECK 4 5x x− 0 1

20 204 5

− 1

5 − 4

1

5. a. The LCM of 3 and 4 is 3 ⋅ 4 = 12.

2 26

3 4

4 3

12

x x+ −+ =

212

3

x +⋅ +

212 6

44( 2) 3( 2) 72

4 8 3 6 72

7 2 72

7 70

10

x

x x

x x

x

x

x

−⋅ = ⋅

+ + − =+ + − =

+ ===

The solution is 10.

CHECK 2 23 4

x x+ −+ 0 6

10 2 10 23 4+ −+ 6

8123 4

+

4 + 2

6

b. The LCM of 5 and 3 is 5 ⋅ 3 = 15.

2 20

5 32 2

15 15 15(0)5 3

3( 2) 5( 2) 0

3 6 5 10 0

2 16 0

2 16

8

x x

x x

x x

x x

x

x

x

+ −− =

+ − − =

+ − − =+ − + =

− + =− = −

=

The solution is 8.

6. Forty percent of 30 is what number?

4030

1002

305

60

512

n

n

n

n

⋅ =

⋅ =

=

=

Thus, 40% of 30 is 12.




80

7. What percent of 30 is 6? 30 6

30 6

30 301 20

5 100

x

x

x

⋅ =⋅

=

= =

Thus, 6 is 20% of 30.

8. 20 is 40% of what number?

10

4020

10040

20100

220

5

5 2 520

2 5 2

n

n

n

n

= ⋅

⋅ =

⋅ =

⋅ ⋅ = ⋅

1

50n =

Thus, 20 is 40% of 50.

9. a. 5.3% of 225 means 0.053 ⋅ 225 = 11.925

or 12 when rounded to the nearest whole number.

b. 10.6% of 226 means

0.106 ⋅ 226 = 23.956 or 24 when

rounded to the nearest whole number.

Exercises 2.2

Checks are left to the student.

1. 57

7 7 57

35

x

x

x

=

⋅ = ⋅

=

The solution is 35.

2. 92

2 2 92

18

y

y

y

=

⋅ = ⋅

=

The solution is 18.

3. 42

2 ( 4) 22

8

x

x

x

− =

⋅ − = ⋅

− =


4. 65

5 5 65

30

a

a

a

=

⋅ = ⋅

=

The solution is 30.

5. 53

3 3 53

15

b

b

b

=−

− ⋅ = − ⋅−

= −


6. 74

4 7 44

28

c

c

c

=−

− ⋅ = − − − =


7. 32

2( 3) 22

6

f

f

f

− =−

− − = − − =

The solution is 6.

8. 64

4 4( 6)4

24

g

g

g

= −−

− = − − − =

The solution is 24.




81

9. 1

4 31

4 44 3

4

3

v

v

v

=

⋅ = ⋅

=

The solution is 4

.3

10. 2

3 72

3 33 7

6

7

w

w

w

=

⋅ = ⋅

=

The solution is 6

.7

11. 3

5 43

5 55 4

15

4

x

x

x

−=

− ⋅ = ⋅

= −

The solution is 15

.4

−

12. 8

9 28

2 29 2

16

9

y

y

y

−=

⋅ − = ⋅

− =

The solution is 16

.9

−

13. 3 33

3 33

3 311

z

z

z

=

=

=

The solution is 11.

14. 4 32

4 32

4 48

y

y

y

=

=

=

The solution is 8.

15. 42 6

42 6

6 67

x

x

x

− =−

=

− =


16. 7 49

7 49

7 77

b

b

b

= −−

=

= −


17. 8 56

8 56

8 87

c

c

c

− =−

=− −

= −


18. 5 45

5 45

5 59

d

d

d

− =−

=− −

= −


19. 5 35

5 35

5 57

x

x

x

− = −− −

=− −

=

The solution is 7.

20. 12 3

12 3

3 34

x

x

x

− = −− −

=− −

=

The solution is 4.




82

21. 3 11

3 11

3 311

3

y

y

y

− =−

=− −

= −

The solution is 11

.3

−

22. 5 17

5 17

5 517

5

z

z

z

− =−

=− −

= −

The solution is 17

.5

−

23. 2 1.2

2 1.2

2 20.6

a

a

a

− =−

=− −

= −


24. 3 1.5

3 1.5

3 30.5

b

b

b

− =−

=− −

= −


25.

3

1

13 4

29

32

1 1 93

3 3 2

3

2

t

t

t

t

=

=

⋅ = ⋅

=

The solution is 3

.2

26. 2

4 63

204

35

3

r

r

r

=

=

=

The solution is 5

.3

27. 1

0.7531

3 3( 0.75)3

2.25

x

x

x

= −

⋅ = −

= −


28. 1

0.2541

4 4 0.254

1.00

y

y

y

=

⋅ = ⋅

=


29.

2

36

4

4( 6)

3

C

−

− =

⋅ −

1

4 3

3 4

8

C

C

= ⋅

− =


30. 2

29

9

F

F

− =

− =


31. 5

106

6 5 610

5 6 5

a

a

=

⋅ = ⋅

1

12a=

The solution is 12.




83

32. 2

247

84

z

z

=

=

The solution is 84.

33. 4

0.45

5 4 5

4 5 4

y

y

− =

− ⋅ − =− 1

0.4⋅0.1

0.5y=−


34. 1

0.54

1 1

2 41

2

x

x

x

−=

−=

=−

The solution is 1

.2

−

35. 2

011

11 2 110

2 11 2

0

p

p

p

−=

− ⋅ = ⋅ − − =

The solution is 0.

36. 4

09

9 4 90

4 9 4

0

q

q

q

−=

− ⋅ = ⋅ − − =

The solution is 0.

37. 3

185

5

3

t− =

1

( 18)⋅ −6

5 3

3 5

30

t

t

−

= ⋅

− =


38. 2

87

28

R

R

− =

− =


39. 7

70.02

0.02 7 0.02( 7)

7 0.02 7

0.02

x

x

x

= −

= ⋅ −

= −


40. 6

60.03

0.03 0.03 6( 6)

6 6 0.030.03

y

y

y

− =

⋅ − = ⋅

− =


41. The LCM of 2 and 3 is 2 ⋅ 3 = 6.

102 3

6 6 6 102 33 2 60

5 60

12

y y

y y

y y

y

y

+ =

⋅ + ⋅ = ⋅

+ ===

The solution is 12.

42. The LCM of 4 and 3 is 4 ⋅ 3 = 12.

144 3

12 12 12 144 3

3 4 168

7 168

24

a a

a a

a a

a

a

+ =

⋅ + ⋅ = ⋅

+ ===

The solution is 24.




84

43. The LCM of 7 and 3 is 7 ⋅ 3 = 21.

107 3

21 21 21 107 3

3 7 210

10 210

21

x x

x x

x x

x

x

+ =

⋅ + ⋅ = ⋅

+ ===

The solution is 21.

44. 2 6 4

3 2

The LCM of 6 and 4 is 2 ⋅ 3 ⋅ 2 = 12.

206 4

12 12 12 206 4

2 3 240

5 240

48

z z

z z

z z

z

z

+ =

⋅ + ⋅ = ⋅

+ ===

The solution is 48.

45. The LCM of 5 and 10 is 10.

65 10

10 10 10 65 10

2 60

3 60

20

x x

x x

x x

x

x

+ =

⋅ + ⋅ = ⋅

+ ===

The solution is 20.


82 6

6 6 6 82 6

3 48

4 48

12

r r

r r

r r

r

r

+ =

⋅ + ⋅ = ⋅

+ ===

The solution is 12.

47. 2 6 8

3 4

The LCM of 6 and 8 is 2 ⋅ 3 ⋅ 4 = 24.

76 8

24 24 24 76 8

4 3 168

7 168

24

t t

t t

t t

t

t

+ =

⋅ + ⋅ = ⋅

+ ===

The solution is 24.

48. 3 9 12

3 4

The LCM of 9 and 12 is 3 ⋅ 3 ⋅ 4 = 36.

149 12

36 36 36 149 12

4 3 504

7 504

72

f f

f f

f f

f

f

+ =

⋅ + ⋅ = ⋅

+ ===

The solution is 72.

49. The LCM of 2 and 5 is 2 ⋅ 5 = 10.

7

2 5 107

10 10 102 5 10

5 2 7

7 7

1

x x

x x

x x

x

x

+ =

⋅ + ⋅ = ⋅

+ ===

The solution is 1.

50. The LCM of 3 and 7 is 3 ⋅ 7 = 21.

20

3 7 2120

21 21 213 7 21

7 3 20

10 20

2

a a

a a

a a

a

a

+ =

⋅ + ⋅ = ⋅

+ ===

The solution is 2.




85

51. The LCM of 3 and 5 is 3 ⋅ 5 = 15.

23 5

15 15 15 23 5

5 3 30

2 30

15

c c

c c

c c

c

c

− =

⋅ − ⋅ = ⋅

− ===

The solution is 15.

52. The LCM of 4 and 7 is 4 ⋅ 7 = 28.

34 7

28 28 28 34 7

7 4 84

3 84

28

F F

F F

F F

F

F

− =

⋅ − ⋅ = ⋅

− ===

The solution is 28.

53. 6 = 2 ⋅ 3

8 = 2 ⋅2 ⋅ 2

12 = 2 ⋅ 2 ⋅ 3

LCM = 2 ⋅ 2 ⋅ 2 ⋅ 3 = 24

5

6 8 125

24 24 246 8 12

4 3 10

10

W W

W W

W W

W

− =

⋅ − ⋅ = ⋅

− ==

The solution is 10.

54. 6 = 2 ⋅ 3

10 = 2 ⋅ 5

3 = 3

LCM = 2 ⋅ 3 ⋅ 5 = 30

4

6 10 34

30 30 306 10 3

5 3 40

2 40

20

m m

m m

m m

m

m

− =

⋅ − ⋅ = ⋅

− ===

The solution is 20.

55. The LCM of 5, 10, and 2 is 10.

3 1

5 10 23 1

10 10 105 10 2

2 3 5

2 8

4

x

x

x

x

x

− =

⋅ − ⋅ = ⋅

− ===

The solution is 4.


3 1 1

7 14 143 1 1

14 14 147 14 14

6 1 1

6 2

2 1

6 3

y

y

y

y

y

− =

⋅ − ⋅ = ⋅

− ==

= =

The solution is 1

.3

57. The LCM of 4, 3, and 2 is 4 ⋅ 3 = 12.

4 2 1

4 3 24 2 1

12 12 124 3 2

3( 4) 4( 2) 6

3 12 4 8 6

4 6

10

10

x x

x x

x x

x x

x

x

x

+ +− = −

+ + ⋅ − ⋅ = ⋅ −

+ − + = −+ − − = −

− + = −− = −

=

The solution is 10.

58. The LCM of 2, 8, and 16 is 16.

1 7 1

2 8 161 7 1

16 16 162 8 16

8( 1) 2 7 1

8 8 2 7 1

10 8 7 1

10 7 9

3 9

3

w w w

w w w

w w w

w w w

w w

w w

w

w

− ++ =

− +⋅ + ⋅ = ⋅

− + = +− + = +

− = += +==

The solution is 3.




86

59. 2 6 4

3 2

The LCM of 6 and 4 is 2 ⋅ 3 ⋅ 2 = 12.

3 7

6 4 43 7

12 12 12 126 4 4

2 9 12 21

2 12 30

10 30

3

xx

xx

x x

x x

x

x

+ = −

⋅ + ⋅ = ⋅ − ⋅

+ = −= −

− = −=

The solution is 3.


4 1

6 3 34 1

6 6 6 66 3 3

8 6 2

6 10

5 10

2

xx

xx

x x

x x

x

x

+ = −

⋅ + ⋅ = ⋅ − ⋅

+ = −= −

− = −=

The solution is 2.

61. 30% of 40 is what number?

3040

1003

4010

12

n

n

n

⋅ =

⋅ =

=

Thus, 30% of 40 is 12.


1780

10068

513.6

n

n

n

⋅ =

=

=

Thus, 17% of 80 is 13.6.


4070

100

2

5

n⋅ =

70⋅14

28

n

n

=

=

Thus, 40% of 70 is 28.


40 8

40 401 20

5 100

x

x

x

⋅ =⋅

=

= =

Thus, 8 is 20% of 40.


30 15

30 301 50

2 10

x

x

x

⋅ =⋅

=

= =

Thus, 15 is 50% of 30.


40 4

40 401 10

10 100

x

x

x

⋅ =⋅

=

= =

Thus, 4 is 10% of 40.


2030

1001

305

15 30 5

5150

n

n

n

n

= ⋅

= ⋅

⋅ = ⋅ ⋅

=

Thus, 30 is 20% of 150.


4020

1002

205

5 5 220

2 2 550

n

n

n

n

= ⋅

= ⋅

⋅ = ⋅ ⋅

=

Thus, 20 is 40% of 50.




87


6012

1003

125

5 5 312

3 3 520

n

n

n

n

= ⋅

= ⋅

⋅ = ⋅ ⋅

=

Thus, 12 is 60% of 20.


5024

1001

242

12 24 2

248

n

n

n

n

= ⋅

= ⋅

⋅ = ⋅ ⋅

=

Thus, 24 is 50% of 48.

71. 16 is what percent of 211? 16 211

16 211

211 2110.076

n

n

n

= ⋅⋅

=

≈

Thus, 7.6% of 211 is 16.


171 7

171 1710.041

n

n

n

⋅ =⋅

=

≈

Thus, 4.1% of the 171 patients died.


2 211 20

211 2110.095

n

n

⋅ =⋅

=

≈

Thus, 9.5% of the 211 patients had heart attacks.


171 7

171 1710.041

n

n

n

⋅ =⋅

=

≈

Thus, 4.1% of the 171 patients had heart attacks.


211 8

211 2110.038

n

n

n

⋅ =⋅

=

≈

Thus, 3.8% of the 211 patients had a stroke.


171 2

171 1710.012

n

n

n

⋅ =⋅

=

≈



171 9

171 1710.053

n

n

n

⋅ =⋅

=

≈

Thus, 5.3% of the 171 patients died.


171 8

171 1710.047

n

n

n

⋅ =⋅

=

≈

Thus, 4.7% of the 171 patients had a heart attack.


211 8

211 2110.038

n

n

n

⋅ =⋅

=

=



171 3

171 1710.018

n

n

n

⋅ =⋅

=

≈


81. a. 85 3500h=

b. 85 3500

3500 341

85 17

h

h

=

= =




88

82. a. 160 3500h=

b. 160 3500

3500 721

160 8

h

h

=

= =

83. a. 260 3500h=

b. 260 3500

3500 613

260 13

h

h

=

= =

84. a. 280 3500h=

b. 280 3500

3500 112

280 2

h

h

=

= =

85. a. 450 3500h=

b. 450 3500

3500 77

450 9

h

h

=

= =

86. 8 × 250 × 20 = 40,000

Her yearly salary is $40,000.

87. 8 × 100 × 20 = 16,000

Her lost wages are $16,000.

88. Let x = multiple of the old wages needed to make up the loss within 1 year. 40,000 56,000

40,000 56,000

40,000 40,000

1.40 or 140%

x

x

x

⋅ =⋅

=

=

Thus, the employee needs a 40% increase to recuperate her lost wages within 1 year.

89. a. Answers may vary.

b. Answers may vary.

90. Answers may vary. Sample answer: It

would be easier to divide by −3.

3 18

3 18

3 36

x

x

x

− =−

=− −

= −

91. Answers may vary. Sample answer: It would be easier to multiply by the

reciprocal of 3

.4

−

315

4

4 3 4

3 4 3

x

x

− =

− ⋅ − =− 1

15⋅5

20x=−

92. ac bc=

93. a b

c c=

94. LCM

95. LCM


5

3015

1003

1510

1015

3

n

n

= ⋅

= ⋅

⋅10 3

3 1050

n

n

= ⋅ ⋅

=

Thus, 15 is 30% of 50.


45 9

45 451 20

5 100

x

x

x

⋅ =⋅

=

= =

Thus, 9 is 20% of 45.




89


12

4060

100

260

5

x⋅ =

⋅

24

x

x

=

=

Thus, 40% of 60 is 24.

99. The LCM of 4 and 5 is 4 ⋅ 5 = 20.

2 13

4 52 1

20 20 20 34 5

5( 2) 4( 1) 60

5 10 4 4 60

9 6 60

9 54

6

x x

x x

x x

x x

x

x

x

+ −+ =

+ −⋅ + ⋅ = ⋅

+ + − =+ + − =

+ ===

The solution is 6.

100. The LCM of 2 and 3 is 2 ⋅ 3 = 6.

3 25

2 33 2

6 6 6 52 3

3( 3) 2( 2) 30

3 9 2 4 30

13 30

17

x x

x x

x x

x x

x

x

+ −− =

+ −⋅ − ⋅ = ⋅

+ − − =+ − + =

+ ==

The solution is 17.

101. The LCM of 6 and 10 is 2 ⋅ 3 ⋅ 5 = 30.

86 10

30 30 30 86 10

5 3 240

8 240

30

y y

y y

y y

y

y

+ =

⋅ + ⋅ = ⋅

+ ===

102. The LCM of 5 and 8 is 5 ⋅ 8 = 40.

35 8

40 40 40 35 8

8 5 120

3 120

40

y y

y y

y y

y

y

− =

⋅ − ⋅ = ⋅

− ===

The solution is 40.

103.

1

48

55 4 5

84 5 4

10

y

y

y

=

⋅ = ⋅

=

The solution is 10.

104.

1

36

44 3 4

63 4 3

8

y

y

y

− =

− ⋅ − =− ⋅

=−


105. 2

47

7 2 7

2 7 2

y

y

− =−

− ⋅ − = −( 4)⋅ −

2

14y=

The solution is 14.

106. 7 16

16

7

y

y

− =

=−

The solution is 16

.7

−

107. 72

14

x

x

=−

=−

108. 34

12

y

y

− =−

=

The solution is 12.




90

109. 4( 6) 4 4(6) 4 24x x x− = − = −

110. 3(6 ) 3 6 3 18 3x x x− = ⋅ − ⋅ = −

111. 5(8 ) 5 8 5 40 5y y y− = ⋅ − ⋅ = −

112. 6(8 2 ) 6 8 6 2 48 12y y y− = ⋅ − ⋅ = −

113. 9(6 3 ) 9 6 9 3 54 27y y y− = ⋅ − ⋅ = −

114. −3(4x − 2) = −3 ⋅ 4x − 3 ⋅ (−2) = −12x + 6

115. −5(3x − 4) = −5 ⋅ 3x − 5 ⋅ (−4) = −15x + 20

116.

5

3 2020

4⋅ =

3 1515

1 4 1⋅ = =

117.

41 24

246⋅ =

1 44

1 6 1⋅ = =

118.

1

4 55

5

−− ⋅ − =

4

1 5⋅−

44

1

= =

119.

1

7 77

3

−− ⋅ − =

3

1 7⋅−

33

1

= =

2.3 Linear Equations

Problems 2.3 1. a. 4 5 17

4 5 5 17 5

4 12

4 12

4 43

x

x

x

x

x

+ =+ − = −

=

=

=

The solution is 3.

CHECK 4x + 5 0 17

4(3) + 5 17

12 + 5

17

b. 3 5 2

3 5 5 2 5

3 7

3 7

3 37

3

x

x

x

x

x

− − =− − + = +

− =−

=− −

= −

The solution is 7

.3

−

CHECK −3x − 5 0 2

( )73

3 5− − − 2

7 − 5

2

2. 7( 1) 4( 2) 5

7 7 4 8 5

7 7 4 13

7 7 7 4 13 7

7 4 6

7 4 4 4 6

3 6

3 6

3 32

x x

x x

x x

x x

x x

x x x x

x

x

x

+ = + ++ = + ++ = +

+ − = + −= +

− = − +=

=

=

The solution is 2.

CHECK 7(x + 1) 0 4(x + 2) + 5

7(2 + 1) 4(2 + 2) + 5

7(3) 4(4) + 5

21 16 + 5

21




91

3. a. The LCM of 3, 7, and 21 is 21.

3 7

20

21 7 3

21

x x= +

20

21⋅ 21= 21

7

x⋅ +

3

20 3 7

20 10

20 10

10 102

x

x x

x

x

x

⋅

= +=

=

=

The solution is 2.

CHECK 2021

0 7 3x x+

2021

2 27 3

+

6 1421 21

+

2021

b. The LCM of 4, 5, and 20 is 20.

5 4

1 17( 4)

4 5 20

20

x x +− =

120

4⋅ − 20

5

x⋅ =

17( 4)

20

x +⋅

5 4 17( 4)

5 4 17 68

5 5 4 17 68 5

4 17 63

4 17 17 17 63

21 63

21 63

21 213

x x

x x

x x

x x

x x x x

x

x

x

− = +− = +

− − = + −− = +

− − = − +− =−

=− −

= −


CHECK 1

4 5x− 0

17( 4)

20

x+

314 5

−− 17( 3 4)

20

− +

5 1220 20

+ 17(1)

20

1720

1720

4. ( )( )

14

61

6 6 46

6 4

6 4 4 4

6 4

6 4

S C

S C

S C

S C

S C

C S

= −

⋅ = ⋅ −

= −

+ = − +

+ == +

5. a. Let S = 12. 3 24

12 3 24

12 24 3 24 24

36 3

36 3

3 312

S L

L

L

L

L

L

= −= −

+ = − +=

=

=

The length of their foot is 12 inches.

b.

3 24

24 3 24 24

24 3

324

3 324

324

3

S L

S L

S L

LS

SL

SL

= −

+ = − +

+ =

+=

+=

+=




92

6. 3 4 7

3 4 4 7 4

3 7 4

3 7 4

3 37 4

3

x y

x y y y

x y

x y

yx

+ =+ − = −

= −−

=

−=

7. a. 2001 2000 1;

992(1) 16,145 17,137

x

C

= − =

= + =

The tuition cost in 2001 was about $17,137.

b. 2005 2000 5;

992(5) 16,145 21,105

x

C

= − =

= + =

The tuition cost in 2001 was about $21,105.

c. 992 16,145

20,000 992 16,145

992 16,14520,000

16,145

C x

x

x

= +

= +

+=

− 16,145−

3855 992

3855

9924

x

x

x

=

=

≈

The cost will be $20,000 4 years after 2000, or in 2004.

Exercises 2.3

Checks are left to the student.

1. 3 12 0

3 12 12 0 12

3 12

3 12

3 34

x

x

x

x

x

− =− + = +

=

=

=

The solution is 4.

2. 5 10 0

5 10 10 0 10

5 10

2

a

a

a

a

+ =

+ − = −

=−=−


3. 2 6 8

2 6 6 8 6

2 2

2 2

2 21

y

y

y

y

y

+ =+ − = −

=

=

=

The solution is 1.

4. 4 5 3

4 5 5 3 5

4 8

4 8

4 42

b

b

b

b

b

− =− + = +

=

=

=

The solution is 2.

5. 3 4 10

3 4 4 10 4

3 6

3 6

3 32

z

z

z

z

z

− − = −− − + = − +

− = −− −

=− −

=

The solution is 2.

6. 4 2 6

4 2 2 6 2

4 8

4 8

4 42

r

r

r

r

r

− − =− − + = +

− =−

=− −

= −


7. 5 1 13

5 1 1 13 1

5 14

5 14

5 514

5

y

y

y

y

y

− + = −− + − = − −

− = −− −

=− −

=

The solution is 14

.5




93

8. 3 1 9

3 1 1 9 1

3 10

3 10

3 310

3

x

x

x

x

x

− + = −− + − = − −

− = −− −

=− −

=

The solution is 10

.3

9. 3 4 10

3 4 4 10 4

3 6

3 6

2 6

2 6

2 23

x x

x x

x x

x x x x

x

x

x

+ = ++ − = + −

= +− = − +

=

=

=

The solution is 3.

10. 4 4 7

4 4 4 7 4

4 3

4 3

3 3

3 3

3 31

x x

x x

x x

x x x x

x

x

x

+ = ++ − = + −

= +− = − +

=

=

=

The solution is 1.

11. 5 12 6 8

5 12 12 6 8 12

5 6 4

5 6 6 6 4

4

4

x x

x x

x x

x x x x

x

x

− = −− + = − +

= +− = − +

− == −


12. 5 7 7 19

5 7 7 7 19 7

5 7 12

5 7 7 7 12

2 12

2 12

2 26

x x

x x

x x

x x x x

x

x

x

+ = ++ − = + −

= +− = − +− =−

=− −

= −


13. 4 7 6 9

4 7 7 6 9 7

4 6 16

4 6 6 6 16

2 16

2 16

2 28

v v

v v

v v

v v v v

v

v

v

− = +− + = + +

= +− = − +− =−

=− −

= −


14. 8 4 15 10

8 4 4 15 10 4

8 15 14

8 15 15 15 14

7 14

7 14

7 72

t t

t t

t t

t t t t

t

t

t

+ = −+ − = − −

= −− = − −

− = −− −

=− −

=

The solution is 2.

15. 6 3 12 0

3 12 0

3 12 12 0 12

3 12

3 12

3 34

m m

m

m

m

m

m

− + =+ =

+ − = −= −

−=

= −


16. 10 15 5 25

5 15 25

5 15 15 25 15

5 10

2

k k

k

k

k

k

+ − =

+ =+ − = −

==

The solution is 2.

17. 10 3 8 6

10 10 3 8 10 6

3 2 6

3 6 2 6 6

3 2

3 2

3 32

3

z z

z z

z z

z z z z

z

z

z

− = −− − = − −

− = − −− + = − − +

= −−

=

= −

The solution is 2

.3

−




94

18. 8 4 10 6

8 8 4 10 8 6

4 2 6

4 6 2 6 6

10 2

10 2

10 101

5

y y

y y

y y

y y y y

y

y

y

− = +− − = − +

− = +− − = + −

− =−

=− −

= −

The solution is 1

.5

−

19. 5( 2) 3( 3) 1

5 10 3 9 1

5 10 3 10

5 10 10 3 10 10

5 3

5 3 3 3

2 0

2 0

2 20

x x

x x

x x

x x

x x

x x x x

x

x

x

+ = + ++ = + ++ = +

+ − = + −=

− = −=

=

=

The solution is 0.

20.

(4 2 ) 7( 1)

4 2 7 7

3 4 7 7

3 4 4 7 7 4

3 7 3

3 7 7 7 3

4 3

4 3

4 43

4

y y y

y y y

y y

y y

y y

y y y y

y

y

y

− − = −− + = −

− = −− + = − +

= −− = − −− = −− −

=− −

=

The solution is 3

.4

21. 5(4 3 ) 7(3 4 )a a− = −

20 15 21 28

20 20 15 21 20 28

15 1 28

15 28 1 28 28

13 1

13 1

13 131

13

a a

a a

a a

a a a a

a

a

a

− = −− − = − −

− = −− + = − +

=

=

=

The solution is 1

.13

22.

3 14.5 1.3

4 43 1

4.5 4.5 1.3 4.54 4

3 15.8

4 43 1 1 1

5.84 4 4 4

25.8

41

5.821

2 2 5.82

11.6

y y

y y

y y

y y y y

y

y

y

y

− = +

− + = + +

= +

− = − +

=

=

⋅ = ⋅

=


23.

7 55.6 3.3

8 87 5

5.6 5.6 3.3 5.68 8

7 58.9

8 87 5 5 5

8.98 8 8 8

28.9

81

8.94

14 4 ( 8.9)

4

35.6

c c

c c

c c

c c c c

c

c

c

c

− + = − −

− + − = − − −

− = − −

− + = − + −

− = −

− = −

− ⋅ − = − ⋅ −

=





95

24.

2

210

33 2

103 3

510

3

3 5 310

5 3 5

x x

x x

x

x

+ =

+ =

=

⋅ = ⋅

1

6x =

The solution is 6.

25.

1 42 2

4 51 1 4 1

2 24 4 5 4

16 52 2

20 2011

2 220

112 2 2 2

2011

420

1 1 11( 4 )

4 4 2011

80

x x

x x

x x

x x

x x x x

x

x

x

− + = +

− + − = + −

− = + −

− = +

− − = − +

− =

− ⋅ − = − ⋅

= −

The solution is 11

.80

−

26.

1 26 2

7 71 1 2 1

6 27 7 7 7

36 2

73

6 2 2 27

34

71 1 3

44 4 7

3

28

x x

x x

x x

x x x x

x

x

x

+ = −

+ − = − −

= −

− = − −

= −

⋅ = ⋅ −

= −

The solution is 3

.28

−

27.

1 23

2 21 2

2 2 2 32 2

1 2 6

2 3 6

2 3 3 6 3

2 9

2 9

2 29

2

x x

x x

x x

x

x

x

x

x

− −+ =

− −⋅ + ⋅ = ⋅

− + − =− =

− + = +=

=

=

The solution is 9

.2

28.

3 5 312

3 33 5 3

3 3 3 123 33 5 3 36

4 8 36

4 8 8 36 8

4 28

4 28

4 47

x x

x x

x x

x

x

x

x

x

+ ++ =

+ +⋅ + ⋅ = ⋅

+ + + =+ =

+ − = −=

=

= The solution is 7.




96

29. 15 4

20 20 20 15 4

4 5 20

20

20

x x

x x

x x

x

x

− =

⋅ − ⋅ = ⋅

− =− =

= −


30. 13 2

6 6 6 13 22 3 6

6

6

x x

x x

x x

x

x

− =

⋅ − ⋅ = ⋅

− =− =

= −


31.

1 2 23

4 31 2 2

12 12 12 34 3

3( 1) 4(2 2) 36

3 3 8 8 36

5 11 36

5 11 11 36 11

5 25

5 25

5 55

x x

x x

x x

x x

x

x

x

x

x

+ −− =

+ −⋅ − ⋅ = ⋅

+ − − =+ − + =

− + =− + − = −

− =−

=− −

= −


32. 4 6

3 44 6

12 123 4

4( 4) 3( 6)

4 16 3 18

4 16 16 3 18 16

4 3 2

4 3 3 3 2

2

z z

z z

z z

z z

z z

z z

z z z z

z

+ +=

+ +⋅ = ⋅

+ = ++ = +

+ − = + −= +

− = − +=

The solution is 2.

33.

2 1 4

3 122 1 4

12 123 12

4(2 1) 4

8 4 4

8 4 4 4 4

8

8

7 0

7 0

7 70

h h

h h

h h

h h

h h

h h

h h h h

h

h

h

− −=

− −⋅ = ⋅

− = −− = −

− + = − +=

− = −=

=

=

The solution is 0.

34.

5 6 7 42

7 35 6 7 4

21 21 21 27 3

3(5 6 ) 7( 7 4 ) 42

15 18 49 28 42

64 10 42

64 64 10 42 64

10 22

10 22

10 1011

5

y y

y y

y y

y y

y

y

y

y

y

− − −− =

− − −⋅ − ⋅ = ⋅

− − − − =− + + =

+ =− + = −

= −−

=

= −

The solution is 11

.5

−

35. 2 3 3 1

12 4

2 3 3 14 4 4 1

2 42(2 3) (3 1) 4

4 6 3 1 4

5 4

5 5 4 5

1

w w

w w

w w

w w

w

w

w

+ +− =

+ +⋅ − ⋅ = ⋅

+ − + =+ − − =

+ =+ − = −

= −





97

36.

7 2 1

6 2 47 2 1

12 12 126 2 4

2(7 2) 6 3

14 4 6 3

14 10 3

14 14 10 3 14

10 11

10 11

11 1110

11

r r

r r

r r

r r

r r

r r r r

r

r

r

++ =

+⋅ + ⋅ = ⋅

+ + =+ + =

+ =− + = −

= −−

=− −

− =

The solution is 10

.11

−

37.

8 23 1 5

6 3 28 23 1 5

6 6 66 3 28 23 2 15

8 21 15

8 8 21 15 8

21 7

21 7

7 73

xx

xx

x x

x x

x x x x

x

x

x

−+ =

−⋅ + ⋅ = ⋅

− + =− =

− − = −− =−

=

− =


38.

1 2 48

2 3 41 2 4

12 12 12 12 ( 8)2 3 4

6( 1) 4( 2) 3( 4) 96

6 6 4 8 3 12 96

13 26 96

13 26 26 96 26

13 122

122

13

x x x

x x x

x x x

x x x

x

x

x

x

+ + ++ + =−

+ + +⋅ + ⋅ + ⋅ = ⋅ −

+ + + + + =−

+ + + + + =−+ =−

+ − =− −

=−

=−

The solution is 122

.13

−

39.

5 4 3( 2)

2 3 25 4 3

6 6 6 6 ( 2)2 3 2

3( 5) 2( 4) 3( 3) 6 12

3 15 2 8 3 9 6 12

7 3 3

7 7 3 3 7

3 10

3 3 3 10

4 10

4 10

4 45

2

x x xx

x x xx

x x x x

x x x x

x x

x x

x x

x x x x

x

x

x

− − −− = − −

− − −⋅ − ⋅ = ⋅ − ⋅ −

− − − = − − +− − + = − − +

− = − +− + = − + +

= − ++ = − + +

=

=

=

The solution is 5

.2

40.

1 2 316

2 3 41 2 3

12 12 12 12 162 3 4

6( 1) 4( 2) 3( 3) 192

6 6 4 8 3 9 192

13 23 192

13 23 23 192 23

13 169

13 169

13 1313

x x x

x x x

x x x

x x x

x

x

x

x

x

+ + ++ + =

+ + +⋅ + ⋅ + ⋅ = ⋅

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

+ =+ − = −

=

=

=

The solution is 13.

41. 1 1

4 42 8

1 14 4

2 2

x x

x x

− + = −

− + = −

This is an identity. The solution is all real numbers.




98

42.

2 16 4

3 5

2 46 4

3 52 4

15( 6 ) 15 15 15 43 5

90 10 12 60

90 10 10 12 10 60

90 2 60

90 60 2 60 60

30 2

1

15

x x

x x

x x

x x

x x

x x

x x x x

x

x

− + = −

− + = −

− + ⋅ = ⋅ − ⋅

− + = −

− + − = − −

− = −− + = − +

− =

=−

The solution is 1

.15

−

43.

1 1(8 4) 5 (4 8) 1

2 44 2 5 2 1

4 3 3

4 3 3 3 3

4 6

4 6

3 6

3 6

3 32

x x

x x

x x

x x

x x

x x x x

x

x

x

+ − = + +

+ − = + +− = +

− + = + += +

− = − +=

=

=

The solution is 2.

44. 1 1

(3 9) 2 (9 18) 33 9

3 2 2 3

5 5

x x

x x

x x

+ + = + +

+ + = + ++ = +

This is an identity. The solution is all real numbers.

45. 3

92 5

310 10 10 10 9

2 510 5 6 90

9 90

9 90

9 910

x xx

x xx

x x x

x

x

x

+ − =

⋅ + ⋅ − ⋅ = ⋅

+ − ==

=

=

The solution is 10.

46.

5 3 110

3 4 65 3 11

12 12 12 12 03 4 6

20 9 22 0

11 22 0

11 22 22 0 22

11 22

11 22

11 112

x x

x x

x x

x

x

x

x

x

− + =

⋅ − ⋅ + ⋅ = ⋅

− + =+ =

+ − = −= −

−=

= −


47.

4 3 5 3

9 2 6 24 3 5 3

18 18 18 189 2 6 2

8 27 15 27

8 27 12

8 8 27 12 8

27 20

27 20

20 2027

20

x x x

x x x

x x x

x x

x x x x

x

x

x

− = −

⋅ − ⋅ = ⋅ − ⋅

− = −− = −

− − = − −− = −− −

=− −

=

The solution is 27

.20

48. 7 4 2 11

2 3 5 67 4 2 11

30 30 30 302 3 5 6

105 40 12 55

77 55

5

7

x x x

x x x

x x x

x

x

− + =−

⋅ − ⋅ + ⋅ = ⋅ −

− + =−=−

=−

The solution is 5

.7

−




99

49.

3 4 1 7 18(19 3) 1

2 8 123 4 1 7 18

24 24 (19 3) 24 1 242 8 12

12(3 4) 3(19 3) 24 2(7 18)

36 48 57 9 24 14 36

21 57 14 12

21 57 57 14 12 57

21 14 69

21 14 14 14 69

7 69

69

7

x xx

x xx

x x x

x x x

x x

x x

x x

x x x x

x

x

+ +− − = −

+ +⋅ − ⋅ − = ⋅ − ⋅

+ − − = − +

+ − + = − −− + =− −

− + − =− − −− =− −

− + =− + −

− =−

=

The solution is 69

.7

50.

11 2 1 17 7 2(3 1) (7 2)

3 2 6 911 2 1 17 7 2

18 18 (3 1) 18 18 (7 2)3 2 6 9

6(11 2) 9(3 1) 3(17 7) 4(7 2)

66 12 27 9 51 21 28 8

39 3 23 29

39 3 3 23 29 3

39 23 32

39 23 23 23 32

16 32

16

16

x xx x

x xx x

x x x x

x x x x

x x

x x

x x

x x x x

x

x

− +− − = − −

− +⋅ − ⋅ − = ⋅ − ⋅ −

− − − = + − −− − + = + − +

− = +− + = + +

= +− = − +

=

=32

162x =

The solution is 2.

51. 2

2

2 2

2

2

C r

rC

Cr

Cr

= π

π=

π π

=π

=π

52. A b h

b hA

b bA

hb

Ah

b

=

=

=

=




100

53. 3 2 6

3 3 2 6 3

2 6 3

2 6 3

2 26 3

2

x y

x x y x

y x

y x

xy

+ =− + = −

= −−

=

−=

54. 5 6 30

5 5 6 30 5

6 30 5

6 30 5

6 65 30

6

x y

x x y x

y x

y x

xy

− =− − = −

− = −− −

=− −

−=

55.

2

2

2 2 2

2

2

2

2

)A r rs

A r rs

A r r r rs

A r rs

A r rs

r r

A rs

r

A rs

r

= π( +

= π + π

− π = π − π + π

− π = π

− π π=

π π− π

=π

− π=

π

56.

2

2

2 2 2

2

2

2

2

2 ( )

2 2

2 2 2 2

2 2

2 2

2 2

2

2

2

2

T r rh

T r rh

T r r r rh

T r rh

T r rh

r r

T rh

r

T rh

r

= π +

= π + π

− π = π − π + π

− π = π

− π π=

π π− π

=π

− π=

π

57. 2 1

1 2

2 11 1

1 2

1 12

2

V P

V P

V PV V

V P

PVV

P

=

⋅ = ⋅

=

58. a c

b da c

b bb d

bca

d

=

⋅ = ⋅

=

59.

( ) ( )

fS

H hf

H h S H hH h

HS hS f

HS hS hS f hS

HS f hS

HS f hS

S Sf Sh

HS

=−

− ⋅ = − ⋅−

− =− + = +

= ++

=

+=

60.

( ) ( )

EI

R nrE

I R nr R nrR nr

IR Inr E

IR Inr Inr E Inr

IR E Inr

IR E Inr

I IE Inr

RI

=+

⋅ + = ⋅ ++

+ =+ − = −

= −−

=

−=




101

61. a. 3 22

22 3 22 22

22 3

22 3

3 322

322

3

S L

S L

S L

S L

SL

SL

= −+ = − ++ =+

=

+=

+=

b. 3 21

21 3 21 21

21 3

21 3

3 321

321

3

S L

S L

S L

S L

SL

SL

= −+ = − ++ =+

=

+=

+=

62. 5 190

190 5 190 190

190 5

190 5

5 5190

5190

5

W H

W H

W H

W H

WH

WH

= −+ = − ++ =+

=

+=

+=

63. 5 200

200 5 200 200

200 5

200 5

5 5200

5200

5

W H

W H

W H

W H

WH

WH

= −+ = − ++ =+

=

+=

+=

64. 172

17 17 172

172

2 ( 17) 22

2 34

34 2

AH

AH

AH

AH

H A

A H

= −

− = − −

− = −

− ⋅ − = − ⋅ −

− + == −

65. S = 3L − 22

or 22

311 22 33

113 3

SL

L

+=

+= = =

Tyrone’s foot is 11 inches long.

66. S = 3L − 21

or 21 7 21 28 1

93 3 3 3

SL

+ += = = =

Maria’s foot is 28 1

93 3

= inches long.

67. First, find the length of a man’s foot for size 7.

22

37 22 29

3 3

SL

L

+=

+= =

Then, solve S = 3L − 21 for women with

29:

3L=

293 21 29 21 8

3S

= − = − =

Sue wears a size 8.

68. Let P = 7.27. 0.37 0.23( 1)

7.27 0.37 0.23( 1)

7.27 0.37 0.23 0.23

7.27 0.14 0.23

7.27 0.14 0.14 0.14 0.23

7.13 0.23

31

P x

x

x

x

x

x

x

= + −

= + −

= + −= +

− = − +=

=

The prices are the same when the weight is 31 ounces.




102

69. Let C = 11.50. 1 0.75( 1)

11.50 1 0.75( 1)

11.50 1 0.75 0.75

11.50 0.25 0.75

11.50 0.25 0.25 0.25 0.75

11.25 0.75

11.25 0.75

0.75 0.7515

C h

h

h

h

h

h

h

h

= + −= + −= + −= +

− = − +=

=

=

After 15 hours, the cost is $11.50.

70. Let N = A and solve for x. N = 0.165x + 4.68

A = −0.185x + 5.38

0.165 4.68 0.185 5.38

0.165 4.68 4.68 0.185 5.38 4.68

0.165 0.185 0.7

0.165 0.185 0.185 0.185 0.7

0.35 0.7

2

x x

x x

x x

x x x x

x

x

+ =− +

+ − =− + −

=− ++ =− + +

==

The run production will be the same when

x = 2, or in 1996 + 2 = 1998.

71. a. 0.423(5) 1.392 0.723 hourH = − =

b. 0.68 hour

72. a. 0.423(6) 1.392 1.146 hoursH = − =

b. 1.19 hours

73. a. 0.423(7) 1.392 0.569 hourH = − =

b. 1.61 hours

74. 1980–1985

75. a. 0.423(8) 1.392 1.992 hourH = − =

b. 1.95 hours

c. younger employees

76. Cost is $30 plus $0.15 per mile. C = 30 + 0.15m

77. Let C = 40 and solve for m. 30 0.15

40 30 0.15

40 30 30 30 0.15

10 0.15

10 0.15

0.15 0.152

663

C m

m

m

m

m

m

= += +

− = − +=

=

=

The mileage rate and flat rate are the same

at 2

663

miles.

78. Flat rate: $40 ⋅ 7 = $280

Mileage rate: 30 7 0.15(300)

210 45

$255

C = ⋅ += +=

The mileage rate is cheaper.

79. a.

0.2

1.2

S C M

S C C

S C

= += +=

b. Let C = 8. S = 1.2C S = 1.2(8) S = 9.6 The selling price is $9.60.

80. a.

0.25

0.25 0.25 0.25

0.75

0.75

S C M

S C S

S S C S S

S C

C S

= += +

− = + −==

b. Let S = 80. C = 0.75(80) C = 60 The article cost $60.



83. The solution to the equation is x = 0. If x = 0, then dividing by x (or 0) is undefined.

84. ax b c+ =

85. literal




103

86.

1 2

1 2

1 2

2 1 2 2

2 1

21

( )2

2 2( )

22

2

2

2

hA b b

hA b b

h hA

b bh

Ab b b b

hA

b bh

A b hb

h

= +

⋅ = ⋅ +

= +

− = + −

− =

−=

87. 50 40 0.2( 100)

50 40 0.2 20

50 20 0.2

50 20 20 20 0.2

30 0.2

30 0.2

0.2 0.2150

150

m

m

m

m

m

m

m

m

= + −= + −= +

− = − +=

=

==

88. 7

12 4 37

12 12 1212 4 3

7 3 4

7 7

1

x x

x x

x x

x

x

= +

⋅ = ⋅ + ⋅

= +

==

The solution is 1.

89.

1 8( 2)

3 5 151 8( 2)

15 15 153 5 15

5 3 8( 2)

5 3 8 16

5 5 3 8 16 5

3 8 11

3 8 8 8 11

11 11

11 11

11 111

x x

x x

x x

x x

x x

x x

x x x x

x

x

x

+− =

+⋅ − ⋅ = ⋅

− = +− = +

− − = + −− = +

− − = − +− =−

=− −

= −


90. 10( 2) 6( 1) 18

10 20 6 6 18

10 20 6 24

10 20 20 6 24 20

10 6 4

10 6 6 6 4

4 4

4 4

4 41

x x

x x

x x

x x

x x

x x x x

x

x

x

+ = + ++ = + ++ = +

+ − = + −= +

− = − +=

=

=

The solution is 1.

91. 5( 2) 3( 1) 9

5 10 3 3 9

5 10 3 12

5 10 10 3 12 10

5 3 2

5 3 3 3 2

2 2

2 2

2 21

x x

x x

x x

x x

x x

x x x x

x

x

x

− + = − + −− − = − − −− − = − −

− − + = − − +− = − −

− + = − + −− = −− −

=− −

=

The solution is 1.

92. 4 5 2

4 5 5 2 5

4 7

4 7

4 47

4

x

x

x

x

x

− − =− − + = +

− =−

=− −

= −

The solution is 7

.4

−

93. 3 8 11

3 8 8 11 8

3 3

3 3

3 31

x

x

x

x

x

+ =+ − = −

=

=

=

The solution is 1.




104

94. 2n n+

95. Quotient means divide: a b

c

+

96. a

b c−

97. Product means multiply and sum means add: a(b + c)

98. (a − b)c

99. Difference means subtract and product means multiply: a bc−

2.4 Problem Solving: Integer, General, and Geometry Problems

Problems 2.4 1. (1) Read the problem.

We are asked to find three consecutive odd integers.

(2) Select the unknown. Let n = first odd integer. Then n + 2 = 2nd odd integer. and n + 4 = 3rd odd integer.

(3) Think of a plan. Translate the sentence into an equation. n + (n + 2) + (n + 4) = 129

(4) Use algebra to solve the resulting

equation.

( 2) ( 4) 129

2 4 129

3 6 129

3 6 6 129 6

3 123

41

n n n

n n n

n

n

n

n

+ + + + =+ + + + =

+ =+ − = −

==

Thus, the three consecutive odd integers are 41, 41 + 2 = 43, and 41 + 4 = 45.

(5) Verify the solution. Since 41 + 43 + 45 = 129, our result is correct.

2. (1) Read the problem. We are asked to find a certain number.

(2) Select the unknown. Let n represent the number.

(3) Think of a plan. Translate the sentence into an equation.

3n − 14 = n + 2

(4) Use algebra to solve the resulting

equation.

3 14 2

3 14 14 2 14

3 16

2 16

8

n n

n n

n n

n

n

− = +

− + = + += +

==

The number is 8.

(5) Verify the solution.

Is 3(8) − 14 = 8 + 2? Yes, since

24 − 14 = 10 is true.

3. (1) Read the problem. We are asked to find the number of calories in the cheeseburger and in the fries.

(2) Select the unknown. Let f = number of calories in the fries. Then f + 120 = number of calories in the cheeseburger.

(3) Think of a plan. Translate the problem. f + (f + 120) = 540

(4) Use algebra to solve the equation. ( 120) 540

120 540

2 120 540

2 420

210

f f

f f

f

f

f

+ + =

+ + =

+ ==

=

Thus, the fries have 210 calories and the cheeseburger has 210 + 120 = 330 calories.

(5) Verify the solution. Does 210 + 330 = 540? Yes, the equation is true.




105

4. (1) Read the problem. We are asked to find the measure of an angle.

(2) Select the unknown. Let m = measure of the angle.

90 − m is its complement.

180 − m is its supplement.

(3) Think of a plan. Translate the problem statement into an equation.

180 − m = 4(90 − m) − 45

(4) Use algebra to solve the equation. 180 4(90 ) 45

180 360 4 45

180 315 4

180 180 315 180 4

135 4

4 135 4 4

3 135

45

m m

m m

m m

m m

m m

m m m m

m

m

− = − −− = − −

− = −− − = − −

− = −− + = − +

==

The measure of the angle is 45°.


Is the supplement (180° − 45° = 135°)

45° less than 4 times its complement

(4 ⋅ (90° − 45°) = 180°)? Yes, since

135° = 180° − 45°.

Exercises 2.4

Students should use the RSTUV method to solve.

An outline of each solution is given.

1. Let n = first even integer. Then n + 2 = 2nd even integer, and n + 4 = 3rd even integer.

( 2) ( 4) 138

2 4 138

3 6 138

3 6 6 138 6

3 132

3 132

3 344

n n n

n n n

n

n

n

n

n

+ + + + =+ + + + =

+ =+ − = −

=

=

=

Thus, the three integers are 44, 44 + 2 = 46, and 44 + 4 = 48.

2. Let n = first odd integer. Then n + 2 = 2nd odd integer, and n + 4 = 3rd odd integer.

( 2) ( 4) 135

2 4 135

3 6 135

3 6 6 135 6

3 129

3 129

3 343

n n n

n n n

n

n

n

n

n

+ + + + =+ + + + =

+ =+ − = −

=

=

=

Thus, the three integers are 43, 43 + 2 = 45, and 43 + 4 = 47.

3. Let n = first even integer. Then n + 2 = 2nd even integer, and n + 4 = 3rd even integer.

( 2) ( 4) 24

2 4 24

3 6 24

3 6 6 24 6

3 30

3 30

3 310

n n n

n n n

n

n

n

n

n

+ + + + = −+ + + + = −

+ = −+ − = − −

= −−

=

= −

Thus, the integers are −10, −10 + 2 = −8,

and −10 + 4 = −6.


( 2) ( 4) 27

2 4 27

3 6 27

3 6 6 27 6

3 33

3 33

3 311

n n n

n n n

n

n

n

n

n

+ + + + = −+ + + + = −

+ = −+ − = − −

= −−

=

= −

Thus, the integers are −11, −11 + 2 = −9,

and −11 + 4 = −7.




106

5. Let n = first integer. Then n + 1 = next consecutive integer.

( 1) 25

1 25

2 1 25

2 1 1 25 1

2 26

2 26

2 213

n n

n n

n

n

n

n

n

+ + = −+ + = −

+ = −+ − = − −

= −−

=

= −

Thus, the integers are −13 and

−13 + 1 = −12.

6. Let n = first integer. Then n + 1 = next consecutive integer.

( 1) 9

1 9

2 1 9

2 10

5

n n

n n

n

n

n

+ + =−

+ + =−+ =−

=−=−

Thus, the integers are −5 and −5 + 1 = −4.

7. Let n = first integer, n + 1 = second integer, n + 2 = last integer.

2 2 23

3 2 23

3 2 2 23 2

3 21

3 21

3 37

n n

n

n

n

n

n

+ + =+ =

+ − = −=

=

=

Thus, the integers are 7, 7 + 1 = 8, and 7 + 2 = 9.

8. Let n = Maria’s number, n + 1 = Latasha’s number n + 2 = Kim’s number.

2 2 47

3 2 47

3 2 2 47 2

3 45

15

n n

n

n

n

n

+ + =

+ =+ − = −

==

Thus, Maria’s apartment number is 15, Latasha’s is 15 + 1 = 16, and Kim’s is 15 + 2 = 17.

9. Let n = first locker number, n + 1 = middle locker number, n + 2 = last locker number.

( 2) 2( 1)

2 2 2

2 2 2 2

n n n

n n n

n n

+ + = ++ + = +

+ = +

This is an identity, so the solution is any consecutive integers.

10. Let e = price of English book. Then e + 27 = price of math book.

( 27) 141

27 141

2 27 141

2 27 27 141 27

2 114

57

e e

e e

e

e

e

e

+ + =+ + =

+ =+ − = −

==

Thus, the English book was $57 and the math book was $57 + $27 = $84.

11. Let p = price of cheaper book. Then p + 24 = price of more expensive book.

( 24) 64

24 64

2 24 64

2 24 24 64 24

2 40

2 40

2 220

p p

p p

p

p

p

p

p

+ + =+ + =

+ =+ − = −

=

=

=

Thus, the books cost $20 and $20 + $24 = $44.

12. Let x = points scored by losing team. Then x + 5 = points scored by winning team.

( 5) 179

5 179

2 5 179

2 5 5 179 5

2 174

2 174

2 287

x x

x x

x

x

x

x

x

+ + =+ + =

+ =+ − = −

=

=

=

Thus, the losing team scored 87 points and the winning team scored 87 + 5 = 92 points.




107

13. Let x = points scored by losing team. Then x + 55 = points scored by winning team.

( 55) 133

55 133

2 55 133

2 55 55 133 55

2 78

39

x x

x x

x

x

x

x

+ + =+ + =

+ =+ − = −

==

Thus, the losing team scored 39 points and the winning team scored 39 + 55 = 94 pts.

14. Let m = amount Mida spent.

Then 2

amount Sandra spent.5

m =

2210

55 2

2105 5

7210

52

150; 150 605

m m

m m

m

m

+ =

+ =

=

= ⋅ =

Mida spent $150 and Sandra spent $60.

15. Let x charges on other card.

Then 3

4 charges on one card.8

x + =

34 147

8

34 147

88 3

4 1478 8

114 147

811

1438

8 11 8143

11 8 118 13

104

x x

x x

x x

x

x

x

x

x

+ + =

+ + =

+ + =

+ =

=

⋅ = ⋅

= ⋅=

Thus, the charges on the cards are $104 and

3(104) 4 3 13 4 39 4 $43.

8+ = ⋅ + = + =

16. Let c = amount of Marcus’s bill. Then c + 2 = amount of Liang’s bill, and (c + 2) + 2 = amount of Mourad’s bill.

( 2) [( 2) 2] 261

2 2 2 261

3 6 261

3 255

85

c c c

c c c

c

c

c

+ + + + + =

+ + + + + =+ =

==

Thus, Marcus’s bill was $85, Liang’s bill was $85 + $2 = $87, and Mourad’s bill was $87 + $2 = $89.

17. Let n = first number. Then 3n = second number,

and 3n − 5 = third number. 3 (3 5) 254

3 3 5 254

7 5 254

7 5 5 254 5

7 259

7 259

7 737

n n n

n n n

n

n

n

n

n

+ + − =+ + − =

− =− + = +

=

=

=

Thus, the three numbers are 37, 3(37) = 111,

and 111 − 5 = 106.

18. Let n = first integer. Then n + 1 = 2nd integer and n + 2 = 3rd integer.

( 1) ( 2) 4 17

1 2 4 17

3 3 4 17

3 20 4

20

n n n n

n n n n

n n

n n

n

+ + + + = −

+ + + + = −+ = −

+ ==

Thus, the integers are 20, 20 + 1 = 21, and 20 + 2 = 22.

19. Let n = smaller number. Then 6n = larger number.

6 147

7 147

7 147

7 721

n n

n

n

n

+ ==

=

=

Thus, the integers are 21 and 6(21) = 126.




108

20. Let x = fraction. 5 3( 1)

5 3 3

5 3 3 3 3

2 3

3

2

x x

x x

x x x x

x

x

= += +

− = − +=

=

The fraction is 3

.2

21. Let w = percent accessing from work. Then w + 25 = percent accessing from home.

( 25) 15 66

25 15 66

2 40 66

2 40 40 66 40

2 26

2 26

2 213

25 38

w w

w w

w

w

w

w

w

w

+ + + =+ + + =

+ =+ − = −

=

=

=+ =

Thus 13% access the Internet from work and 38% from home.

22. Let a = Internet market audience in Los Angeles. Then a + 727 = Internet market audience in New York.

( 727) 7955

727 7955

2 727 7955

2 7228

3614

727 4341

a a

a a

a

a

a

a

+ + =+ + =

+ =

==

+ =

Thus, the Internet market audience in Los Angeles is 3614 (thousand) and in New York is 4341 (thousand).

23. Let x = salary per week for men 16−24

years old. Then x + 330 = salary per week for men over 25 years old.

330 722

330 330 722 330

392

x

x

x

+ =+ − = −

=

Men 16−24 years old make $392 per week.

24. Let y = salary per week for women 16−24

years old. Then y + 188 = salary per week for women over 25 years old.

188 542

188 188 542 188

354

y

y

y

+ =+ − = −

=

Women 16−24 years old make $354 per

week.

25. Let a = height of antenna. 1250 1472

1250 1250 1472 1250

222

a

a

a

+ =− + = −

=

The antenna is 222 feet high.

26. 0.10 10

17.50 0.10 10

17.50 10 0.10 10 10

7.50 0.10

7.50 0.10

0.10 0.1075

C m

m

m

m

m

m

= += +

− = + −=

=

=

Thus 75 miles were traveled.

27. 96 32

0 96 32

0 96 96 96 32

96 32

96 32

32 323

v t

t

t

t

t

t

= −= −

− = − −− = −− −

=− −

=

The rocket will reach its highest point at 3 seconds.

28. 96 32

16 96 32

16 96 96 96 32

80 32

80 32

32 322.5

v t

t

t

t

t

t

= −= −

− = − −− = −− −

=− −

=

The rocket’s velocity decreases to 16 feet per second at 2.5 seconds.




109

29. Let r votes received by loser. Then r + 372 = votes received by winner.

( 372) 980

372 980

2 372 980

2 372 372 980 372

2 608

2 608

2 2304

372 676

r r

r r

r

r

r

r

r

r

+ + =+ + =

+ =+ − = −

=

=

=+ =

Thus, the candidates received 304 votes and 676 votes.

30. Let r = cost of Russian program.

Then r − 19 = cost of U.S. program. ( 19) 71

19 71

2 19 71

2 19 19 71 19

2 90

45

19 26

r r

r r

r

r

r

r

r

+ − =

+ − =

− =− + = +

==

− =

The cost of the Russian program is $45 billion and the cost of the U.S. program is $26 billion.

31. Let m = number of minutes after the first 3 minutes.

3.05 0.7 7.95

3.05 3.05 0.7 7.95 3.05

0.7 4.90

0.7 4.90

0.7 0.77

m

m

m

m

m

+ =− + = −

=

=

=

He talked for 7 minutes after the first 3 minutes. The call was 10 minutes long.

32. a. Let h = number of hours after first hour. 1.25 0.75 7.25

1.25 1.25 0.75 7.25 1.25

0.75 6

0.75 6

0.75 0.758

h

h

h

h

h

+ =− + = −

=

=

=

Since she parked 8 hours at 75¢ per hour, she parked 9 hours total.

b. 1.25 0.75 10

1.25 1.25 0.75 10 1.25

0.75 8.75

0.75 8.75

0.75 0.7511.6 or 11

h

h

h

h

h

+ =− + = −

=

=

=

(round down to the nearest whole number) She can park 11 hours at 75¢ per hour plus the 1st hour at $1.25. So, she can park 12 hours for $10 or less.

33. a. Let m = number of miles. 0.95 1.25 28.45

0.95 0.95 1.25 28.45 0.95

1.25 27.50

1.25 27.50

1.25 1.2522

m

m

m

m

m

+ =− + = −

=

=

=

The ride was 22 miles long.

b. Find the cost of a 12-mile ride using the equation from a. 0.95 1.25 0.95 1.25(12)

0.95 15

15.95

m+ = += +=

The limo is cheaper, since $15 < $15.95.

34. Let x = number in the 18−25 category.

Then x + 844,000 = # in the 35-and-older category.

( 844,000) 7,072,000

844,000 7,072,000

2 844,000 7,072,000

2 6,228,000

2 6,228,000

2 23,144,000

844,000 3,958,000

x x

x x

x

x

x

x

x

+ + =

+ + =

+ ==

=

=

+ =

Thus, 3,114,000 are in the 18−25 group and

3,958,000 are in the 35-and-older group.




110

35. Let m = measure of the angle.

90 − m = complement and

180 − m = supplement. 180 2(90 ) 20

180 180 2 20

180 200 2

180 180 200 180 2

20 2

2 20 2 2

20

m m

m m

m m

m m

m m

m m m m

m

− = − +− = − +− = −

− − = − −− = −

− + = − +=




180 − m = supplement.

180 3(90 ) 10

180 270 3 10

180 280 3

100 3

2 100

50

m m

m m

m m

m m

m

m

− = − +− = − +

− = −− = −

==





180 3(90 )

180 270 3

180 180 270 180 3

90 3

3 90 3 3

2 90

2 90

2 245

m m

m m

m m

m m

m m m m

m

m

m

− = −− = −

− − = − −− = −

− + = − +=

=

=





180 4(90 )

180 360 4

180 4

3 180

60

m m

m m

m m

m

m

− = −

− = −− = −

==





180 4(90 ) 54

180 360 4 54

180 306 4

180 180 306 180 4

126 4

4 126 4 4

3 126

3 126

3 342

m m

m m

m m

m m

m m

m m m m

m

m

m

− = − −− = − −− = −

− − = − −− = −

− + = − +=

=

=





180 3(90 ) 41

180 270 3 41

180 229 3

49 3

2 49

24.5

m m

m m

m m

m m

m

m

− = − −− = − −

− = −− = −

==

The measure of the angle is 24.5°.

41. Let A = amount of garbage.

0.5 236

1.5 236

1.5 236

1.5 1.5157.333

A A

A

A

A

+ =

=

=

=

The amount, A, of garbage is about 157 million tons.

42. Let n = number of landfills.

6157 1767

7924

n

n

− =

=

The number n of landfills is 7924.

43. Let g = amount of garbage sent to landfills.

9 131

9 9 131 9

140

g

g

g

− =

− + = +

=

The amount, g, of garbage sent to landfills is 140 million tons.




111

44. Let p = pounds of garbage discarded in 2002.

3.09 0.05

3.14

p

p

= −

=

The amount, p, of garbage discarded in 2002 was 3.14 lb per person per day.

45. Let R = amount of recycled materials.

66.7 72.3

66.7 66.7 72.3 66.7

5.6

R

R

R

+ =

+ − = −

=

The total amount, R, of recycled materials is 5.6 million tons.

46. Let M = the amount of material recovered for recycling.

49.8 55.4

5.6

M

M

+ =

=

The amount, M, of material recovered for recycling is 5.6 million tons.

47. Let g = amount of garbage burned.

33 0.14

33 0.14

0.14 0.14235.714

g

g

g

=

=

≈

The amount, g, of garbage burned was about 236 million tons.

48. Let g = amount of garbage gone to landfills.

131 0.55

238.182

g

g

=

=

The amount, g, of garbage gone to landfills is about 2387 million tons.

49. Let t = amount of yard trimmings.

35.2% 23.1%

35.2% 23.1% 23.1% 23.1%

12.1%

t

t

t

= +

− = + −

=

The portion, p, of yard trimmings in 2003 was 12.1%.

50. Let b = amount of lead.

0.93 2

2.1505

b

b

=

≈

The amount, b, of lead in recycled batteries is about 2.15 billion pounds.

51. Let x = the number of years Diophantus lived.

1youth years

6x =

1grew beard

12x =

1married

7x =

1one-half life span of father's age.

2x =

1 1 1 15 4

6 12 7 214 7 12 42

984 84 84 84

759

8475 75 84 75

984 84 84 84

99

8484 84 9

99 9 84

84

x x x x x

x x x x x

x x

x x x x

x

x

x

+ + + + + =

+ + + + =

+ =

− + = −

=

⋅ = ⋅

=

Diophantus lived 84 years.

52. Answers may vary. Sample answer: Determine what the problem is asking for.



55. RSTUV

56. consecutive

57. complementary

58. supplementary




112


( 2) ( 4) 249

2 4 249

3 6 249

3 6 6 249 6

3 243

3 243

3 381

2 83

4 85

n n n

n n n

n

n

n

n

n

n

n

+ + + + =+ + + + =

+ =+ − = −

=

=

=+ =+ =

The integers are 81, 83, and 85.

60. Let n = number. 4 3 9

4 12

3 12

4

n n

n n

n

n

− = +

= +=

=

The number is 4.

61. Let p = calories in the pizza. Then p + 70 = calories in the shake.

( 70) 530

70 530

2 70 530

2 70 70 530 70

2 460

2 460

2 2230

70 300

p p

p p

p

p

p

p

p

p

+ + =+ + =

+ =+ − = −

=

=

=+ =

The pizza had 230 calories and the shake had 300 calories.

62. Let m = the measure of the angle.



180 3(90 ) 47

180 270 3 47

180 223 3

43 3

2 43

21.5

m m

m m

m m

m m

m

m

− = − −− = − −

− = −− = −

==

The measure of the angle is 21.5°.

63. 55 100

55 100

55 5520

11

T

T

T

=

=

=

64. 88 3240

88 3240

88 88405

11

R

R

R

=

=

=

65. 15 120

15 120

15 158

T

T

T

=

=

=

66. 81 3240

40

T

T

=

=

67. 75 600

75 600

75 758

x

x

x

− = −− −

=− −

=

68. 45 900

20

x

x

− =−

=

69. 0.02 70

0.02 70

0.02 0.023500

P

P

P

− = −− −

=− −

=

70. 0.05 100

2000

R

R

− =−

=

71. 0.04 40

0.04 40

0.04 0.041000

x

x

x

− = −− −

=− −

=

72. 0.03 30

1000

x

x

− =

=




113

2.5 Problem Solving: Motion, Mixture, and Investment Problems

Problems 2.5

1. (1) Read the problem.

We are asked to find the average speed of the bus.

(2) Select the unknown. Let R = average speed.

(3) Think of a plan.

T = 214 − 12 − 24 = 178

178 6000

R T D

R

× =× =

(4) Use algebra to solve the equation. 178 6000

178 6000

178 17833.7

R

R

R

=

=

≈

The average speed of the bus is 33.7 miles per hour.


33.7 178 5998.6

R T D× =× =

2. (1) Read the problem. We are asked to find the number of hours if the car is going 60 mph.

(2) Select the unknown. Let T = number of hours.


R × T = D

car 60 T 60T

bus 40 T + 3 40(T + 3)

60T = 40(T + 3)

(4) Use algebra to solve the equation. 60 40( 3)

60 40 120

60 40 40 40 120

20 120

20 120

20 206

T T

T T

T T T T

T

T

T

= += +

− = − +=

=

=

It takes the car 6 hours to overtake the bus.


car: 60 × 6 = 360

bus: 40 × 9 = 360

3. (1) Read the problem. We are asked to find how many hours to meet if the faster bus travels at 50 mph.

(2) Select the unknown. Let T = hours until they meet.


R × T = D

Supercruiser 40 T 40T

bus 50 T 50T

40T + 50T = 3240

(4) Use algebra to solve the equation. 40 50 3240

90 3240

90 3240

90 9036

T T

T

T

T

+ ==

=

=

Each bus traveled 36 hours before they met.


40 × 36 = 1440

50 × 36 = 1800

1440 + 1800 = 3240




114

4. (1) Read the problem. We are asked to find the number of ounces of the 50% solution that should be added to make a 30% solution.

(2) Select the unknown. Let x = number of ounces of 50% solution to added.


%

×

oz

=

Amt of Pure acid in Final Mix

50% sol. 0.50 x 0.50x

5% sol. 0.05 32 1.60

30% sol. 0.30 x + 32 0.30(x + 32)

0.50x + 1.60 = 0.30(x + 32)

(4) Use algebra to solve the equation. 10 0.50 10 1.60 10 [0.30( 32)]

5 16 3( 32)

5 16 3 96

5 16 16 3 96 16

5 3 80

5 3 3 3 80

2 80

2 80

2 240

x x

x x

x x

x x

x x

x x x x

x

x

x

⋅ + ⋅ = ⋅ ++ = ++ = +

+ − = + −= +

− = − +=

=

=

The photographer must add 40 ounces of the 30% solution.

(5) Verify the solution. 0.50(40) + 1.60 = 21.6 0.30(40 + 32) = 21.6

5. (1) Read the problem. We are asked to find how much is invested if her income is only $400.

(2) Select the unknown. Let s = amount invested in stocks.

Then 6000 − s = amount invested in

bonds.


P × r = I

stocks s 0.05 0.05s

bonds 6000 − s 0.10 0.1(6000 − s)

0.05s + 0.10(6000 − s) = 400

(4) Use algebra to solve the equation. 0.05 0.10(6000 ) 400

0.05 600 0.10 400

600 0.05 400

0.05 200

0.05 200

0.05 0.054000

s s

s s

s

s

s

s

+ − =

+ − =− =

− =−− −

=− −

=

The woman invested $4000 in stocks and $2000 in bonds.

(5) Verify the solution. 5% of $4000 is $200 and 10% of $2000 is $200, so the total interest is $400.

Exercises 2.5

Students should use the RSTUV method to solve.

An outline of each solution is given.

1. Let D = 400 and T = 8.

8 400

50

R T D

R

R

× =

× ==

The bus’ average speed is 50 miles per hour.

2. Let T = 12 − 7 = 5 and D = 260.

5 260

52

R T D

R

R

× =

× ==

The car’s speed is 52 miles per hour.

3. Let D = 246 and T = 120.

120 246

120 246

120 1202.05 2

R T D

R

R

R

× =× =

=

= ≈

The rate of play of the tape is 2 meters per minute.




115

4. a. Let D = 12 and T = 6.

6 12

2

R T D

R

R

× =

× ==

or let D = 1 and T = 6.

6 1

1

6

R T D

R

R

× =× =

=

The rate of output is 2 inches per second

or 1

6 page per second.

b. Let 1

6R = and D = 60.

160

6360

R T D

T

T

× =

× =

=

It would take 360 seconds or

3606 minutes.

60=

5. Let D = 200 and R = 400.

400 200

400 200

400 4001

2

R T D

T

T

T

× =× =

=

=

It takes 1

2 hour to go from Tampa to

Miami.

6. R × T = D

freight 30 T + 1 30(T + 1)

passenger 60 T 60T

30( 1) 60

30 30 60

30 30

1

T T

T T

T

T

+ =+ =

==

It will take the passenger train 1 hour to overtake the freight train.

7. R × T = D

bus 60 T + 2 60(T + 2)

car 90 T 90T

60( 2) 90

60 120 90

60 60 120 90 60

120 30

120 30

30 304

T T

T T

T T T T

T

T

T

+ =+ =

− + = −=

=

=

It will take her 4 hours to catch the bus.

8. R × T = D

train 50 T 50T

car 60 T − 1 60(T − 1)

50 60( 1)

50 60 60

10 60

6

T T

T T

T

T

= −= −

− =−=

Find either the train’s distance or the car’s distance.

50 6 300

R T D× =× =

The towns are 300 miles apart.

9. R × T = D

bicycle 15 T 15T

car 60 12

T − ( )12

60 T +

115 60

2

15 60 30

45 30

2

3

T T

T T

T

T

= −

= −

− =−

=

Find either the bicycle’s distance or the car’s distance.

215 10

3

R T D× =

× =

Their house is 10 miles from the school.




116

10. R × T = D

jet 480 T − 1 480(T − 1)

plane 480 T 480T

480( 1) 480 1632

480 480 480 1632

960 480 1632

960 2112

2.2

T T

T T

T

T

T

− + =

− + =− =

==

They will pass each other in 2.2 hours on 2 hours 12 minutes.

11. R × T = D

car 50 T 50T

another car 55 T 55T

50 55 630

105 630

105 630

105 1056

T T

T

T

T

+ ==

=

=

The cars will meet in 6 hours.

12. R × T = D

crew 70 T 70T

her 65 T–2 55(T–2)

50 55 630

105 630

6

T T

T

T

+ =

==

The cars will meet in 6 hours. 70 65( 2) 275

70 65 130 275

135 130 275

135 130 130 275 130

135 405

135 405

135 1353

T T

T T

T

T

T

T

T

+ − =+ − =

− =− + = +

=

=

=

Find the distance the crew traveled.

70 3 210

R T D× =× =

The first crew traveled 210 kilometers.

13. R × T = D

out 480 T 480T

back 640 7 − T 640(7 − T)

480 640(7 )

480 4480 640

480 640 4480 640 640

1120 4480

1120 4480

1120 11204

T T

T T

T T T T

T

T

T

= −= −

+ = − +=

=

=

Find the distance out or the distance back.

480 4 1920

R T D× =× =

The base is 1920 miles from the target.

14. R × T = D

driving 40 T 40T

walking 5 14

2 T− ( )14

5 2 T−

140 5 2

4

940 5

4

4540 5

445

40 5 5 5445

454

1 45 145

45 4 451

4

T T

T T

T T

T T T T

T

T

T

= − = −

= −

+ = − +

=

⋅ = ⋅

=

Find the distance he drove.

140 10

4

R T D× =

⋅ =

His car broke down 10 miles from his house.




117

15. R × T = D

first seconds 20R 180 20R(180)

last seconds R 60 R(60)

20 (180) (60) 366

3600 60 366

3660 366

3660 366

3660 36600.1

R R

R R

R

R

R

+ =+ =

=

=

=

During the first 180 seconds, the shuttle was traveling at 20(0.1) = 2 feet per second or 24 inches per second. During the last 60 seconds, the shuttle was traveling at 0.1 feet per second or 1.2 inches per second.

16. R × T = D

first seconds 13R 180 13R(180)

last seconds R 60 R(60)

13 (180) (60) 366

2340 60 366

2400 366

2400 366

2400 24000.1525

R R

R R

R

R

R

+ =+ =

=

=

=

The shuttle’s final rate of approach is 0.1525 feet per second or 12(0.1525) = 1.83 inches per second.

17. % × liters = Amount of glycerin in final mixture

40% solution 0.40 x 0.40x

80% solution 0.80 10 8

65% solution 0.65 x + 10 0.65(x + 10)

0.40 8 0.65( 10)

0.40 8 0.65 6.5

0.40 0.65 1.5

0.25 1.5

6

x x

x x

x x

x

x

+ = ++ = +

= −− = −

=

Mix 6 liters of the 40% glycerin solution.




118

18. % × parts = Amount of acetic acid

99.5% acid 0.995 x 0.995x

10% acid 0.10 100 10

28% acid 0.28 x + 100 0.28(x + 100)

0.995 10 0.28( 100)

0.995 10 0.28 28

0.995 0.28 18

0.715 18

0.715 18

0.715 0.71525

x x

x x

x

x

x

x

+ = ++ = +

= +=

=

≈

Add 25 parts of glacial acetic acid.

19. price per pound × pounds = price

copper 0.65 x 0.65x

zinc 0.30 70 − x 0.30(70 − x)

brass 0.45 70 0.45(70)

0.65 0.30(70 ) 0.45(70)

100 0.65 100 0.30(70 ) 100 0.45(70)

65 30(70 ) 45(70)

65 2100 30 3150

35 2100 3150

35 1050

30

70 40

x x

x x

x x

x x

x

x

x

x

+ − =⋅ + ⋅ − = ⋅

+ − =+ − =

+ ===

− =

Mix 30 pounds of copper with 40 pounds of zinc.


Oolong 19 x 19x

Another 4 50 − x 4(50 − x)

Mixture 7 50 7(50)

19 4(50 ) 7(50)

19 200 4 350

15 200 350

15 150

10

x x

x x

x

x

x

+ − =+ − =

+ ===

Mix 10 pounds of Oolong tea.




119


Blue Jamaican 5 x 5x

Regular 2 80 2(80)

Mixture 2.60 x + 80 2.6(x + 80)

5 2(80) 2.6( 80)

5 160 2.6 208

5 2.6 48

2.4 48

20

x x

x x

x x

x

x

+ = ++ = +

= +==

Mix 20 pounds of Blue Jamaican coffee.

22. % × ounces = Amount of alcohol in final mixture

vermouth 0.10 x 0.1x

gin 0.60 20 0.6(20)

martini 0.30 x + 20 0.3(x + 20)

0.1 0.6(20) 0.3( 20)

10 0.1 10 0.6(20) 10 0.3( 20)

6(20) 3( 20)

120 3 60

60 3

2 60

30

x x

x x

x x

x x

x x

x

x

+ = +⋅ + ⋅ = ⋅ +

+ = ++ = +

= − +− = −

=

Mix 30 ounces of vermouth.

23. % × ounces = amount of alcohol in final mixture

40% 0.4 x 0.4x

20% 0.2 64 − x 0.2(64 − x)

30% 0.3 64 0.3(64)

0.4 0.2(64 ) 0.3(64)

10 0.4 10 0.2(64 ) 10 0.3(64)

4 2(64 ) 3(64)

4 128 2 192

2 128 192

2 64

32

64 32

x x

x x

x x

x x

x

x

x

x

+ − =⋅ + ⋅ − = ⋅

+ − =+ − =

+ ===

− =

Mix 32 ounces of each.




120

24. % × quarts = Amount of antifreeze in final solution

50% 0.5 30 − x 0.5(30 − x)

100% 1 x x

70% 0.7 30 0.7(30)

0.5(30 ) 0.7(30)

15 0.5 21

15 0.5 21

0.5 6

12

x x

x x

x

x

x

− + =− + =

+ ===

Drain 12 quarts of 50% solution and replace with 12 quarts of pure antifreeze.

25. % × quarts = Amount of antifreeze in final solution

50% 0.5 30 − x 0.5(30 − x)

0% 0 x 0

30% 0.3 30 0.3(30)

0.5(30 ) 0 0.3(30)

15 0.5 9

0.5 6

12

x

x

x

x

− + =− =− = −

=

Drain 12 quarts of 50% solution and replace with 12 quarts of water.

26. % × ounces = Amount of juice in mixture

Concentrate x 12 12x

Water 0 3(12) = 36 0 ⋅ 36 = 0

10% juice 0.1 12 + 3(12) = 48 0.1(48)

12 0 0.1(48)

12 4.8

0.4

x

x

x

+ ===

The concentrate is 40% juice.




121

27. % × ounces = Amount of juice in mixture

Welch’s x 12 12x

Water 0 3(12) = 36 0 ⋅ 36 = 0

30% juice 0.3 12 + 3(12) = 48 0.3(48)

12 0 0.3(48)

12 14.4

1.2

x

x

x

+ ===

This is impossible, since it would need 120% of juice in the concentrate.

28. P × r = I

5% x 0.05 0.05x

7% 15,000 − x 0.07 0.07(15,000 − x)

0.05 0.07(15,000 ) 870

100 0.05 100 0.07(15,000 ) 100 870

5 7(15,000 ) 87,000

5 105,000 7 87,000

105,000 2 87,000

2 18,000

9000

15,000 6000

x x

x x

x x

x x

x

x

x

x

+ − =⋅ + ⋅ − = ⋅

+ − =+ − =

− =− = −

=− =

$9000 is invested at 5% and $6000 is invested at 7%.

29. P × r = I

6% x 0.06 0.06x

8% 20,000 − x 0.08 0.08(20,000 − x)

0.06 0.08(20,000 ) 1500

100 0.06 100 0.08(20,000 ) 100 1500

6 8(20,000 ) 150,000

6 160,000 8 150,000

160,000 2 150,000

2 10,000

5000

20,000 15,000

x x

x x

x x

x x

x

x

x

x

+ − =⋅ + ⋅ − = ⋅

+ − =+ − =

− =− = −

=− =

$5000 is invested at 6% and $15,000 is invested at 8%.




122

30. P × r = I

7.5% x 0.075 0.075x

6% 25,000 − x 0.06 0.06(25,000 − x)

0.075 0.06(25,000 ) 1620

0.075 1500 0.06 1620

1500 0.015 1620

0.015 120

8000

25,000 17,000

x x

x x

x

x

x

x

+ − =+ − =

+ ===

− =

$8000 is invested at 7.5% and $17,000 is invested at 6%.

31. P × r = I

5% x 0.05 0.05x

7% 18,000 − x 0.07 0.07(18,000 − x)

0.05 0.07(18,000 ) 1100

0.05 1260 0.07 1100

1260 0.02 1100

0.02 160

8000

x x

x x

x

x

x

+ − =+ − =

− =− = −

=

$8000 is invested at 5% in the savings account.

32. P × r = I

8% 20,000 0.08 0.08(20,000)

6% x 0.06 0.06x

0.08(20,000) 0.06 2200

1600 0.06 2200

0.06 600

10,000

x

x

x

x

+ =+ =

==

She must invest $10,000 at 6%.




123

33. P × r = I

5% x 0.05 0.05x

6% 10,000 − x 0.06 0.06(10,000 − x)

0.05 0.06(10,000 ) 60

0.05 600 0.06 60

0.05 660 0.06

0.11 660

6000

10,000 4000

x x

x x

x x

x

x

x

= − += − += −==

− =


34. P × r = I

8% x + 500 0.08 0.08(x + 500)

6% x 0.06 0.06x

0.08( 500) 0.06 600

0.08 40 0.06 600

40 0.14 600

0.14 560

4000

500 4500

x x

x x

x

x

x

x

+ + =+ + =

+ ===

+ =


35. P × r = I

6% x 0.06 0.06x

10% 40,000 − x 0.10 0.10(40,000 − x)

0.06 0.10(40,000 )

0.06 4000 0.1

0.16 4000

25,000

40,000 15,000

x x

x x

x

x

x

= −= −==

− =

$25,000 is invested at 6% and $15,000 is invested at 10%.




124

36. P × r = I

lost 5% x 0.05 0.05x

gained 11% 10,000 − x 0.11 0.11(10,000 − x)

0.05 0.11(10,000 )

0.05 1100 0.11

0.16 1100

6875

10,000 3125

x x

x x

x

x

x

= −= −==

− =

$6875 was invested at 5% and $3125 was invested at 11%.




40. D RT=

41. I Pr=

42. P × r = I

5% 8000 − x 0.05 0.05(8000 − x)

10% x 0.10 0.10x

0.05(8000 ) 0.1 650

400 0.05 0.1 650

400 0.05 650

0.05 250

5000; 8000 3000

x x

x x

x

x

x x

− + =− + =

+ =

== − =


43. % × gallons = Amount of salt in mixture

10% 0.1 x 0.1x

20% 0.2 15 0.2(15)

16% 0.16 x + 15 0.16(x + 15)

0.1 0.2(15) 0.16( 15)

0.1 3 0.16 2.4

0.1 0.16 0.6

0.06 0.6

10

x x

x x

x x

x

x

+ = ++ = +

= −− = −

=

Thus 10 gallons of 10% salt solution should be added.




125

44. R × T = D

1st train 40 T 40T

2nd train 35 T 35T

40 35 300

75 300

4

T T

T

T

+ ===

They meet after 4 hours.

45. R × T = D

bus 50 T + 1 50(T + 1)

car 60 T 60T

50( 1) 60

50 50 60

50 10

5

T T

T T

T

T

+ =+ =

==

The car overtakes the bus in 5 hours.

46. Let D = 250 and T = 5.

5 250

5 250

5 550

R T D

R

R

R

× =× =

=

=

The average speed is 50 miles per hour.

47. 2.9(30) 71 87 71 158+ = + =

48. 5 5

(95 32) (63) 359 9

− = =

49. 1

(20)(10) 10(10) 1002

= =

50. 2(38) 10 76 10 66− = − =

51. 22 1 23 2

73 3 3

+= =

52. 21 1 22 1

73 3 3

+= =

2.6 Formulas and Geometry Applications

Problems 2.6

1. a. Let h = 15.

2.8 28.1

2.8(15) 28.1

42 28.1

70.1

H h= += += +=

Thus a woman with a 15-inch humerus should be about 70.1 inches tall.

b. 2.8 28.1

28.1 2.8

2.828.1

2.8 2.828.1

2.828.1

2.8

H h

H h

hH

Hh

Hh

= +

− =

−=

−=

−=

c. Let H = 61.7.

61.7 28.112

2.8h

−= =

Thus the length of the humerus of a 61.7-inch-tall woman is 12 inches.

2. a. Let C = 22.

1( 4)

61 1

(22 4) (18) 36 6

S C= −

= − = =

The ant’s speed is 3 cm/sec.

b. ( )

( )

14

61

6 6 46

6 4

6 4

6 4

S C

S C

S C

S C

C S

= −

⋅ = ⋅ −

= −

+ =

= +




126

c. Let S = 2. 6(2) 4

12 4

16

C = += +=

Thus, the temperature is 16°C.

3. a. The final cost is obtained by adding the original cost C and the tax T. F = C + T

b. Let C = 10.

10

10

10

F C T

F T

F T

T F

= += +

− == −

4. a. 2(2.4 in.) (1.2 in.) 2.88 in.

A LW=

= ⋅ =

b. 2 2

2(2.4 in.) 2(1.2 in.)

4.8 in. 2.4 in.

7.2 in.

P L W= += += +=

c. Let P = 100 and L = 20. 2 2

100 2(20) 2

100 40 2

100 40 2

60 2

30

P L W

W

W

W

W

W

= += += +

− ===

The width is 30 inches.

5. a.

2

1

21

(30 in.)(15 in.)2

225 in.

A bh=

=

=

b. Let A = 300 and b = 20.

1

21

300 (20)2

300 10

30

A bh

h

h

h

=

=

==

Thus, the height of the sail is 30 feet.

6. a. Let C = 9π.

2

2

9

24.5

C r

r

r

r

= π9π = π

=

=

The radius is 4.5 inches.

b. Find the area of the entire CD. 2 2(4.5) 20.25A r= π = π = π

Find the area of the nonrecorded part. 2 2(0.75) 0.5625A r= π = π = π

The area of the recorded region is

20.25 π − 0.5625π = 19.6875π 2in.

7. a. (2 10) 8 180

10 10 180

10 190

19

x x

x

x

x

− + =− =

==

2(19) − 10 = 28 and 8(19) = 152

The angles measure 28° and 152°.

b. 8 15 4 3

8 4 12

4 12

3

x x

x x

x

x

− = −= +==

8(3) − 15 = 9 and 4(3) − 3 = 9

Both measure 9°.

c. (3 3) ( 17) 90

4 14 90

4 76

19

x x

x

x

x

− + + =+ =

==

3(19) − 3 = 54 and 19 + 17 = 36


8. a. Let S = 2.

22

322 2 24

83 3

SL

+=

+= = =

The foot is 8 inches long.




127

b. Let S = 3.

21

321 3 24

83 3

SL

+=

+= = =

The foot is 8 inches long.

c. Let L = 17. 3 22

3(17) 22

51 22

29

S L= −= −= −=

His right foot needs a size 29.

Exercises 2.6

1. a. Let R = 30 and T = 4. D = RT = 30(4) = 120

b. Let R = 55 and T = 5. D = RT = 55(5) = 275 The car traveled 275 miles.

c.

or

D RT

D RT

T TD D

R RT T

=

=

= =

d. Let D = 180 and T = 3.

18060

3

DR

T= = =

The rate is 60 miles per hour.

2. a. Let D = 240 and T = 4.

24060

4

DR

T= = =


b. Let D = 140 and T = 4.

14035

4

DR

T= = =


c. D

RT

DT R T

TTR D

DT

R

=

⋅ = ⋅

=

=

d. Let R = 35 and D = 105.

1053

35

DT

R= = =

It would take the train 3 hours.

3. a. Let H = 60. 5 190

5(60) 190

300 190

110

W H= −= −= −=

He should weigh 110 pounds.

b. 5 190

190 5

190

5190

5

W H

W H

WH

WH

= −+ =+

=

+=

c. Let W = 200.

200 190 390 178 in. 6 ft

5 5 2H

+= = = =

He should be 78 inches or 6 feet 6 inches tall.

4. a. Let A = 6.

617 17 17 3 14

2 2

AH = − = − = − =

Thus, a 6-year-old should sleep 14 hours.

b. 172

172

2( 17) 22

2 34

34 2

AH

AH

AH

H A

A H

= −

− = −

− − = − ⋅ −

− + == −




128

c. Let H = 11. 34 2

34 2(11) 34 22 12

A H= −

= − = − =

Thus, a child should sleep 11 hours at 12 years of age.

5. a. Let C = 15. 9

3259

(15) 32527 32

59

F C= +

= +

= +=

The corresponding temperature is 59°F.

b. 9

3259

325

5 5 9( 32)

9 9 55 5

( 32) or ( 32)9 9

F C

F C

F C

F C C F

= +

− =

⋅ − = ⋅

− = = −

c. Let F = 50.

5( 32)

95

(50 32)95

(18)910

C F= −

= −

=

=

The corresponding temperature is 10°C.

6. a. P = I − E

b. Let I = 2700 and E = 347.

P = I − E = 2700 − 347 = 2353

The profit is $2353.

c. P I E

P I I I E

P I E

E I P

= −− = − −− = −

= −

d. Let P = 750 and I = 1300.

E = I − P = 1300 − 750 = 550

The expenses are $550.

7. a. Btu

EERw

=

b. Let Btu = 9000 and w = 1000.

Btu 9000EER 9

1000w= = =

The EER is 9.

c. Btu

EER

Btu(EER)

(EER)( ) Btu

Btu (EER)( )

w

w ww

w

w

=

=

==

d. Let w = 2000 and EER = 10. Btu = (EER)(w) = 10(2000) = 20,000 It produces 20,000 Btu per hour.

8. a. C = A − L

b. Let A = 4800 and L = 2300.

C = A − L = 4800 − 2300 = 2500

The capital is $2500.

c. C A L

C A L

L A C

= −− = −

= −

d. Let C = 18,200 and A = 30,000.

L = 30,000 − 18,200 = 11,800

Its liabilities are $11,800.

9. a. S = C + M

b. Let M = 15 and C = 52. S = C + M = 52 + 15 = 67 The selling price should be $67.

c. S C M

S C M

M S C

= +− =

= −

d. Let S = 18.75 and C = 10.50.

M = S − C = 18.75 − 10.50 = 8.25

The markup is $8.25.




129

10. a. TS dN= π

b. Let d = 2 and N = 100.

T (3.14)(2)(100) 628S dN= π = =

The tip speed is 628 meters per second.

c. T

T

T

T

S dN

S dN

d dS

Nd

SN

d

= ππ

=π π

=π

=π

d. Let T 275S = π and d = 2.

T 275137.5

SN

d

π= = =

π π(2)

The number of revolutions per second is 137.5.

11. a. Let W = 10 and L = 20. 2 2

2(20) 2(10)

40 20

60

P L W= += += +=

The perimeter is 60 cm.

b. Let P = 220 and W = 20. 2 2

220 2 2(20)

220 2 40

180 2

90

P L W

L

L

L

L

= += += +==

The length is 90 cm.

12. a. Let W = 15 and L = 30. 2 2

2(30) 2(15) 60 30 90

P L W= +

= + = + =

The perimeter is 90 cm.

b. Let P = 180 and L = 60. 2 2

180 2(60) 2

180 120 2

60 2

30

P L W

W

W

W

W

= += += +==

The width is 30 cm.

13. a. Let r = 10.

C = 2πr = 2(3.14)(10) = 62.8

The circumference is 62.8 inches.

b. 2

2

2 2

2

2

C r

C r

Cr

Cr

= ππ

=π π

=π

=π

c. Let C = 20π.

2010

2

Cr

π= = =

π 2π

The radius is 10 inches.

14. Let C = 26π. 2

26 2

26 2

213

C r

r

r

r

= ππ = ππ π

=2π π

=


15. a. Let L = 4.2 and W = 3.1. A = LW = (4.2)(3.1) = 13.02

The area is 13.02 2m .

b.

or

A LW

A LW

L LA A

W WL L

=

=

= =

c. Let A = 60 and L = 10.

606

10

AW

L= = =

The width is 6 meters.

16. Vertical angles must be equal. 3 5 2 25

3 2 30

30

x x

x x

x

− = += +=

3(30) − 5 = 85 and 2(30) + 25 = 85

The angles each measure 85°.




130


4 10 2

2 10

5

x x

x x

x

x

− = −− = −− =

= −

15 − 4(−5) = 35 and 25 − 2(−5) = 35



3 40 5

2 40

20

x x

x x

x

x

− = −− = − −

= −= −

80 − 3(−20) = 140 and 40 − 5(−20) = 140



3 40 5

2 40

20

x x

x x

x

x

+ = += − +

− = −=

80 + 3(20) = 140 and 40 + 5(20) = 140



4 10 2

2 10

5

x x

x x

x

x

− = −− = −− =

= −

17 − 4(−5) = 37 and 27 − 2(−5) = 37



6 5 30

30

x x

x x

x

− = += +=

6(30) − 5 = 175 and 5(30) + 25 = 175


22. The angles are complementary. 2 (3 10) 90

5 10 90

5 100

20

x x

x

x

x

+ − =− =

==

2(20) = 40 and 3(20) − 10 = 50


23. The angles are complementary. 7 (5 30) 90

12 30 90

12 120

10

x x

x

x

x

+ − =− =

==

7(10) = 70 and 5(10) − 30 = 20


24. The angles are complementary. (2 15) 3 90

5 15 90

5 105

21

x x

x

x

x

− + =− =

==

2(21) − 15 = 27 and 3(21) = 63


25. The angles are complementary. (4 25) 90

5 25 90

5 115

23

x x

x

x

x

− + =− =

==

4(23) − 25 = 67 and 23


26. The angles are supplementary. (3 20) 7 180

10 20 180

10 160

16

x x

x

x

x

+ + =+ =

==

3(16) + 20 = 68 and 7(16) = 112


27. The angles are supplementary. (8 30) (2 10) 180

10 40 180

10 140

14

x x

x

x

x

+ + + =+ =

==

8(14) + 30 = 142 and 2(14) + 10 = 38



8 20 180

8 160

20

x x

x

x

x

+ + + =+ =

==

3(20) + 9 = 69 and 5(20) + 11 = 111





131


8 12 180

8 168

21

x x

x

x

x

+ + + =+ =

==

3(21) + 6 = 69 and 5(21) + 6 = 111


30. Let P = 80 and L = 30. 2 2

80 2(30) 2

80 60 2

20 2

10

P L W

W

W

W

W

= += += +==

It was 10 feet wide.

31. Let P = 1110 and L = 480. 2 2

1110 2(480) 2

1110 960 2

150 2

75

P L W

W

W

W

W

= += += +==

The pool is 75 meters wide.

32. Let L = 120 and P = 346. 2 2

346 2(120) 2

346 240 2

106 2

53

P L W

W

W

W

W

= += += +==

The football field is 53 yards wide.

33. Let C = 14.13. 2

14.13 2(3.14)

2.25

C r

r

r

= π==

The diameter is 2(2.25) = 4.5 inches.

34. Let C = 251.2 2

251.2 2(3.14)

40

C r

r

r

= π==

The diameter is 2(40) = 80 feet.

35. a. 10

10

10

C A

C A

A C

= +− =

= −

b. Let C = 50.

A = C − 10 = 50 − 10 = 40

The American size is 40.

36. a. 30

30

30

C A

C A

A C

= +− =

= −

b. Let C = 42.

A = C − 30 = 42 − 30 = 12

The American size is 12.

37. a. Let t = 1985 − 1975 = 10.

9.74 0.40

9.74 0.40(10)

9.74 4

13.74

N t= += += +=

In 1985, there were 13.74 million recreational boats.

b. 9.74 0.40

9.74 0.40

9.74

0.409.74 5( 9.74)

or 0.40 2

N t

N t

Nt

N Nt t

= +− =−

=

− −= =

c. Let N = 17.74.

9.74 17.74 9.74 820

0.40 0.40 0.40

Nt

− −= = = =

1975 + 20 = 1995 In 1995, the expected number of recreational boats is 17.74 million.

38. a. Let t = 2000 − 1980 = 20.

720 5

720 5(20)

720 100

820

N t= += += +=

You would expect 820 teams in 2000.

b. 720 5

720 5

720

5720

5

N t

N t

Nt

Nt

= +− =−

=

−=




132

c. Let N = 795.

720 795 720 7515

5 5 5

Nt

− −= = = =

1980 + 15 = 1995 In 1995, the expected number of teams is 795.

39. a. 2010 2000 10;

28 420

28(10) 420

280 420

700

x

A x

= − =

= += +

= +=

They would pay $700.

b. 28 420

420 28 420 420

420 28

420 28

28 28420

28

A x

A x

A x

A x

Ax

= +

− = + −

− =−

=

−=

c. 420

281000 420 580

20.71428 28

Ax

−=

−= = ≈

Vehicle insurance will be $1000 in about 21 years (2021).

40. a. 2010 2000 10;

40 786

40(10) 786

1186

x

A x

= − =

= +

= +=

They would pay $1186.

b. 40 786

786 40

786

40

A x

A x

Ax

= +

− =

−=

c. 786

401000 786

5.3540

Ax

−=

−= =

Vehicle insurance will be $1000 in about 5 years (2005).

41. a. 3.3 34

3.3(10) 34 33 34 67

H r= +

= + = + =

He would be 67 in. tall.

b. 3.3 34

34 3.3

34 3.3

3.3 3.334

3.3

H r

H r

H r

Hr

= +

− =

−=

−=

c. 34

3.370.3 34 36.3

113.3 3.3

Hr

−=

−= = =

His radius is 11 in. long.

42. a. 3.3 32 3.3(10) 32 65H r= + = + =

She would be 65 in. tall.

b. 3.3 32

32 3.3

32

3.3

H r

H r

Hr

= +

− =

−=

c. 32 61.7 32

93.3 3.3

Hr

− −= = =

Her radius is 9 in. long.

43. a. 2.9 62

2.9(30) 62

87 62

149

H t= +

= +

= +=

She would be 149 cm tall.

b. 2.9 62

62 2.9

62 2.9

2.9 2.962

2.9

H t

H t

H t

Ht

= +

− =

−=

−=

c. 62

2.9151.9 62 89.9

312.9 2.9

Ht

−=

−= = =

Her tibia is 31 cm long.




133

44. The room is 320

408

= square yards or

40 ⋅ 9 = 360 square feet.

Let A = 360 and L = 20.

360 20

18

A LW

W

W

===

The room is 18 feet wide.

45. Let A = 5400 and W = 60.

5400 (60)

90

A LW

L

L

===

The sod area can be 90 feet long.

46. Let L = 70 and P = 240. 2 2

240 2 2(70)

240 2 140

100 2

50

P W L

W

W

W

W

= += += +==

The yard is 50 feet wide.




50. Answers may vary. Sample answer: Compare volumes since two hamburgers could have the same area or the same circumference and still have different volumes.

51. Answers may vary. Sample answer: C = 2πr

Let 4

22

r = = and 6

32

r = = and compare.

2 (3) 31.5

(2) 2

π= =

2π or 150%

Using this method, the Monster burger is 50% bigger than the Lite burger.

52. LW

53. 2L + 2W

54. 1

2bh

55. 2

rπ

56. 2 rπ

57. equal


5 15 90

5 105

21

x x

x

x

x

− + =− =

==

3(21) − 15 = 48 and 2(21) = 42



5 25 90

5 115

23

x x

x

x

x

− + =− =

==

3(23) − 25 = 44 and 2(23) = 46


60. The angles are supplementary. 6 (4 20) 180

10 20 180

10 160

16

x x

x

x

x

+ + =+ =

==

6(16) = 96 and 4(16) + 20 = 84



10 40 180

10 140

14

x x

x

x

x

+ + + =+ =

==

7(14) + 30 = 128 and 3(14) + 10 = 52



8 20 180

8 160

20

x x

x

x

x

+ + + =+ =

==

4(20) + 9 = 89 and 4(20) + 11 = 91





134


8 20 180

8 160

20

x x

x

x

x

+ + + =+ =

==

2(20) + 6 = 46 and 6(20) + 14 = 134


64. a. Let r = 10. 2

2(10) 100 100(3.14) 314

A r=π

=π = π= =

The area is 100π 2in. or 314 2in. .

2

2 (10) 20

C r= π= π = π=20(3.14)= 62.8

The circumference is 20π in. or 62.8 in.

b. Let C = 40π. 2

40 2

40 2

220

C r

r

r

r

= ππ = ππ π

=2π π

=


65. a. Let h = 20. 2.75 71.48

2.75(20) 71.48

126.48

H h= += +=

The woman is 126.48 centimeters.

b. 2.75 71.48

71.48 2.75

71.48

2.7571.48

2.75

H h

H h

Hh

Hh

= +− =−

=

−=

c. Let H = 140.23.

71.48 140.23 71.4825

2.75 2.75

Hh

− −= = =

The woman’s humerus is 25 centimeters.

61. a. T = C + t

b. Let T = 8.48 and C = 8.

8.48 8

0.48

T C t

t

t

= += +=

The tax is $0.48.

67. a. Let b = 10 and h = 15.

1

21

(10)(15)275

A bh=

=

=

The area is 75 2in. .

b. 1

22

2 2 or

A bh

A bh

A Ab b

h h

=

=

= =

c. Let A = 18 and h = 9.

2 2(18)4

9

Ab

h= = =

The base is 4 inches.

68. a. Let t = 1985 − 1975 = 10.

15 0.60

15 0.60(10) 15 6 21

N t= +

= + = + =

There were 21 thousand theaters in 1985.

b. 15 0.60

15 0.60

15

0.6015 5( 15)

or 0.60 3

N t

N t

Nt

N Nt t

= +− =−

=

− −= =

c. Let N = 27.

27 15

0.6020

t

t

−=

=

1975 + 20 = 1995 In 1995, the number of theaters totaled 27,000.




135

69. a. Let F = 41.

5 160

9 95 160

(41)9 9205 160

9 945

95

C F= −

= −

= −

=

=

The temperature is 5°C.

b. 5 160

9 99 9 5 9 160

5 5 9 5 99

325

9 932 or 32

5 5

C F

C F

C F

C F F C

= −

⋅ = ⋅ − ⋅

= −

+ = = +

c. Let C = 20. 9

3259

(20) 32536 32

68

F C= +

= +

= +=

The temperature is 68°F.

70. 3 2 2( 2)

3 2 2 4

3 2 2

2

x x

x x

x x

x

− = −

− = −

= −=−

71. 2 1 3

2 1 1 3 1

2 4

2 4

4

x x

x x

x x

x x x x

x

− = +− + = + +

= +− = − +

=

72. 3 2 2( 1)

3 2 2 2

3 2

0

x x

x x

x x

x

− = −

− = −

==

73. 4( 1) 3 7

4 4 3 7

4 4 4 3 7 4

4 3 3

4 3 3 3 3

3

x x

x x

x x

x x

x x x x

x

+ = ++ = +

+ − = + −= +

− = − +=

74.

3

4 6 63

12 12 124 6 6

3 2 2( 3)

2 6

2 2 2 6

3 6

3 6

3 32

x x x

x x x

x x x

x x

x x x x

x

x

x

− −+ =

− − ⋅ + ⋅ = ⋅

− + = −− = −

− − = − −− = −− −

=− −

=

75. 13 2

6 6 6 13 22 3 6

6

6

x x

x x

x x

x

x

− =

⋅ − ⋅ = ⋅

− =− =

= −

2.7 Properties of Inequalities

Problems 2.7 1. a. Since 5 is to the right of 3, 5 > 3.

b. Since −1 is to the right of −4, −1 > −4.

c. Since −5 is to the left of −4, −5 < −4.

2. a. For x ≤ −2, any number less than or

equal to −2 is a solution.




136

b. For x > −3, any number greater than −3

is a solution.

3. a. 4 3 3( 2)

4 3 3 6

4 3 3 3 6 3

4 3 3

4 3 3 3 3

3

x x

x x

x x

x x

x x x x

x

− < −− < −

− + < − +< −

− < − −< −

Any number less than −3 is a solution.

b. 3( 2) 2 5

3 6 2 5

3 6 6 2 5 6

3 2 1

3 2 2 2 1

1

x x

x x

x x

x x

x x x x

x

+ ≥ ++ ≥ +

+ − ≥ + −≥ −

− ≥ − −≥ −

Any number greater than or equal to −1

is a solution.

4. a. 4 3 2 5

4 3 3 2 5 3

4 2 2

4 2 2 2 2

2 2

2 2

2 21

x x

x x

x x

x x x x

x

x

x

+ ≤ ++ − ≤ + −

≤ +− ≤ − +

≤

≤

≤

Any number less than or equal to 1 is a solution.

b. 5( 1) 3 1

5 5 3 1

5 5 5 3 1 5

5 3 6

5 3 3 3 6

2 6

2 6

2 23

x x

x x

x x

x x

x x x x

x

x

x

− > +− > +

− + > + +> +

− > − +>

>

>

Any number greater than 3 is a solution.

5. a. 4 20

4 20

4 45

x

x

x

− <−

>− −

> −

Any number greater than −5 is a

solution.

b. 23

3 3 23

6

x

x

x

−>

− − < − ⋅

< −

Any number less than −6 is a solution.

c. 2( 1) 4 1

2 2 4 1

2 2 2 4 1 2

2 4 3

2 4 4 4 3

2 3

2 3

2 23

2

x x

x x

x x

x x

x x x x

x

x

x

− ≤ +− ≤ +

− + ≤ + +≤ +

− ≤ − +− ≤−

≥− −

≥ −

Any number greater than or equal to

3

2− is a solution.




137

6.

8

3 4 48

12 12 123 4 4

4 3 3( 8)

3 24

3 3 3 24

4 24

4 24

4 46

x x x

x x x

x x x

x x

x x x x

x

x

x

− −+ <

− − ⋅ + ⋅ < ⋅

− + < −− < −

− − < − −− < −− −

>− −

>


7. a. 2 < x and x < 4 can be written as 2 < x < 4. The solution consists of all numbers between 2 and 4.

b. 3

1 3 1 ( )

3

x

x

x

≥ −− ⋅ ≤ − ⋅ −

− ≤

Now −3 ≤ x and x ≤ −1 can be written as

−3 ≤ x ≤ −1. The solution consists of all

numbers between −3 and −1, inclusive.

c. 2 6

2 2 6 2

4

x

x

x

+ ≤+ − ≤ −

≤

3 6

3 6

3 32

x

x

x

− <−

>− −

> −

Rearranging, we have −2 < x ≤ 4. The

solution consists of all numbers

between −2 and 4 and the number 4.

8. P = 20 − 2x < 10

20 2 10

20 20 2 10 20

2 10

2 10

2 25

x

x

x

x

x

− <− − < −

− < −− −

>− −

>

This means that more than 5 years after 1997, or in 2003, the percent of smokers in this group is less than 10%.

Exercises 2.7

1. Since 8 is to the left of 9, 8 < 9.

2. Since −8 is to the right of −9, −8 > −9.


4. Since 7 is to the right of 3, 7 > 3.

5. Since 1

4 is to the left of

1,

3 1 1

.4 3

<

6. Since 1

5 is to the left of

1,

2 1 1

.5 2

<

7. Since 2

3− is to the right of −1,

21.

3− > −

8. Since 1

5− is to the right of −1,

11.

5− > −

9. Since 1

34

− is to the left of −3, 1

3 3.4

− < −

10. Since 1

45

− is to the left of −4, 1

4 4.5

− < −

11. 2 6 8

2 6 6 8 6

2 2

2 2

2 21

x

x

x

x

x

+ ≤+ − ≤ −

≤

≤

≤


12. 4 5 3

4 8

2

y

y

y

− ≤

≤

≤





138

13. 3 4 10

3 4 4 10 4

3 6

3 6

3 32

y

y

y

y

y

− − ≥ −− − + ≥ − +

− ≥ −− −

≤− −

≤


14. 4 2 6

4 8

2

z

z

z

− − ≥

− ≥

≤−

Any number less than or equal to −2 is a

solution.

15. 5 1 14

5 1 1 14 1

5 15

5 15

5 53

x

x

x

x

x

− + < −− + − < − −

− < −−

>− −

>


16. 3 1 8

3 9

3

x

x

x

− + <−

− <−>


17. 3 4 10

3 4 4 10 4

3 6

3 6

2 6

2 6

2 23

a a

a a

a a

a a a a

a

a

a

+ ≤ ++ − ≤ + −

≤ +− ≤ − +

≤

≤

≤


18. 4 4 7

4 3

3 3

1

b b

b b

b

b

+ ≤ +

≤ +≤

≤


19. 5 12 6 8

5 12 12 6 8 12

5 6 4

5 6 6 6 4

4

4

z z

z z

z z

z z z z

z

z

− ≥ −− + ≥ − +

≥ +− ≥ − +

− ≥≤ −


solution.

20. 5 7 7 19

5 7 12

2 12

6

z z

z z

z

z

+ ≥ +

≥ +− ≥

≤−


solution.

21. 10 3 7 6

10 10 3 7 10 6

3 3 6

3 6 3 6 6

3 3

3 3

3 31

x x

x x

x x

x x x x

x

x

x

− ≤ −− − ≤ − −

− ≤ − −− + ≤ − − +

≤ −−

≤

≤ −


solution.




139

22. 8 4 12 6

4 20 6

10 20

2

y y

y y

y

y

− ≤− +

− ≤− +

− ≤−≥

Any number greater than or equal to 2 is a solution.

23. 5( 2) 3( 3) 1

5 10 3 9 1

5 3

5 3 3 3

2 0

2 0

2 20

x x

x x

x x

x x x x

x

x

x

+ < + +

+ < + +<

− < −<

<

<

Any number less than 0 is a solution.

24. 5(4 3 ) 7(3 4 ) 12

20 15 21 28 12

15 13 28

13 13

1

x x

x x

x x

x

x

− < − +

− < − +− < −

<<


25.

1 42 2

4 51 4

20( 2 ) 20 20(2 ) 204 5

40 5 40 16

40 40 11

80 11

80 11

80 8011

80

x x

x x

x x

x x

x

x

x

− + ≥ +

− + ≥ +

− + ≥ +

− ≥ +

− ≥−

≤− −

≤−

Any number less than or equal to 11

80− is a

solution.

26.

1 26 2

7 71 2

7 6 7 7 2 77 7

42 1 14 2

42 1 1 14 2 1

42 14 3

42 14 14 14 3

28 3

28 3

28 283

28

x x

x x

x x

x x

x x

x x x x

x

x

x

+ ≥ −

⋅ + ⋅ ≥ ⋅ − ⋅

+ ≥ −+ − ≥ − −

≥ −− ≥ − −

≥ −−

≥

≥ −

Any number greater than or equal to 3

28−

is a solution.

27. 15 4

20 20 20 15 4

4 5 20

20

20

x x

x x

x x

x

x

− ≤

⋅ − ⋅ ≤ ⋅

− ≤− ≤

≥ −

Any number greater than or equal to −20 is

a solution.

28. 13 2

6 6 6 13 22 3 6

6

6

x x

x x

x x

x

x

− ≤

⋅ − ⋅ ≤ ⋅

− ≤− ≤

≥ −

Any number greater than or equal to −6 is a

solution.




140

29.

7 2 1 3

6 2 47 2 1 3

12 12 126 2 4

2(7 2) 6 3 3

14 4 6 9

14 14 10 9 14

10 5

10 5

5 52

2

xx

xx

x x

x x

x x x x

x

x

x

x

++ ≥

+⋅ + ⋅ ≥ ⋅

+ + ≥ ⋅+ + ≥

− + ≥ −≥ −

−≤

− −− ≤

≥ −


solution.

30.

8 23 1 5

6 3 28 23 1 5

6 6 66 3 28 23 2 15

8 8 21 15 8

21 7

21 7

7 73

3

xx

xx

x x

x x x x

x

x

x

x

−+ ≥

−⋅ + ⋅ ≥ ⋅

− + ≥− − ≥ −

− ≥−

≥

− ≥≤ −


solution.

31. x < 3 and −x < −2 (or x > 2) can be rewritten

as 2 < x < 3. The solution consists of all numbers between 2 and 3.

32. −x < 5 (or x > −5) and x < 2 can be rewritten

as −5 < x < 2. The solution consists of all

numbers between −5 and 2.

33. 1 4

1 1 4 1

3

x

x

x

+ <+ − < −

<

and 1

1 ( ) 1 ( 1)

1

x

x

x

− < −− ⋅ − > − ⋅ −

>

Now x < 3 and x > 1 can be rewritten as 1 < x < 3. The solution consists of all numbers between 1 and 3.

34. 2 1

2 2 1 2

3

x

x

x

− <− + < +

<

and 2

1( ) 1(2)

2

x

x

x

− <− − > −

> −

Now x < 3 and x > −2 can be rewritten as

−2 < x < 3. The solution consists of all


35. 2 3

2 2 3 2

5

x

x

x

− <− + < +

<

and 2

1 2 1 ( )

2

x

x

x

> −− ⋅ < − ⋅ −

− <

Now x < 5 and −2 < x can be rewritten as

−2 < x < 5. The solution consists of all the


36. 3 1

3 3 1 3

4

x

x

x

− <− + < +

<

and 1

1 1 1 ( )

1

x

x

x

> −− ⋅ < − ⋅ −

− <


−1 < x < 4. The solution consists of all the


37. 2 3

1

x

x

+ <<

and 4 1

5

x

x

− < +− <







141

38. 4 5

1

x

x

+ <<

and 1 2

3

x

x

− < +− <




39. 1 2

3

x

x

− >>

and 7 12

5

x

x

+ <<

Now x > 3 and x < 5 can be rewritten as 3 < x < 5. The solution consists of all numbers between 3 and 5.

40. 2 1

3

x

x

− >>

and 5

5

1 15

x

x

x

− > −− −

<− −

<

Now x > 3 and x < 5 can be rewritten as 3 < x < 5. The solution consists of all numbers between 3 and 5.

41. 20°F < t < 40°F

42. h ≤ 29,029 ft

43. $12,000 < s < $13,000

44. 18 < m < 22

45. 2 ≤ e ≤ 7

46. a > $150 billion

47. $3.50 < c < $4.00

48. r < 19,000 mi

49. a < 41 ft

50. d ≤ 370

51. N = 720 + 5t > 800 720 5 800

720 720 5 800 720

5 80

5 80

5 516

t

t

t

t

t

+ >− + > −

>

>

>

1980 + 16 = 1996 After 1996, you would expect that the NCAA would have more than 800 teams.

52. D = −10t + 260 < 200

10 260 200

10 260 260 200 260

10 60

10 60

10 106

t

t

t

t

t

− + <− + − < −

− < −− −

>− −

>

1999 + 6 = 2005 After 2005, you would expect the number of deaths per 100,000 to be less than 200.

53. C = 127 + 17t > 300 127 17 300

127 127 17 300 127

17 173

17 173

17 1710.2 or 11

t

t

t

t

t

+ >− + > −

>

>

>

1980 + 11 = 1991 After 1991, you would expect the average hospital room rate to surpass the $300 per day mark.

54. C = 45 + 2t > 60 45 2 60

45 45 2 60 45

2 15

2 15

2 27.5 or 8

t

t

t

t

t

+ >− + > −

>

>

>

1985 + 8 = 1993 In 1993, you would expect the per capita consumption of poultry products to exceed 60 pounds per year.

55. J = 5 ft = 60 in.

56. B > F




142

57. F = S − 3

58. F > J

59. S = 6 ft 5 in. = 77 in.

60. B > F and F > J can be rewritten as B > F > J.

61. F = S − 3 = 77 − 3 = 74

74 in.

or 6 ft 2 in.

B F

B

B

>>>


63. You have to change the direction of the inequality when multiplying or dividing by a negative number.

64. Answers may vary. Sample answer: A number cannot simultaneously be less than

−5 and greater than 2.


66. x a y a+ < +

67. x a y a− < −

68. ax ay<

69. ax ay>

70. x y

a a<

71. x y

a a>

72. 2 6

2 2 6 2

4

x

x

x

+ ≤+ − ≤ −

≤

and 3 9

3 9

3 33

x

x

x

− ≤−

≥− −

≥ −

Now x ≤ 4 and x ≥ −3 can be rewritten as

−3 ≤ x ≤ 4. The solution consists of all

numbers between −3 and 4, inclusive.

73. 3

1 3 1 ( )

3

x

x

x

≥ −− ⋅ ≤ − ⋅ −

− ≤

and x ≤ −1

Now −3 ≤ x and x ≤ −1 can be rewritten as

−3 ≤ x ≤ −1. The solution consists of all

numbers between −3 and −1, inclusive.

74. 2 < x and x < 4 can be rewritten as 2 < x < 4. The solution consists of all numbers between 2 and 4.

75.

4

3 4 44

12 12 123 4 4

4 3 3( 4)

3 12

3 3 3 12

4 12

4 12

4 43

x x x

x x x

x x x

x x

x x x x

x

x

x

− −+ <

− − ⋅ + ⋅ < ⋅

− + < −− < −

− − < − −− < −− −

>− −

>


76. 4 5 11

4 5 5 11 5

4 6

4 6

3 6

3 6

3 32

x x

x x

x x

x x x x

x

x

x

+ < ++ − < + −

< +− < − +

<

<

<





143

77. 3( 1) 3

3 3 3

3 3 3 3 3

3 6

3 6

2 6

2 6

2 23

x x

x x

x x

x x

x x x x

x

x

x

− > +− > +

− + > + +> +

− > − +>

>

>


78. 4 7 3( 2)

4 7 3 6

4 7 7 3 6 7

4 3 1

4 3 3 3 1

1

x x

x x

x x

x x

x x x x

x

− < −− < −

− + < − +< +

− < − +<


79. 3( 1) 2 5

3 3 2 5

3 3 3 2 5 3

3 2 2

3 2 2 2 2

2

x x

x x

x x

x x

x x x x

x

+ ≥ ++ ≥ +

+ − ≥ + −≥ +

− ≥ − +≥


80. x ≥ −2


solution.

81. x < 1 Any number less than 1 is a solution.

82. Since 5 is to the left of 7, 5 < 7.

83. Since −2 is to the left of −1, −2 < −1.


85. Since 1

3 is to the right of −3,

13.

3> −

86. F = 181.5 + 0.8t > 189.5 181.5 0.8 189.5

0.8 8

10

t

t

t

+ >>

>

2000 + 10 = 2010 After 2010, you would expect the daily consumption of grams of fat to exceed 189.5.

87.

88.

89.

90.

91.

92.

93.

94.

95.




144

Collaborative Learning

Answers may vary. Sample answers:

1. Let t = 2001 − 1985 = 16.

45 2

45 2(16)

45 32

77

C t= += += +=

The per capita consumption of poultry in 2001 is estimated to be 77 pounds per year. The table value for 2001 is approximately 83 pounds per year.


3. C is highest in 2001 and lowest in 1980.



Review Exercises

1. a. If x = 5, then 7 = 14 − x becomes

7 = 14 − 5, which is a false statement.


b. If x = 4, then 13 = 17 − x becomes

13 = 17 − 4 which is a true statement.

Thus 4 is a solution of the equation.

c. If x = −2, then 8 = 6 − x becomes

8 = 6 − (−2), which is a true statement.

Thus −2 is a solution of the equation.

2. a. 1 1

3 31 1 1 1

3 3 3 32

3

x

x

x

− =

− + = +

=

The solution is 2

.3

b. 5 2

7 75 5 2 5

7 7 7 77

71

x

x

x

x

− =

− + = +

=

=

The solution is 1.

c. 5 1

9 95 5 1 5

9 9 9 96

92

3

x

x

x

x

− =

− + = +

=

=

The solution is 2

.3

3. a. 5 2 5

3 49 9 9

3 5

9 93 3 5 3

9 9 9 92

9

x x

x

x

x

− + + − =

+ =

+ − = −

=

The solution is 2

.9

b. 4 2 6

2 37 7 7

2 6

7 72 2 6 2

7 7 7 74

7

x x

x

x

x

− + + − =

+ =

+ − = −

=

The solution is 4

.7




145

c. 5 1 5

4 56 6 6

4 5

6 64 4 5 4

6 6 6 61

6

x x

x

x

x

− + + − =

+ =

+ − = −

=

The solution is 1

.6

4. a. 3 4( 1) 2 3

3 4 4 2 3

3 2

3 2 2 2

5 or 5

x x

x x

x

x

x x

= − + −= − + −= −

+ = − += =

The solution is 5.

b. 4 5( 1) 9 4

4 5 5 9 4

4 4

4 4 4 4

0 or 0

x x

x x

x

x

x x

= − + −= − + −= +

− = + −= =

The solution is 0.

c. 5 6( 1) 8 5

5 6 6 8 5

5 2

5 2 2 2

3 or 3

x x

x x

x

x

x x

= − + −= − + −= +

− = + −= =

The solution is 3.

5. a. 6 3( 1) 2 3

6 3 3 2 3

9 3 2 3

9 3 3 2 3 3

9 2

x x

x x

x x

x x x x

+ + = ++ + = +

+ = ++ − = + −

=


b.

2 4( 1) 7 4

2 4 4 7 4

6 4 7 4

6 6 4 7 6 4

4 1 4

4 4 1 4 4

8 1

8 1

8 81

8

x x

x x

x x

x x

x x

x x x x

x

x

x

− + − = − −− + − = − −

− + = − −− + + = − + −

= − −+ = − − +

= −−

=

= −

The solution is 1

.8

−

c. 1 2( 1) 3 2

1 2 2 3 2

3 2 3 2

3 2 2 3 2 2

3 3

x x

x x

x x

x x x x

− − + = −− − − = −

− − = −− − + = − +

− =


6. a. 5 2( 1) 2 7

5 2 2 2 7

2 7 2 7

x x

x x

x x

+ + = ++ + = +

+ = +

Since both sides are identical, the equation is an identity. The solution is all real numbers.

b. 2 3( 1) 5 3

2 3 3 5 3

5 3 5 3

x x

x x

x x

− + − = − +− + − = − +

− + = − +


c. 3 4( 1) 1 4

3 4 4 1 4

1 4 1 4

x x

x x

x x

− − − = −− − + = −

− = −





146

7. a. 1

351

5 5 ( 3)5

15

x

x

x

= −

⋅ = ⋅ −

= −


b. 1

271

7 7 ( 2)7

14

x

x

x

= −

⋅ = ⋅ −

= −


c. 5 10

5 10

5 52

x

x

x

= −−

=

= −


8. a. 3

94

4 3 4( 9)

3 4 3

12

x

x

x

− = −

− ⋅ − = − ⋅ −

=

The solution is 12.

b. 3

95

5 3 5( 9)

3 5 3

15

x

x

x

− = −

− ⋅ − = − ⋅ −

=

The solution is 15.

c. 2

63

3 2 3( 6)

2 3 2

9

x

x

x

− = −

− ⋅ − = − ⋅ −

=

The solution is 9.

9. a. 2

53 4

212 12 12 5

3 44 6 60

10 60

10 60

10 106

x x

x x

x x

x

x

x

+ =

⋅ + ⋅ = ⋅

+ ==

=

=

The solution is 6.

b. 3

64 2

34 4 4 6

4 26 24

7 24

7 24

7 7

x x

x x

x x

x

x

+ =

⋅ + ⋅ = ⋅

+ ==

=

The solution is 24

.7

c. 3

105 10

310 10 10 10

5 102 3 100

5 100

5 100

5 520

x x

x x

x x

x

x

x

+ =

⋅ + ⋅ = ⋅

+ ==

=

=

The solution is 20.

10. a. 13 4

12 12 12 13 4

4 3 12

12

x x

x x

x x

x

− =

⋅ − ⋅ = ⋅

− ==

The solution is 12.




147

b. 102 7

14 14 14 102 7

7 2 140

5 140

5 140

5 528

x x

x x

x x

x

x

x

− =

⋅ − ⋅ = ⋅

− ==

=

=

The solution is 28.

c. 24 5

20 20 20 24 5

5 4 40

40

x x

x x

x x

x

− =

⋅ − ⋅ = ⋅

− ==

The solution is 40.

11. a. 1 1

14 6

1 112 12 12 1

4 63( 1) 2( 1) 12

3 3 2 2 12

5 12

5 5 12 5

17

x x

x x

x x

x x

x

x

x

− +− =

− +⋅ − ⋅ = ⋅

− − + =− − − =

− =− + = +

=

The solution is 17.

b. 1 1

06 8

1 124 24 24 0

6 84( 1) 3( 1) 0

4 4 3 3 0

7 0

7 7 0 7

7

x x

x x

x x

x x

x

x

x

− +− =

− +⋅ − ⋅ = ⋅

− − + =− − − =

− =− + = +

=

The solution is 7.

c. 1 1

08 10

1 140 40 40 0

8 105( 1) 4( 1) 0

5 5 4 4 0

9 0

9 9 0 9

9

x x

x x

x x

x x

x

x

x

− +− =

− +⋅ − ⋅ = ⋅

− − + =− − − =

− =− + = +

=

The solution is 9.

12. a. What percent of 30 is 6? 30 6

30 6

30 301 20

5 100

x

x

x

⋅ =⋅

=

= =

Thus, 6 is 20% of 30.

b. What percent of 40 is 4? 40 4

40 4

40 401 10

10 100

x

x

x

⋅ =⋅

=

= =

Thus, 4 is 10% of 40.

c. What percent of 50 is 10? 50 10

50 10

50 501 20

5 100

x

x

x

⋅ =⋅

=

= =

Thus, 10 is 20% of 50.

13. a. 20 is 40% of what number?

4020

1002

205

5 5 220

2 2 550

n

n

n

n

= ⋅

= ⋅

⋅ = ⋅ ⋅

=

Thus 20 is 40% of 50.




148

b. 30 is 90% of what number?

9030

1009

3010

10 10 930

9 9 10100

31

333

n

n

n

n

n

= ⋅

= ⋅

⋅ = ⋅ ⋅

=

=

Thus, 30 is 90% of 1

33 .3

c. 25 is 75% of what number?

7525

1003

254

4 4 325

3 3 4100

31

333

n

n

n

n

n

= ⋅

= ⋅

⋅ = ⋅ ⋅

=

=

Thus, 25 is 75% of 1

33 .3

14. a.

1 19( 4)

5 4 201 19( 4)

20 20 205 4 20

4 5 19( 4)

4 5 19 76

4 4 5 19 76 4

5 19 72

5 19 19 19 72

24 72

24 72

24 243

x x

x x

x x

x x

x x

x x

x x x x

x

x

x

+− =

+⋅ − ⋅ = ⋅

− = +− = +

− − = + −− = +

− − = − +− =−

=− −

= −


b.

1 6( 5)

5 4 51 6( 5)

20 20 205 4 5

4 5 24( 5)

4 5 24 120

5 24 116

29 116

29 116

29 294

x x

x x

x x

x x

x x

x

x

x

+− =

+⋅ − ⋅ = ⋅

− = +

− = +− = +

− =

−=

− −=−


c.

1 29( 6)

5 4 201 29( 6)

20 20 205 4 20

4 5 29( 6)

4 5 29 174

5 29 170

34 170

34 170

34 345

x x

x x

x x

x x

x x

x

x

x

+− =

+⋅ − ⋅ = ⋅

− = +

− = +− = +

− =

−=

− −=−


15. a. 1

21

2 22

2

2

2 2 or

A bh

A bh

A bh

A bh

b bA A

h hb b

=

⋅ = ⋅

=

=

= =

b. 2

2

2 2

or 2 2

C r

C r

C Cr r

= ππ

=π π

= =π π




149

c. 3

3 33

3

3

3 3 or

bhV

bhV

V bh

V bh

h hV V

b bh h

=

⋅ = ⋅

=

=

= =

16. a. Let n = first number. Then n + 20 = other number.

( 20) 84

20 84

2 20 84

2 64

32

20 52

n n

n n

n

n

n

n

+ + =

+ + =+ =

==

+ =

The numbers are 32 and 52.

b. Let n = first number. Then n + 19 = other number.

( 19) 47

19 47

2 19 47

2 28

14

19 33

n n

n n

n

n

n

n

+ + =+ + =

+ =

==

+ =


c. Let n = first number. Then n + 23 = second number.

( 23) 81

23 81

2 23 81

2 58

29

23 52

n n

n n

n

n

n

n

+ + =+ + =

+ =

==

+ =


17. a. Let c = calories in chicken breast. Then c + 22 = calories in pie.

( 22) 578

2 22 578

2 556

2 556

2 2278

22 300

c c

c

c

c

c

c

+ + =+ =

=

=

=

+ =

The chicken breast has 278 calories and the pie has 300 calories.

b. Let c = calories in chicken breast. Then c + 38 = calories in pie.

( 38) 620

2 38 620

2 582

2 582

2 2291

38 329

c c

c

c

c

c

c

+ + =+ =

=

=

=

+ =


c. Let c = calories in chicken breast. Then c + 42 = calories in pie.

( 42) 650

2 42 650

2 608

2 608

2 2304

42 346

c c

c

c

c

c

c

+ + =+ =

=

=

=

+ =





150

18. Let m = measure of angle.



a. 180 3(90 ) 20

180 270 3 20

180 250 3

180 180 250 180 3

70 3

3 70 3 3

2 70

2 70

2 235

m m

m m

m m

m m

m m

m m m m

m

m

m

− = − −− = − −− = −

− − = − −− = −

− + = − +=

=

=


b. 180 3(90 ) 30

180 270 3 30

180 240 3

180 180 240 180 3

60 3

3 60 3 3

2 60

2 60

2 230

m m

m m

m m

m m

m m

m m m m

m

m

m

− = − −− = − −− = −

− − = − −− = −

− + = − +=

=

=


c. 180 3(90 ) 40

180 270 3 40

180 230 3

180 180 230 180 3

50 3

3 50 3 3

2 50

2 50

2 225

m m

m m

m m

m m

m m

m m m m

m

m

m

− = − −− = − −− = −

− − = − −− = −

− + = − +=

=

=


19. a. R × T = D

car 1 40 T + 1 40(T + 1)

car 2 50 T 50T

40( 1) 50

40 40 50

40 10

40 10

10 104

T T

T T

T

T

T

+ =+ =

=

=

=

It takes the second car 4 hours to overtake the first car.

b. R × T = D

car 1 30 T + 1 30(T + 1)

car 2 50 T 50T

30( 1) 50

30 30 50

30 20

30 20

20 203 1

or 12 2

T T

T T

T

T

T T

+ =+ =

=

=

= =

It takes the second car 1

12

hours to

overtake the first car.

c. R × T = D

car 1 40 T + 1 40(T + 1)

car 2 60 T 60T

40( 1) 60

40 40 60

40 20

40 20

20 202

T T

T T

T

T

T

+ =+ =

=

=

=

It takes the second car 2 hours to overtake the first car.




151

20. a. price per pound × pounds = price

product 1.50 x 1.5x

another product 3.00 15 3(15)

mixture 2.40 15 + x 2.4(15 + x)

1.5 3(15) 2.4(15 )

1.5 45 36 2.4

1.5 9 2.4

9 0.9

10

x x

x x

x x

x

x

+ = ++ = +

+ ===

10 pounds should be mixed.

b. price per pound × pounds = price

product 2.00 x 2x


mixture 2.50 15 + x 2.5(15 + x)

2 3(15) 2.5(15 )

2 45 37.5 2.5

2 7.5 2.5

7.5 0.5

15

x x

x x

x x

x

x

+ = ++ = ++ =

==


c. price per pound × pounds = price

product 6.00 x 6x


mixture 4.50 15 + x 4.5(15 + x)

6 2(15) 4.5(15 )

6 30 67.5 4.5

6 37.5 4.5

1.5 37.5

25

x x

x x

x x

x

x

+ = ++ = +

= +==





152

21. a. P × r = I

5% x 0.05 0.05x

6% 30,000 − x 0.06 0.06(30,000 − x)

0.05 0.06(30,000 ) 1600

0.05 1800 0.06 1600

1800 0.01 1600

0.01 200

20,000

30,000 10,000

x x

x x

x

x

x

x

+ − =+ − =

− =− = −

=− =


b. P × r = I

7% x 0.07 0.07x

9% 30,000 − x 0.09 0.09(30,000 − x)

0.07 0.09(30,000 ) 2300

0.07 2700 0.09 2300

2700 0.02 2300

0.02 400

20,000

30,000 10,000

x x

x x

x

x

x

x

+ − =+ − =

− =− = −

=− =


c. P × r = I

6% x 0.06 0.06x

10%

30,000 − x 0.10 0.10(30,000 − x)

0.06 0.10(30,000 ) 2000

100 0.06 100 0.10(30,000 ) 100 2000

6 10(30,000 ) 200,000

6 300,000 10 200,000

300,000 4 200,000

4 100,000

25,000

30,000 5000

x x

x x

x x

x x

x

x

x

x

+ − =⋅ + ⋅ − = ⋅

+ − =+ − =

− =− = −

=− =

$25,000 is invested at 6% and $5000 is invested at 10%.



Cumulative Review Chapters 1−2

153

22. a. 3.05 3

3 3.05

3 3 or

3.05 3.05

C m

C m

C Cm m

= +− =− −

= =

Let C = 27.40.

27.40 38

3.05m

−= =

The call lasted 8 minutes.

b. 3.15 3

3 3.15

3 3 or

3.15 3.15

C m

C m

C Cm m

= +− =− −

= =

Let C = 34.50.

34.50 310

3.15m

−= =


c. 3.25 2

2 3.25

2 2 or

3.25 3.25

C m

C m

C Cm m

= +− =− −

= =

Let C = 21.50.

21.50 26

3.25m

−= =


23. a. The angles are supplementary. (3 20) 2 180

5 20 180

5 200

40

x x

x

x

x

− + =− =

==

3(40) − 20 = 100 and 2(40) = 80


b. Vertical angles must be equal. 7 10 3 30

7 3 40

4 40

10

x x

x x

x

x

− = += +==

7(10) − 10 = 60 and 3(10) + 30 = 60


c. The angles are complementary. (5 15) (2 5) 90

7 20 90

7 70

10

x x

x

x

x

+ + + =+ =

==

5(10) + 15 = 65 and 2(10) + 5 = 25


24. a. Since −8 is to the left of −7, −8 < −7.

b. Since 1

2 is to the right of −3,

13.

2> −

c. Since 4 is to the left of 1

4 ,3

1

4 4 .3

<

25. a. 4 2 2( 2)

4 2 2 4

4 2 2 2 4 2

4 2 6

4 2 2 2 6

2 6

2 6

2 23

x x

x x

x x

x x

x x x x

x

x

x

− < +− < +

− + < + +< +

− < − +<

<

<


b. 5 4 2( 1)

5 4 2 2

5 4 4 2 2 4

5 2 6

5 2 2 2 6

3 6

3 6

3 32

x x

x x

x x

x x

x x x x

x

x

x

− < +− < +

− + < + +< +

− < − +<

<

<





154

c. 7 1 3( 1)

7 1 3 3

7 1 1 3 3 1

7 3 4

7 3 3 3 4

4 4

4 4

4 41

x x

x x

x x

x x

x x x x

x

x

x

− < +− < +

− + < + +< +

− < − +<

<

<


26. a. 6( 1) 4 2

6 6 4 2

6 6 6 4 2 6

6 4 8

6 4 4 4 8

2 8

2 8

2 24

x x

x x

x x

x x

x x x x

x

x

x

− ≥ +− ≥ +

− + ≥ + +≥ +

− ≥ − +≥

≥

≥


b. 5( 1) 2 1

5 5 2 1

5 5 5 2 1 5

5 2 6

5 2 2 2 6

3 6

3 6

3 32

x x

x x

x x

x x

x x x x

x

x

x

− ≥ +− ≥ +

− + ≥ + +≥ +

− ≥ − +≥

≥

≥


c. 4( 2) 2 2

4 8 2 2

4 8 8 2 2 8

4 2 10

4 2 2 2 10

2 10

2 10

2 25

x x

x x

x x

x x

x x x x

x

x

x

− ≥ +− ≥ +

− + ≥ + +≥ +

− ≥ − +≥

≥

≥


27. a.

1

3 6 61

6 6 63 6 6

2 1

1

1

2 1

2 1

2 21

2

x x x

x x x

x x x

x x

x x x x

x

x

x

−− + ≤

− ⋅ − + ⋅ ≤ ⋅

− + ≤ −− ≤ −

− − ≤ − −− ≤ −− −

≥− −

≥


2

is a solution.




155

b.

1

4 7 71

28 28 284 7 7

7 4 4( 1)

3 4 4

3 4 4 4 4

7 4

7 4

7 74

7

x x x

x x x

x x x

x x

x x x x

x

x

x

−− + ≤

− ⋅ − + ⋅ ≤ ⋅

− + ≤ −− ≤ −

− − ≤ − −− ≤ −− −

≥− −

≥


7

is a solution.

c.

1

5 3 31

15 15 155 3 3

3 5 5( 1)

2 5 5

2 5 5 5 5

3 5

3 5

3 35

3

x x x

x x x

x x x

x x

x x x x

x

x

x

−− + ≤

− ⋅ − + ⋅ ≤ ⋅

− + ≤ −≤ −

− < − −− ≤ −− −

≥− −

≥


3

is a solution.

28. a. 2 4

2

x

x

+ ≤≤

and 2 6

2 6

2 23

x

x

x

− <−

>− −

> −

Now x ≤ 2 and x > −3 can be rewritten

as −3 < x ≤ 2. The solution consists of

all numbers between −3 and 2,

including 2.

b. 3 5

2

x

x

+ ≤≤

and 3 9

3 9

3 33

x

x

x

− <−

>− −

> −




including 2.

c. 1 2

1

x

x

+ ≤≤

and 4 8

4 8

4 42

x

x

x

− <−

>− −

> −




including 1.

Cumulative Review Chapters 1−−−−2

1. The additive inverse of −7 is 7.

2. 9 9

9 910 10

− =

3. The LCD is 63.

2 18

7 63− = − and

2 14

9 63− = −

2 2 18 14 32

7 9 63 63 63

− + − = − + − = −




156

4. −0.7 − (−8.9) = −0.7 + 8.9 = 8.2

5. (−2.4)(3.6) = −8.64

6. 4(2 ) (2 2 2 2) (16) 16− = − ⋅ ⋅ ⋅ = − = −

7. 7 5 7 24 168 21

8 24 8 5 40 5

− ÷ − = − ⋅ − = =

8. 5 60 5 6 3

12 6 3

72 3

69

y x z÷ ⋅ − = ÷ ⋅ −= ⋅ −= −=

9. In 9 ⋅ (8 ⋅ 5) = 9 ⋅ (5 ⋅ 8), we changed the

order of multiplication. The commutative property of multiplication was used.

10. 6(5 7) 6(5 ) 6(7)

30 42

x x

x

+ = += +

11. 5 ( 6 ) 5 6

( 5 6)

1

cd cd cd cd

cd

cd

cd

− − − = − += − +==

12. 2 2( 4) 3( 1)

2 2 8 3 3

(2 2 3 ) ( 8 3)

3 ( 11)

3 11

x x x

x x x

x x x

x

x

− + − += − − − −= − − + − −= − + −= − −

13. The quotient of (a − 4b) and c is written as

4.

a b

c

−

14. If x = 4, then 11 = 15 − x becomes

11 = 15 − 4 which is a true statement. Thus

4 is a solution of the equation.

15. 5 4( 3) 4 3

5 4 12 4 3

5 8

5 8 8 8

13 or 13

x x

x x

x

x

x x

= − + −= − + −= −

+ = − += =

The solution is 13.

16. 7

213

3 7 3( 21)

7 3 7

3 3

9

x

x

x

x

− = −

− ⋅ − = − ⋅ −

= ⋅=

The solution is 9.

17. 23 5

15 15 15 23 5

5 3 30

2 30

2 30

2 215

x x

x x

x x

x

x

x

− =

⋅ − ⋅ = ⋅

− ==

=

=

The solution is 15.

18.

2( 1)4

4 92( 1)

36 4 36 364 9

144 9 4 2( 1)

144 9 8( 1)

144 9 8 8

144 144 9 8 8 144

9 8 136

9 8 8 8 136

17 136

17 136

17 178

x x

x x

x x

x x

x x

x x

x x

x x x x

x

x

x

+− =

+⋅ − ⋅ = ⋅

− = ⋅ +− = +− = +

− − = + −− = −

− − = − −− = −− −

=− −

=

The solution is 8.

19. 2

2

2 2

2 2

6

6

6 6

or 6 6

S a b

S a b

a aS S

b ba a

=

=

= =

20. Let n = first number. Then n + 35 = second number.




157

( 35) 155

2 35 155

2 35 35 155 35

2 120

2 120

2 260

35 95

n n

n

n

n

n

n

n

+ + =+ =

+ − = −=

=

=+ =


21. Let b = annual return from bonds. Then b + 105 = annual return from stocks.

( 105) 595

2 105 595

2 105 105 595 105

2 490

2 490

2 2245

105 350

b b

b

b

b

b

b

b

+ + =+ =

+ − = −=

=

=+ =

The bonds return $245 and the stocks return $350.

22. R × T = D

Train A 40 T + 6 40(T + 6)

Train B 50 T 50T

40( 6) 50

40 240 50

240 10

24

T T

T T

T

T

+ =+ =

==

It takes 24 hours for train B to catch train A.

23. P × r = I

bonds x 0.12 0.12x

certificates

5000 − x 0.14 0.14(5000 − x)

0.12 0.14(5000 ) 660

0.12 700 0.14 660

700 0.02 660

0.02 40

2000

5000 3000

x x

x x

x

x

x

x

+ − =+ − =

− =− = −

=− =

Arlene invested $2000 in bonds and $3000 in certificates of deposit.




158

24.

5

6 5 55

30 30 306 5 5

5 6 6( 5)

6 30

6 6 6 30

5 30

5 30

5 56

x x x

x x x

x x x

x x

x x x x

x

x

x

−− + ≤

− ⋅ − + ⋅ ≤ ⋅

− + ≤ −≤ −

− ≤ − −− ≤ −− −

≥− −

≥





159



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testbankcollege.eutestbankcollege.eu/sample/Solution-Manual... · 63 Chapter 2 Equations, Problem Solving, and Inequa lities 2.1 The Addition and Subtraction Properties of Equa lity