Lehmann IA SSM Ch8

8/20/2019 Lehmann IA SSM Ch8

1/39

236

Chapter 8

Rational Functions

Homework 8.1

1. The only value that will make the function

undefined (i.e., the denominator 0) is 0.

Therefore, the domain is the set of all real

numbers except 0.

3. Since there are no values that will make the

function undefined, the domain is the set of all

real numbers.

5. 3 0

3

x

x

+ =

= −

Since 3− is the only value that will make the

function undefined, the domain is the set of all

real numbers except 3− .

7. 2 0

2

x

x

− =

=

The domain is the set of all real numbers except

2.

9. 2 1 0

2 1

1

2

x

x

x

+ =

= −

= −

The domain is the set of all real numbers

except1

2− .

11. ( 8)( 4) 0

8 0 4 0

8 4

x x

x x

x x

+ − =

+ = − =

= − =


8− and 4.

13.2

3 10 0

( 5)( 2) 0

5 0 2 0

5 2

x x

x x

x x

x x

− − =

− + =

− = + =

= = −


5 and 2− .

15.

2

9 0( 3)( 3) 0 x

x x

− =

− + =

3 0 3 0 x x − = + =

3 3 x x = = −


3 and 3.−

17.2

4 25 0

(2 5)(2 5) 0

2 5 0 or 2 5 0

2 5 2 5

5 5

2 2

x

x x

x x

x x

x x

− =

− + =

− = + =

= = −

= = −


5 5

and .2 2

− −

19.2

2

1 0

1

x

x

+ =

= −

Since a real number squared is nonnegative,

there is no real solution. The domain is the set of

all real numbers.

21.22 7 15 0

( 5)(2 3) 0

5 0 2 3 0

5 2 3

3

2

x x

x x

x x

x x

x

− − =

− + =

− = + =

= = −

= −


5 and3

.2

−

23.2

3 6 0

3 9 4(1)(6)

2

3 9 24

2

3 15

2

x x

x

x

x

− + =

± −=

± −=

± −=


2/39

SSM: Intermediate Algebra Homework 8.1

237

Since the square root of a negative number is not

real, there is no real solution. The domain is the

set of all real numbers.

25.2

3 2 7 0 x x − − =

2 4 4(3)( 7)

6

2 4 84

6

x

x

± − −=

± +=

2 88

6

2 4 22

6

2 2 22

6

1 223

x

x

x

x

±=

± ⋅=

±=

±=


1 22 1 22 and .

3 3

+ −

27.3 2

4 8 9 18 0 x x x − − + = 2

2

4 ( 2) 9( 2) 0

( 2)(4 9) 0

( 2)(2 3)(2 3) 0

2 0 or 2 3 0 or 2 3 0

2 2 3 2 3

3

2

x x x

x x

x x x

x x x

x x x

x

− − − =

− − =

− − + =

− = − = + =

= = = −

=3

2

x = −


2,3 3

and .2 2

−

29. 2( 1) 3( 2) 0 x x − + − =

2 2 3 6 0

5 8 0

5 8

85

x x

x

x

x

− + − =

− =

=

=


8.

5

31. ( 2)( 3) ( 1)( 5) 0 x x x x + − + + + =

2 2

2

6 6 5 0

2 5 1 0

x x x x

x x

− − + + + =

+ − =

5 25 4(2)( 1)

4 x

− ± − −

=

5 25 8

4 x

− ± +=

5 33

4 x

− ±=


5 33 5 33 and .

4 4

− + − −

33.

2

6

x x

x

= x ⋅

x x ⋅4

1

x x x x x

=⋅ ⋅ ⋅ ⋅

35.

7

4

20 5

15

x

x

=4 x ⋅ ⋅ x ⋅ x ⋅ x ⋅

5

x x x ⋅ ⋅ ⋅

3 x ⋅ ⋅ x ⋅ x ⋅ x ⋅

34

3

x

=

37.4 ( 7)4 28

5 35

x x

x

−−=

− 5 ( 7) x −

4

5=

39.

2

2

( 5) ( 2)7 10

7 18

x x x x

x x

+ ++ +=

− − ( 9) ( 2) x x − +

5

9

x

x

+=

−

41.2

5 ( 2)5 104

x x

x

++ =

− ( 2) x +5

2( 2) x x =

−−

43.2

3 (4 5)12 15

16 25

x x

x

++=

− (4 5) x +

3

4 5(4 5) x x =

−−

45.

2

2

( 7)49

14 49

x x

x x

−−=

− +

( 7)

( 7)

x

x

+

−

7

7( 7)

x

x x

+=

−−

47.

2 2

2 2

2 50 2( 25)

6 24 30 6( 4 5)2

x x

x x x x

− −=

− − − −

=( 5) x − ( 5)

2

x +

3 ( 5) x ⋅ −

5

3( 1)( 1)

x

x x

+=

++

49.

2

2

( 4) (2 1)2 7 4

2 9 4

x x x x

x x

− +− −=

+ + ( 4) (2 1) x x + +

4

4

x

x

−=

+


3/39

Homework 8.2 SSM : Intermediate Algebra

238

51.5 5

5

x x

x

− −=

− 1( 5) x− −

11

1= = −

−

53.

4 ( 3)4 12

18 6

x x

x

−−

=− 6 ( 3) x− −

4 2

6 3= = −−

55.2

6 ( 3)6 18 6( 3)

(3 )(3 )9

x x x

x x x

−− −= =

− +− 1( 3) x− − ( 3)

6=

3

x

x

+

−

+

57.

2 2

2 2

2 35 2 35

3 10 1( 3 10)

( 5)

x x x x

x x x x

x

+ − + −=

− + + − − −

−

=

( 7)

1( 5)

x

x

+

− −

7

( 2)( 2)

x

x x

+= −

++

59.

2 (3 2) (2 3)6 5 6

15 10

x x x x

x

+ −− −=

− 5 (2 3) x− −

3 2

5

x += −

61.

3 2 2

2

3 21 36 3 ( 7 12)

( 3)( 3)9

3 ( 4) ( 3)

x x x x x x

x x x

x x x

+ + + +=

− +−

+ +

=

( 3) ( 3) x x− +

3 ( 4)

3

x x

x

+=

−

63.

3 2 2

2 2

2

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

( 1)( 1)( 4)

3( 1)( 2)

x x x x x x x x x x

x x x

x x

− − + − − −=

+ − + −

−− −= =

− +

( 2) x + ( 2)

3 ( 1)

x

x

−

− ( 2) x +

2

3

x −=

65. a.420 2

(50) 8 or 8.4 hours50 5

T = =

It will take 8.4 hours if you drive 50 mph.

b. 420 7(55) 7 7.64 hours55 11

420(60) 7 hours

60

420 6(65) 6 6.46 hours

65 13

420(70) 6 hours

70

T

T

T

T

= = ≈

= =

= = ≈

= =

c. T is a decreasing function. This makes

sense in terms of the trip because the faster

you drive, the shorter the time of the trip; i.e.

as speed increases, time decreases.

67. Answers may vary.

69. domain = { }3, 2, 1,0,1,2,3− − −

range = { }50.4,25.2,16.8,12.6,10.08,8.4,7.2

71. original student’s

2(3 4) 3 5

(3 4)(3 1) 3 1

2(7) 3 5

7(2) 2

14 3

1417

14

+ +

+ − −

+

+

The results are not the same, so the original

expression is not equivalent to the student’s

expression.

73.(1 1)(1 2) 0( 1) 0

(1) 0(1 3)(1 4) ( 2)( 3) 6

f − − −= = = =

− − − −

75.(3 1)(3 2) 2(1) 2

(3) undefined(3 3)(3 4) 0( 1) 0

f − −= = = =

− − −

77. I would tell the student that 2 and 4 are in the

domain because they do not make the

denominator 0. The only values that must be

excluded are 5 and 1 because when we put these

into the function, the denominator equals 0.

79. Written response

Homework 8.2

1.2

5 2 10

x x x⋅ =

3. 5 3 5 3

3

7 5 7 6 7

2 6 2 5

x x x x

x

⋅÷ = ⋅ =

2

2

x⋅ 2⋅

3

3

5 x

⋅

⋅

221

5

x=

5. 5 3 6 5

4 8 10

x

x

−⋅ =

− 4 ( 2) x −

3 ( 2) x −⋅

2 5⋅

3

8=


4/39

SSM : Intermediate Algebra Homework 8.2

239

7.

3

3

6 18 5 15 6 18

5 5 5 15

x x x x

x x x x

− − −÷ = ⋅

−

6 ( 3) x −=

5 x

x

⋅

2

5 ( 3)

x

x

⋅

−

26

25

x

=

9. ( 3) x − ( 8)

( 2)

x

x

+

+

4 ( 2)

( 3)

x

x

+

⋅

+ 3 ( 3) x −

4( 8)

3( 3)

x

x

+=

+

11. ( 8)( 9) 2( 8)

3( 3)( 5) 12( 5)

( 8)

x x x

x x x

x

+ + +÷

− − −

+

=

( 9)

3( 3) ( 5)

x

x x

+

− −

12 ( 5) x −⋅

2 ( 8) x +

12( 9)

6( 3)

2( 9)

3

x

x

x

x

+=

−

+=

−

13.

2

7 2 7

3 15 5 3 15

5

3 ( 5)

x x x x

x x x x

x

+ + +÷ = ⋅

+

+

=2

x

2

5

x

x

⋅

⋅ 5 x + 5

3

x

=

15.

2

2

10 21 2 18

9 9

( 3)

x x x

x x

x

+ + −⋅

− −

+

=

( 7)

9

x

x

+

−

2 ( 9) x −⋅

( 3) x + ( 3)

2( 7)3

x

x

x

−

+=

−

17.

2 2

2

2

3 2 6

3 3 6 6

3 2 6 6

3 3 6

( 2)

x x x x

x x

x x x

x x x

x

+ + − −÷

− −

+ + −= ⋅

− − −

+

=

( 1)

3 ( 1)

x

x

+

−

6 ( 1) x −⋅

( 2) x + ( 3)

6( 1) 2( 1)

3( 3) 3

x

x x

x x

−

+ += =

− −

19.2 12 18 3 2 12 4 4

1 4 4 1 18 3

2 ( 6)

x x x x

x x x x

x

− − − +÷ = ⋅

+ + + −

−

=

1 x +

4 ( 1) x +⋅

3( 6) x − −

8

3= −

21.

2 2

2

2

2 32 7 6

62 24

2( 16) ( 6)( 1)

( 6)( 4) 6

2 ( 4)

x x x

x x x

x x x

x x x

x

− − +⋅

+− −

− − −= ⋅

− + +

+

=

( 4)

( 6)

x

x

−

− ( 4) x +

( 6) x −⋅

( 1)

6

2( 4)( 1)

6

x

x

x x

x

−

+

− −=

+

23.

2

2 2

2

2

4 25

10 25 3 9 12

1( 4) 1( 25)

( 5)( 5) 3( 3 4)

1( 4)

x x

x x x x

x x

x x x x

x

− −⋅

+ + − −

− − − −= ⋅

+ + − −

− −

=

( 5) x +

1( 5)

( 5)

x

x

− +

⋅

+

( 5)

3 ( 4)

x

x

−

− ( 1)

5

3( 5)( 1)

x

x

x x

+

−=

+ +

25.

2

2 2

3 8 16

16 2 3

1( 3)

x x x

x x x

x

− + + +⋅

− − −

− −

=

( 4) x +

( 4)

( 4)

x

x

+

⋅

−

( 4)

( 3)

x

x

+

− ( 1)

4

( 4)( 1)

x

x

x x

+

+= −

− +

27.

2 2

2 2

7 10 4

2 5 12 8 18

x x x

x x x

− + − − +÷

+ − −

2 2

2 2

2 2

2

7 10 8 18

2 5 12 4

1( 7 10) 2(4 9)

(2 3)( 4) 1( 4)

1

x x x

x x x

x x x

x x x

− + − −= ⋅

+ − − +

− − + −= ⋅

− + − −

−

=

( 5) ( 2) x x − −

(2 3) x −

2 (2 3)

( 4)

x

x

−

⋅

+

(2 3)

1

x +

− ( 2) x − ( 2)

2( 5)(2 3)

( 4)( 2)

x

x x

x x

+

− +=

+ +


5/39


240

29.2 2

2

4 6 4 24

36 6 12

2(2 3) 4( 6)

(3 4)(2 3)1( 36)

x x

x x x

x x

x x x

− − +⋅

− + −− + +

= ⋅− +− −

−= 2 (2 3) x +

−1( 6) x +

4 ( 6)

( 6)

x

x

+⋅− (3 4) (2 3) x x − +

8

( 6)(3 4) x x =

− −

31.

2 ( 4)16 3

2 6 4

x x x

x x

−− +⋅ =

+ −

( 4)

2 ( 3)

x

x

+

+3 x +

⋅4 x −

4

2

x +=

33.

2 2 2

2

25 15 3 25 12

4 20 4 20 15 312

( 5)

x x x x

x x x x

x

− − −÷ = ⋅

+ + −

+= ( 5) x −

4 ( 5) x +4⋅ 3⋅

2

3

x ⋅− ( 5) x −

2 2

1

x

x = − = −

35.

2 2

2 2

8 16 4 4

8 16 16

( 4)

x x x x

x x x

x

− + − +⋅

− + −−

= ( 4)

8 ( 2)

x

x x

−

− −

( 2) x −⋅

( 2)

( 4)

x

x

−

− ( 4)

( 4)( 2)

8 ( 4)

x

x x

x x

+

− −= −

+

37.

2

28 14 5 4 8

6 36 24

(4 5) (2 1)

x x x

x x

x x

+ + −⋅ − −++ +

=2

4( 2)

3 (2 1)6( 4)

x

x x

−⋅− ++

2 2

4(4 5)( 2) 2(4 5)( 2)

18( 4) 9( 4)

x x x x

x x

+ − + −= = −

− + +

39.

2 5 4 2

2 2 5 4

9 3 9 2

2 3

( 3)

x x x x x

x x x x x

x

− + − +÷ = ⋅

+ ++

=2 4

( 3) 2

( 3)

x x

x x x

− +⋅

+ 6

( 3)( 2) x x

x

− +=

41.

22 9 12

(3 5 12)2

x x

x x

x

−+ − ÷

+

2

2

3 5 12 2

1 9 12

x x x

x x

+ − += ⋅

−

( 3) (3 4) x x + −=

2

1 3 (3 4)

x

x x

+⋅

−( 3)( 2)

3

x x

x

+ +=

43.

3 2

2

2

12 20 8 5 15

2 6 4

4 (3 5 2) 5( 3)

2( 3) ( 2)( 2)

2

x x x x

x x

x x x x

x x x

− − −⋅

− −− − −

= ⋅− − +

= 2 ( 2) x x ⋅ − (3 1)

2

x +

( 3) x −

5 ( 3) x −⋅

( 2) x − ( 2)

10 (3 1)

2

x

x x

x

+

+= +

45.

2 2

3 2 2

4 5 8 12

6 4 24 10 25

x x x x

x x x x x

+ − + +⋅

+ − − + +

2

2

( 1)( 5) ( 6)( 2)

( 5)( 5)( 6) 4( 6)

( 1)( 5) ( 6)( 2)

( 5)( 5)( 6)( 4)

( 1) ( 5)

x x x x

x x x x x

x x x x

x x x x

x x

− + + += ⋅

+ ++ − +− + + +

= ⋅+ ++ −

− +=

( 6) x + ( 2) x +

( 6)

( 2)

x

x

+⋅

−

( 2) x +

( 5) x + ( 5)

1

( 2)( 5)

x

x

x x

+

−=− +

47.

7 2 2

2 13

20 14 24 6

5 159 8

x x x x x

x x x

− + + −÷ ⋅ −−

7 2

2 2 13

20 5 15 6

9 14 24 8

4

x x x x

x x x x

− + −= ⋅ ⋅ − − +

=7

5 x ⋅( 3) x + ( 3) x −

5 ( 3) x −⋅( 12) ( 2) x x − −

( 2) x −⋅

( 3) x +

4 7

2 x ⋅ ⋅ 6

6

25

2 ( 12)

x

x x

⋅

=−


6/39


241

49.

3 6

2

12 22

6 12 11 224

x x x

x x x

÷ ⋅ − + +−

3

2

12 11

4

x

x= ÷

−

62

6( 2) 11

x x

x

⋅⋅

− − ( 2) x

+

3 7

2

12 2

6( 2)( 2)4

x x

x x x= ÷

− − +−

3

2 7

12 6( 2)( 2)

4 2

x x x

x x

− − += ⋅

−

312 x

=( 2) x − ( 2) x +

2−⋅

3 ( 2) x⋅ − ( 2) x +

2 3 x⋅

44

36

x x

−=

⋅

51.

2 2 2

2 2 2

2

4 5 4

5 1 1

4 5 1

5 1 4

( 4)

x x x

x x x

x x x

x x x

x

− + − ⋅ ÷ + − −

− + − = ⋅ ⋅ + − −

−=

2( 5) x +

2( 5) x +⋅

2( 1) x −

2( 1) x −⋅

2( 4) x −1=

53.

2 2

2 2

6 16 64( ) ( )

3 40 3 10

( 8) ( 2)

x x x f x g x

x x x x

x x

− − −= ⋅

+ − − −− +

=( 8) x +

( 8)

( 5)

x

x

+⋅

−

( 8)

( 5) ( 2)

x

x x

−

− +

2

2

( 8)

( 5)

x

x

−=

−

55.

2 2

2 2

2 2

2 2

64 6 16( ) ( )

3 10 3 40

64 3 40

3 10 6 16

( 8) ( 8)

x x xg x f x

x x x x

x x x

x x x x

x x

− − −÷ = ÷

− − + −− + −

= ⋅− − − −+ −

=( 5) x −

( 8) ( 5)

( 2)

x x

x

+ −⋅

+ ( 8) x −2

2

( 2)

( 8)

( 2)

x

x

x

+

+=

+

57.

2 2

2 2

2 2

2 2

1 8 7( ) ( )

3 28 5 4

1( 1) 5 4

3 28 8 7

1( 1)

x x x f x g x

x x x x

x x x

x x x x

x

− − +÷ = ÷

− − + +− − + +

= ⋅− − − +

− −=

( 1)

( 7) ( 4)

x

x x

+

− +

( 4) x +⋅

( 1)

( 7) ( 1)

x

x x

+

− −2

2

( 1)

( 7)

x

x

+= −

−

59. Answers may vary.

Ex: Let 3. x = 2?

2

3 2 3 2 3 4

3 5 3 5 3 25

+ + +⋅ =

+ − −

?5 5 9 4

8 2 9 25

+⋅ =

− −

25 13

16 16≠

− −

Correct product:2

2

2 2 4 4

5 5 25

x x x x

x x x

+ + + +⋅ =

+ − −

61. a.2

(19) 0.077(19) 4.06(19) 17.38

0.077(361) 77.14 17.38

27.797 77.14 17.38

66.723

c = − + +

= − + +

= − + +

=

Approximately 66,723,000 households had

cable television in 1999.

b. (19) 1.20(19) 80.92 103.72h = + =

There were approximately 103,720,000

households in 1999.

c. 66,723,000

100 64.3%103,720,000

⋅ ≈

Approximately 64.3% of all households hadcable television in 1999. In order to get the

percentage, the ratio must be multiplied by

100.


7/39


242

d. 2

2

0.077 4.06 17.38( ) 100

1.2 80.92

7.7 406 1738

1.2 80.92

t t p t

t

t t

t

− + += ⋅

+− + +

=+

In order to get the percentage, the ratio must bemultiplied by 100.

e. 27.7(19) 406(19) 1738

(19)1.2(19) 80.92

64.3%

p− + +

=+

≈

This is the same as the result in part c.

f. 27.7(18) 406(18) 1738

(18)1.2(18) 80.92

63.9%

p− + +

=+

≈

Approximately 63.9% of all households had

cable television in 1998.

g. The first screen shows the equations.

The screens below show the window and the

two graphs.

63. Answers may vary. Ex:5 5

and5 5

x x

x x

+ +− −


Homework 8.3

1. 3 2 1 5 11 1 1

x x x x x x

+ ++ =− − −

3.2 2

2 2

2 2 2

2 2

3 5 2 7 15

10 21 10 21

3 5 2 7 15 2 15

10 21 10 21

( 5)( 3) 5

( 7)( 3) 7

x x x x

x x x x

x x x x x x

x x x x

x x x

x x x

+ + +−

+ + + ++ − − − − −

= =+ + + +

− + −= =+ + +

5.2 2 2 2 2

2

2 4 2 4 2 4 2 4

2( 2)

x x x

x x x x x x x x

x

x

−− = ⋅ − = − =

−=

7.2

6 4 6 4 2

2 2 2

6 6 6 6

2

6

5 3 5 6 3 5

610 12 10 12 5

30 15 30 15 15(2 )

60 60 60 60

2

4

x

x x x x x

x x x

x x x x

x

x

+ = ⋅ + ⋅

+ +

= + = =+

=

9.3 4 3 2 4 1

1 2 1 2 2 1

x x

x x x x x x

− + + = ⋅ + ⋅ + − + − − +

3 6 4 4 7 2

( 1)( 2) ( 1)( 2) ( 1)( 2)

x x x

x x x x x x

− + −= + =

+ − + − + −

11.6 4

( 4)( 6) ( 1)( 4)

6 1 4 6

( 4)( 6) 1 ( 1)( 4) 6

x x x x

x x

x x x x x x

−+ − − +

− − = ⋅ − ⋅ + − − − + −

6 6 4 24

( 4)( 6)( 1) ( 4)( 6)( 1)

6 6 4 24 2 18

( 4)( 6)( 1) ( 4)( 6)( 1)

2( 9)

( 4)( 6)( 1)

x x

x x x x x x

x x x

x x x x x x

x

x x x

− −= −

+ − − + − −− − + +

= =+ − − + − −

+=

+ − −


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243

13.5 2 5 2

3 6 5 15 3( 2) 5( 3)

5 5( 3) 2 3( 2)

3( 2) 5( 3) 5( 3) 3( 2)

25( 3) 6( 2)

15( 2)( 3) 15( 2)( 3)

25 75 6 12

15( 2)( 3) 15( 2)( 3)

x x x x

x x

x x x x

x x

x x x x

x x

x x x x

− = −− + − +

+ −= ⋅ − ⋅

− + + −

+ −= −− + − +

+ −= −

− + − +

25 75 6 12 19 87

15( 2)( 3) 15( 2)( 3)

x x x

x x x x

+ − + += =

− + − +

15.2 2

3 5 3 5

( 5)( 5) ( 5)25 5 x x x x x x x + = +

+ − −− −

3 5 5

( 5)( 5) ( 5) 5

3 5 25

( 5)( 5) ( 5)( 5)

8 25

( 5)( 5)

x x

x x x x x x

x x

x x x x x x

x

x x x

+ = ⋅ + ⋅ − + − + +

= +− + − +

+=

− +

17.2 2

2 3

9 7 12

2 3

( 3)( 3) ( 4)( 3)

x x x

x x x x

+− − +

= ++ − − −

2 4 3 3

( 3)( 3) 4 ( 4)( 3) 3

x x

x x x x x x

− + = ⋅ + ⋅

+ − − − − +

2 8 3 9

( 3)( 3)( 4) ( 3)( 3)( 4)

5 1

( 3)( 3)( 4)

x x

x x x x x x

x

x x x

− += +

+ − − + − −+

=+ − −

19.3 2 1 3 2 2 3

21 1 1 1 1 1

x x x x x

x x x x x

− + − + − + = ⋅ + = + + + + + +

3 1

1

x

x

−=

+

21.2

2 ( 2)2 4 22

13 2

x x

x x

++− = −

+ + ( 2) x + ( 1) x +

2 2 2 1 2

1 1 1 1 1

2 2 2 2

1 1 1

x

x x x

x x

x x x

+ = − = ⋅ − + + + +

= − =+ + +

23.8 4 8 4 12

6 6 6 6 6 x x x x x − = + =

− − − − −

25.2

2 1 3

14 24 212 1 3

( 7)( 3) 2( 7)

x

x x x

x

x x x

++

−− −+= +

− + − −

2 1 3

( 7)( 3) 2( 7)

2 1 2 3 3

( 7)( 3) 2 2( 7) 3

4 2 3 9

2( 7)( 3) 2( 7)( 3)

4 2 3 9 7 1

2( 7)( 3) 2( 7)( 3) 2( 3)

x

x x x

x x

x x x x

x x

x x x x

x x x

x x x x x

+= −

− + −

+ + = ⋅ − ⋅ − + − + + +

= −− + − +

+ − − −= = =

− + − + +

27.2 2

2

2 1 2 1

7 2 2 74 49 4 49

2 1

2 7 (2 7)(2 7)

2 2 7 1

2 7 2 7 (2 7)(2 7)

4 14 1

(2 7)(2 7) (2 7)(2 7)

x x x x

x x x x

x x

x x x

x x x

x x x x

x x x

x x x x

− + +− = −

− −− −+

= −− + −

+ + = ⋅ − − + + −

+ += −

− + − +

2 24 14 1 4 13 1

(2 7)(2 7) (2 7)(2 7)

x x x x x

x x x x

+ − − + −= =

− + − +

29.

2 2 2

1 2 1 1 2 2

2 1 2 1 1 2

2 1 4 4 2 2 5

( 2)( 1) ( 2)( 1) ( 2)( 1)

x x x x x x

x x x x x x

x x x x x x

x x x x x x

− + − − + + + = + + − + − − +

− + + + + += + =

+ − + − + −

31.

2 2 2

3 1 3 4 1 5

5 4 5 4 4 5

7 12 6 5 2 17

( 5)( 4) ( 5)( 4) ( 5)( 4)

x x x x x x

x x x x x x

x x x x x x

x x x x x x

+ − + + − − + = + − + − + + −

+ + − + + += + =

− + − + − +


9/39


244

33.2

2

2

4 5 4 5

2 10 2( 5) ( 5)( 5)25

4 5 5 2

2( 5) 5 ( 5)( 5) 2

20 10

2( 5)( 5) 2( 5)( 5)

30 ( 6)( 5) 6

2( 5)( 5) 2( 5)( 5) 2( 5)

x x

x x x x x

x x

x x x x

x x

x x x x

x x x x x

x x x x x

+ +− = −

+ + − +−

+ − = − + − − +

− −= −

+ − + −

− − − + −= = =

+ − + − −

35.2

2

2 1

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

2 1

1( 4)( 3)

1 3

( 4)( 1)( 3) 3

x x

x x x x x

x x

x x x

x x

x x x x

+ −+

− + +− +

+ + = +− + − +

+ − + + +

2 2

2 2

2

2

3 2 2 3

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

2 5 1

( 4)( 1)( 3)

x x x x

x x x x x x

x x

x x x

+ + + −= +

− + + − + +

+ −=

− + +

37.2 2

2 3 2 3

( 2)( 2) ( 2)4 2

1 3 4

2 2 2

x x x x

x x x x x x x

x x x

+ ++ = +

+ − −− −

= + =− − −

39.2 2

1 4

4 20 25 6 17 5

x x

x x x x

− +−

+ + + +

1 4

(2 5)(2 5) ( 2 5)(3 1)

x x

x x x x

− += −

+ + + +

1 3 1

(2 5)(2 5) 3 1

4 2 5

(2 5)(3 1) 2 5

x x

x x x

x x

x x x

− + = − + + + + + + + +

2 2

2 2

3 2 1 2 13 20

(3 1)(2 5) (3 1)(2 5)

x x x x

x x x x

− − + += −

+ + + +

2 2

2

2

2

3 2 1 2 13 20

(3 1)(2 5)

15 21

(3 1)(2 5)

x x x x

x x

x x

x x

− − − − −=

+ +

− −=

+ +

41.2 2

3 1 2 1

4 4 3 5 2

x x

x x x x

− +−

+ + + −

2 2

2 2

2 2 2

2 2

3 1 2 1

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

3 1 3 1

( 2)( 2) 3 1

2 1 2

(3 1)( 2) 2

9 6 1 2 5 2

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

9 6 1 2 5 2 7 11 1

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

x x

x x x x

x x

x x x

x x

x x x

x x x x

x x x x

x x x x x x

x x x x

− += −

+ + − +

− − = + + − + + − − + +

− + + += −

− + − +

− + − − − − −= =

− + − +

43.2 3 2

1 5

6 24 3 6 24

x

x x x x x

−+

− − −

2

1 5

6 ( 4) 3 ( 2 8)

1 5

6 ( 4) 3 ( 4)( 2)

1 2 5 2

6 ( 4) 2 3 ( 4)( 2) 2

x

x x x x x

x

x x x x x

x x

x x x x x x

−= +

− − −−

= +− − +

− + = + − + − + 2

2

2 10

6 ( 4)( 2) 6 ( 4)( 2)

8

6 ( 4)( 2)

x x

x x x x x x

x x

x x x

+ −= +

− + − +

+ +=

− +

45.2

2 3 1

2 2 44

2 3 1

( 2)( 2) 2 2( 2)

2 3 2 1

( 2)( 2) 2 2 2( 2)

2 3 6 1

( 2)( 2) ( 2)( 2) 2( 2)

x x x

x x x x

x

x x x x x

x

x x x x x

+ − + −−

= + − + − + − − = + − + − + − − −

= + − + − + − −

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

3 4 2 1 2

( 2)( 2) 2 2( 2) 2

x

x x x

x x

x x x x

− −+ − −

− + = − + − − +


10/39


245

6 8 2

2( 2)( 2) 2( 2)( 2)

5 10 5( 2) 5

2( 2)( 2) 2( 2)( 2) 2( 2)

x x

x x x x

x x

x x x x x

− += −

+ − + −− −

= = =+ − + − +

47.2

3 2 3 2

1 56 5

3 2 3 2

1 ( 5)( 1) 5

3 2 3 2 1

1 ( 5)( 1) 5 1

3 2 3 2 2

1 ( 5)( 1) ( 5)( 1)

x

x x x x

x

x x x x

x x

x x x x x

x x

x x x x x

− − + + ++ + −

= − + + + + + − + = − + + + + + + − +

= − + + + + + +

3 4 1

1 ( 5)( 1)3 5 4 1

1 5 ( 5)( 1)

3 15 4 1

( 1)( 5) ( 1)( 5)

3 15 4 1 16

( 1)( 5) ( 1)( 5)

x

x x x x x

x x x x

x x

x x x x

x x x

x x x x

−= −

+ + ++ − = − + + + +

+ −= −

+ + + ++ − + − +

= =+ + + +

49.

2

2 2

2 2

2 2

2 2 2

4 5 3 15 2

2 4 7 10

4 5 3( 5) ( 2)2 ( 2)( 2) ( 5)( 2)

4 5 3 4 5 2 3

2 2 2( 2) ( 2)

4 13 10 3 4 16 10

( 2) ( 2) ( 2)

x x x x

x x x x

x x x x

x x x x x

x x x x x

x x x x x

x x x x x

x x x

+ + −+ ⋅ + − + +

+ + −= + ⋅ + + − + + + + + = + = + + + ++ +

+ + + += + =

+ + +

2

2

2(2 8 5)

( 2)

x x

x

+ +=

+

51.

2

2 2

5 5 4 4

3 6 2 1 2 1

x x x

x x x x x

+ +⋅ + +

+ + + +

2

2

5( 1) 4 4

3( 2) 2 1

5( 1) ( 2)( 2) 5( 2)

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

x x x

x x x

x x x x

x x x x

+ + += ⋅ + + +

+ + + += ⋅ =

+ + + +

53.

2 2

2

3 4( ) ( )

4 3

3 3 4 4

4 3 3 4

9 16

( 4)( 3) ( 4)( 3)

2 25

( 4)( 3)

x x f x g x

x x

x x x x

x x x x

x x

x x x x

x

x x

+ ++ = +

− −+ − + − = + − − − −

− −= +− + − −

−=

− −

55.4 3

( ) ( )3 4

x xg x f x

x x

+ +− = −

− −

2 2

2 2

4 4 3 3

3 4 4 3

16 9

( 3)( 4) ( 3)( 4)

16 9 7

( 3)( 4) ( 3)( 4)

x x x x

x x x x

x x

x x x x

x x

x x x x

+ − + − = − − − − −

− −= −

− − − −− − + −= =

− − − −

57.2

2 1( ) ( )

3 62 8

2 1

( 4)( 2) 3( 2)

2 3 1 4

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

x x f x g x

x x x

x x

x x x

x x x

x x x x

− +− = −

+− −− +

= −− + +

− + − = − − + + −

2

2 2

3 6 3 4

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

3 6 3 4 6 2

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

x x x

x x x x

x x x x x

x x x x

− − −

= −− + − +− − + + − + −

= =− + − +

59. The student did not multiply each expression by

“1”. The expression2

1 x + should have been

multiplied by2

2

x

x

++

not1

2 x + and the expression

3

2 x + should have been multiplied by

1

1

x

x

++

not

1

1 x + .


11/39


246

The correct addition is below:

2 3

1 2

2 2 3 1

1 2 2 12 4 3 3

( 1)( 2) ( 1)( 2)

5 7

( 1)( 2)

x x

x x

x x x x

x x

x x x x

x

x x

++ +

+ + = + + + + +

+ += +

+ + + ++

=+ +

61. The student did not subtract the entire numerator

in the second expression. The student only

subtracted 5 x and not 1. The correct addition is

below:

9 5 1 9 5 1 4 1

3 3 3 3

x x x x x

x x x x

+ − − −− = =

− − − −

63. Answers may vary. Ex:3

and1 1

x x

x x − −


Homework 8.4

1.

2

2 3 2 2

3 3 3

x x

x x x

x

= ÷ = ⋅ =

3. 5 32

2 5 2

5

77 21 7

21 21 3

x x x

x x x

x

= ÷ = ⋅ =

5.

3

3 7 3

7 7

3

7 4

9

9 12 9 2016

16 20 1612 12

20

3 3 4 5 15

4 4 3 4 16

x

x x x

x x

x

x x

= ÷ = ⋅

⋅ ⋅= ⋅ =

⋅ ⋅

7.

( )

2

2 2

2

2 1

2 1 11 3055 5 5 5 5

11 30

( 1) 6( 1)( 1) ( 5)( 6)

5 5( 1) 5

x x

x x x x x

x x x

x x

x x x x x x

x x

+ ++ + − +− = ⋅

+ − +− +

+ −+ + − −= ⋅ =

− +

9.

2

22

2 2 2

49

49 7 213 9

5 14 3 9 5 14

7 21

( 7)( 7) 7( 3) 7( 7)3 ( 3) ( 7)( 2) 3 ( 2)

x

x x x x

x x x x x x

x

x x x x

x x x x x x

−− −− = ⋅

− − − − −−

+ − − += ⋅ =− − + +

11. 32 2 2

2

3 3 3

6 1 5

5

2 8 10 10 2

x x x x x

x

x x x

−= = ⋅ =

+

13. 3 233

3

3 2 3 2

2 32 3

2 3

5 4 5 4 5 4

x x x x x x

x x

x x x x

−− − = ⋅ =+ + +

15. 2 2 2

2

4 6 2

2 5 53 3 39 7 2 2 33

5 5 5

x x x x

x x

x x x

−−−

= = ⋅ = −−

17.

33 444 3

2 2 2 33 3

x x x x

x x

x x

++ + = ⋅ =− − −

19. 2 2

2 2

2 2

11 22 2

1 1 4 14 4

(2 1)

(2 1)(2 1) 2 1

x x x x x

x x

x x

x x x

x x x

−− − = ⋅ = −− − −

= =− + +

21.

( )

32 32 3

3

2 2

2

2

1 8 151 8 15

1 5 1 5

5 ( 3)8 15 3

( 5)5

x x x x x x x

x

x x x x

x x x x x

x x x x x

− +− + = ⋅

− −

− −− + −= = =−−

23.

22

( 4)( 1)4 14 12 2 ( 4)( 1)

4 1 4 1

x x x x

x x x x x x

x x x x x x

x x x x

−− − +− + − + = ⋅− + − − − + − +


12/39


247

2 2

2 2

2

2

( 1) 2 ( 4) 2 8

( 4) 2 ( 1) 4 2 2

9 ( 9) 9

( 6) 66

x x x x x x x x

x x x x x x x x

x x x x x

x x x x x

+ − − + − += = =

− − + − − −

− + − − −= =

− + +− −

25.

22

( 4)44

2 2 ( 4)

4 4

x x x x x

x x x

x x

++ −− − = ⋅− + + − −

2

2

( 4) 2 4 2

( 4) 3 4 3

x x x x

x x x x

− + − += =

− − − −

27.

1 1 1 1

( 3) ( 3)3 3

3 3 ( 3) 3 ( 3)

3 1

3 ( 3) ( 3)

x x x x x x x x

x x x x

x x x x

− − + − ++ += ⋅ = + +

− −= =

+ +

29.

2 22 2 2 2

2 2

2 2 2 2

2 2 2 2

2 2 2 2

1 1 1 1

( 2)( 2) ( 2)

2 2 ( 2)

( 2) ( 4 4)

2 ( 2) 2 ( 2)

4 4 4( 1)

2 ( 2) 2 ( 2)

x x x x x x

x x

x x x x x

x x x x

x x

x x x x

− − ++ + = ⋅ +

− + − + += =

+ +− − − +

= =+ +

31.2

2 2

6 10

2 8 41 2

12 16

6 10

2( 4) ( 4)

1 2

( 4)( 3) ( 4)( 4)

x x x

x x x

x x x

x x x x

+− −

−− − −

+− −

=−

− + − +

6 10 2

2( 4) ( 4) 2

1 ( 4) 2 ( 3)

( 4)( 3) ( 4) ( 4)( 4) ( 3)

6 20

2 ( 4)

4 2 6

( 4)( 3)( 4) ( 4)( 3)( 4)

x

x x x x

x x

x x x x x x

x

x x

x x

x x x x x x

⋅ + ⋅− −=+ +⋅ − ⋅

− + + − + ++

−=+ +

−− + + − + +

6 20

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

4 2 6 2 ( 4)( 3)( 4)

( 4)( 3)( 4)

(6 20)( 3)( 4)( 2) 2

2(3 10)( 3)( 4)

2 ( 2)

(3 10)( 3)( 4)

( 2)

x

x x x x x x

x x x x x x

x x x

x x x

x x

x x x

x x

x x x

x x

+− + +−= ⋅

+ − − − + +− + +

+ + +=− − ⋅

+ + +=

− ++ + +

= −+

33.

2

2

2 2

2 42 2( )3 2 3 2 2 6 4

410

x x

x x x xh x

x

x x x x

x

+ + + = = ⋅ = + + +

+=

35.

( )( )

2

2

5( 2)5 10

( 3)( 3)6 9( )4 8 4( 2)

( 3)( 1)4 3

3 15( 2) 5( 1)

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

x x

x x x xh x

x x

x x x x

x x x x

x x x x

++− −− += =

+ +− −− +

− −+ −= ⋅ =

− − + −

37.

2

2

2

2

1111

( 1)( 1)111 1 ( 1)( 1)

1 11 1

( 1)( 1) ( 1) 1 1

( 1)( 1) ( 1) 1 1

2 ( 2)( 1)

( 2)( 1)2

x x x x

x x

x x

x x x x x

x x x x x

x x x x

x x x x

−−

− +− − = ⋅ − + + + + +

− + − + − − −= =

− + + − − + −

− − − += =

+ −+ −

39.

1 2 222

2 1 2

2 2

1 11 1

1 1 1 1

1

1

x x x x x x x

x x x

x x x x

x

x

− −

− −

++ + = = ⋅ − − −

+=

−

41.

1 2 3

2 2 3

2

1

1 1

x x x x x x x

x x x x x

x

−

−

− − − = ⋅ = + ++


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248

2

2 2

( 1) ( 1)( 1)

( 1)( 1) ( 1)( 1)

x x x x x

x x x x x x

− + −= =

+ − + + − +

2

( 1)

1

x x

x x

−=

− +

43. The student must add the two expressions in the

parentheses before taking the reciprocals. The

correct simplification is below:

2

1 1 1 2 1

1 1 2 2 2

2

2 2 2 2

2 2 2 2 2

x x

x x

x x x

x

x x x x

x x x

x x x x x

= ÷ + = ÷ ⋅ + ⋅ +

+ = ÷ + = ÷ = ⋅ = + +

45. Method 1

2 2 2 2

2 2 2 2

22

2 2 2

2

6 5 6 5 6 5

2 3 2 2 3 4 3

2 2 2 2 2

6 5

6 5 4 3 6 5 2

4 3 4 32

2

2(6 5 )

4 3

x x

x x x x x x x

x x

x x x x x x x

x

x x x x x

x x x x x

x

x

x

− − ⋅ −= =

+ ⋅ + ⋅ +

−− + −

= = ÷ = ⋅+ +

−=

+

Method 2

222

2

2 2

6 56 52 12 10

2 3 2 3 4 32

2 2

2(6 5 )

4 3

x x x x x x

x x

x x x x

x

x

−− − = ⋅ =+ + +

−

=+

Written responses may vary.

Homework 8.5

1.7 2

1

7 2 1

7 2

5

x x

x x

x x

x

x

= +

⋅ = +

= +=

Check.?7 2

1(5) (5)

= +

7 7

5 5=

The solution is 5.

3. 2 3 4

8 8

x x

x x

− +=

− −

2 3 4( 8) ( 8)

8 8

2 3 4

7

x x

x x

x x

x x

x

− +− = −

− −− = +

=

Check.?

2(7) 3 7 47 8 7 8

11 11

− +=− −− = −

The solution is 7.

5.2 5

7 7

x

x x

−=

− −

2 5( 7) ( 7)

7 7

2 5

7

x

x x

x x

x

x

−− = −

− −− =

=

Check.

?(7) 2 5

(7) 7 (7) 7

5 5

0 0

− =− −

=

Division by zero is undefined. Empty solution

set.

7.5 1 7

3 x x + =

5 1 73 3

3

15 216

x x

x x

x

x

⋅ + = ⋅

+ ==

Check.?5 1 7

(6) 3 (6)

7 7

6 6

+ =

=

The solution is 6.


14/39


249

9.2 3

1 1 x x =

− +

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

1 1

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

5

x x x x

x x

x x

x x

x

− + = − +− +

+ = −+ = −

=

Check.?2 3

(5) 1 (5) 1

1 1

2 2

=− +

=

The solution is 5.

11.3 2

11 5 x

+ =+

3 25( 1) 5( 1) 1

1 5

15 2( 1) 5 5

15 2 2 5 5

2 17 5 5

12 3

4

x x

x

x x

x x

x x

x

x

+ + = + ⋅ + + + = +

+ + = ++ = +

==

Check.?3 21

(4) 1 5

1 1

+ =+

=

The solution is 4.

13.2

1 1 4

2 2 4

1 1 4

2 2 ( 2)( 2)

x x x

x x x x

+ =− + −

+ =− + − +

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

2 2 ( 2)( 2) x x x x

x x x x

− + + = − +− + − +

2 2 4

2 4

2

x x

x

x

+ + − ==

=

Check.?

2

1 1 4

(2) 2 (2) 2 (2) 4

1 1 4

0 4 0

+ =− + −

+ =


set.

15.2

4 82

2 2

4 82

2 ( 2)

4 8( 2) 2 ( 2)

2 ( 2)

x x x

x x x

x x x x

x x x

+ =− −

+ =− −

− + = − − −

2

2

2

2 ( 2) 4 8

2 4 4 8

2 8

4

2

x x x

x x x

x

x

x

− + =

− + =

=

== ±

Check 2 x = ?

2

4 82

(2) 2 (2) 2(2)

8 40

+ =− −

=

Division by zero is undefined.

Check. 2 x = − ?

2

4 82

( 2) 2 ( 2) 2( 2)

1 1

+ =− − − − −

=

The solution is 2.−

17.2

48 6 7

3 52 15

48 6 7

( 5)( 3) 3 5

x x x x

x x x x

+ =+ −− −

+ =− + + −

48 6 7( 5)( 3) ( 5)( 3)

( 5)( 3) 3 5

48 6( 5) 7( 3)

48 6 30 7 21

6 18 7 21

3

x x x x

x x x x

x x

x x

x x

x

− + + = − +− + + −

+ − = ++ − = +

+ = +− =

Check.?

2

48 6 7

( 3) 3 ( 3) 5( 3) 2( 3) 15

48 6 7 0 0 8

+ =− + − −− − − −

+ = −


set.


15/39


250

19.2

2 1 16

5 5 25 x x x + =

+ − −

2 1 16

5 5 ( 5)( 5) x x x x + =

+ − + −

( 5)( 5)

16( 5)( 5)

( 5)( 5)

2 1

5 5

x x

x x

x x

x x

+ −

= + −+ −

++ −

2( 5) 5 16

2 10 5 16

3 5 16

3 21

7

x x

x x

x

x

x

− + + =− + + =

− ===

Check.?

2

2 1 16

(7) 5 (7) 5 (7) 25

2 2

3 3

+ =+ − −

=

The solution is 7.

21.2

2 2

5 6 11 30

2 2

5 6 ( 5)( 6)

x

x x x x

x

x x x x

+ =− − − +

+ =− − − −

2

2

2

2 2( 5)( 6) ( 5)( 6)

5 6 ( 5)( 6)

( 6) 2( 5) 2

6 2 10 2

4 10 2

4 12 0

x

x x x x

x x x x

x x x

x x x

x x

x x

− − + = − −− − − −

− + − =

− + − =

− − =

− − =

4 16 4(1)( 12)

2

4 16 48

2

4 64

24 8

2

x

± − −=

± +=

±

=±

=

4 8 4 8 or

2 2

6 or 2

x x

x x

+ −= =

= = −

Check. 6 x = ?

2

(6) 2 2

(6) 5 (6) 6 (6) 11(6) 30

2 2 6

0 0

+ =− − − +

+ =

Division by zero is undefined.

Check. 2 x = − ?

2

( 2) 2 2

( 2) 5 ( 2) 6 ( 2) 11( 2) 30

1 1

28 28

−+ =

− − − − − − − +

=

The solution is 2.−

23.2

1 103

x x

+ =

2 2

2

2

2

1 103

3 10

3 10 0

(3 5)( 2) 0

x x x x

x x

x x

x x

+ =

+ =

+ − =− + =

3 5 0 or 2 0

3 5 2

5

3

x x

x x

x

− = + == = −

=

Check.5

3 x =

( )

?

255

3 3

1 103

18 18

5 5

+ =

=

Check. 2 x = −

( )

?

2

1 103

2 2

5 5

2 2

+ =− −

=

The solutions are5

3

and 2.−

25. 2 2

2 1 32 5

x x x

− = + +

2 2

2 2

2 2

2

2 1 32 5

2 2 5 3

0 3 5

x x

x x x

x x x

x x

− = + +

− = + +

= + +


16/39


251

1 1 4(3)(5)

6 x

− ± −=

1 1 60

6

− ± −=

1 59

6

− ± −=

No real solutions. Empty solution set.

27. 3

43 3

34

3 3

x

x x

x

x x

= −− −

= +− −

3( 3) ( 3) 4

3 3

4( 3) 3

4 12 3

4 9

3 9

3

x

x x

x x

x x

x x

x x

x

x

− = − + − − = − +

= − += −

− = −=

Check.?(3) 3

4(3) 3 3 (3)

3 3 4

0 0

= −− −

= −


set.

29. 2

4 2

34 21

4 2

( 7)( 3) 3

x x

x x x

x x

x x x

+ −=

−+ −+ −

= −+ − −

2

2

2

4 2( 7)( 3) ( 7)( 3)

( 7)( 3) 3

4 ( 7)( 2)

4 ( 5 14)

4 5 14

6 10 0

x x

x x x x

x x x

x x x

x x x

x x x

x x

+ −+ − = + − −

+ − −

+ = − + −

+ = − + −

+ = − − +

+ − =

6 36 4(1)( 10)

2

6 36 40

2

6 76

2

6 4 19

2

6 2 19

2

3 19

x

− ± − −=

− ± +=

− ±=

− ± ⋅=

− ±=

= − ±

Check 3 19 x = − ± using a calculator. Thesolutions are 3 19 and 3 19− + − − .

31. 1 2

1 2 2

x

x x

−+ =+ +

2 2

2

2

1 22( 1)( 2) 2( 1)( 2)

1 2 2

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

2 4 3 2 4 4

3 7 2 4 4

3 11 6 0

( 3)(3 2) 0

x

x x x x

x x

x x x x x

x x x x x

x x x

x x

x x

−+ + + = + +

+ +

+ + + + = − +

+ + + + = − −

+ + = − −

+ + =+ + =

3 0 or 3 2 0

3 3 2

2

3

x x

x x

x

+ = + == − = −

= −

Check. 3 x = − ?( 3) 1 2

( 3) 1 2 ( 3) 2

2 2

− −+ =

− + − +=

Check.2

3 x = −

( )

( ) ( )

?2

3

2 23 3

1 2

21 2

3 3 2 2

−

− −

−+ =

+ +

− = −

The solutions are2

3 and - .3

−


17/39


252

33. 4 5 6

5 2 3 6

4 5 6

5 2 3( 2)

x

x x x

x

x x x

++ =

− − −+

+ =− − −

2

2

2

4 53( 5)( 2)

5 2

6 3( 5)( 2)

3( 2)

12( 2) 15( 5) ( 5)( 6)

12 24 15 75 30

27 99 30

0 26 69

x x

x x

x

x x

x

x x x x

x x x x

x x x

x x

− − + − − +

= − − − − + − = − +

− + − = + −

− = + −

= − +2

26 26 4(1)(69)

2

26 676 276

2

26 400

2

26 20

2

x

± −=

± −=

±=

±=

26 20 26 20 or

2 2

46 6

2 2

23 3

x x

x x

x x

+ −= =

= =

= =

Check. 23 x = ?4 5 (23) 6

(23) 5 (23) 2 3(23) 6

29 29

63 63

++ =

− − −

=

Check. 3 x = ?4 5 (3) 6

(3) 5 (3) 2 3(3) 6

3 3

++ =

− − −=

The solutions are 3 and 23.

35. 2 2

2 6 01

2 60

( 1) ( 1)( 1)

x

x x x

x

x x x x

+ − =− −

+− =

− + −

2 6( 1)( 1) 0

( 1) ( 1)( 1)

x

x x x

x x x x

+− + − = − + −

2

2

( 1)( 2) 6 0

3 2 6 0

3 2 0

( 2)( 1) 0

2 0 or 1 0

2 1

x x x

x x x

x x

x x

x x

x x

+ + − =

+ + − =

− + =− − =

− = − == =

Check. 2 x = ?

2 2

(2) 2 60

(2) (2) (2) 1

0 0

+− =

− −=

Check. 1 x = ?

2 2

(1) 2 60

(1) (1) (1) 1

+− =

− −

3 60

0 0− =

Division by zero is undefined. The solution is 2.

37. 2

3 5 2

4 62 24

3 5 2

( 4)( 6) 4 6

x x

x x x x

x x

x x x x

+ ++ =

− ++ −+ +

− =− + − +

3 5( 4)( 6)

( 4)( 6) 4

2 ( 4)( 6)

6

x

x x

x x x

x

x x

x

+− + − − + −

+= − +

+

2

2

2

3 5( 6) ( 4)( 2)

3 5 30 2 8

4 27 2 8

0 2 19

x x x x

x x x x

x x x

x x

+ − + = − +

+ − − = − −

− − = − −

= + +

2 4 4(1)19

2

2 72

2

x

− ± −=

− ± −=


39. 2

1 1 1

1 1 11 1 1

1 1 ( 1)( 1)

x x x

x x x

x x x

x x x x

+ − −+ =

− + −+ − −+ =

− + − +

1 1 1( 1)( 1) ( 1)( 1)

1 1 ( 1)( 1)

x x x

x x x x

x x x x

+ − −− + + = − +

− + − +


18/39


253

2 2

2

2

( 1)( 1) ( 1)( 1) 1

2 1 2 1 1

2 2 1

2 1 0

x x x x x

x x x x x

x x

x x

+ + + − − = −

+ + + − + = −

+ = −

+ + =

1 1 4(2)(1)

4

1 7

4

x

− ± −=

− ± −=


41. 2 2 2

3 6 2 1

2 3 2 4

3 6 2 1

( 2)( 1) ( 2)( 1) ( 2)( 2)

x x x

x x x x x

x x x

x x x x x x

+ + −+ =

− − + + −+ + −

+ =− + + + + −

3 6( 2)( 1)( 2) ( 2)( 1) ( 2)( 1)

2 1 ( 2)( 1)( 2)

( 2)( 2)

x x

x x x

x x x x

x

x x x

x x

+ +− + + +− + + +

−= − + +

+ −

2 2 2

2 2

( 2)( 3) ( 2)( 6) ( 1)(2 1)

5 6 4 12 2 1

2 9 6 2 1

9 6 1

8 5

5

8

x x x x x x

x x x x x x

x x x x

x x

x

x

+ + + − + = + −

+ + + + − = + −

+ − = + −− = −

=

=

Check.

( ) ( ) ( )

( )

( )

55 5 ?88 8

2 2 25 5 5 5 58 8 8 8 8

2 13 6

2 3 2 4

16 16

231 231

−+ ++ =

− − + + −

− = −

The solution is5

.8

43.

2

2

2

5 4

2 1 3 2 6 2

5 42 1 3 2 (2 1)(3 2)

x x

x x x x

x x

x x x x

−− =

+ − − −

−− =+ − + −

2

5(2 1)(3 2)

2 1 3 2

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

4

x

x x

x x

x

x x

x x

+ − −+ −

= + −+ −

−

2

2 2

2 2

2

5(3 2) (2 1) 4

15 10 2 4

14 10 2 4

0 3 14 6

x x x x

x x x x

x x x

x x

− − + = −

− − − = −

− − = −

= − +

14 196 4(3)(6)

6

14 196 72

6

14 124

6

14 4 31

6

14 2 31

6

x

± −=

± −=

±=

± ⋅=

±=

7 31

3

±=

Check7 31

3 x

±= using a calculator. The

solutions are7 31

3

±.

45.3

45

3( 5)4 ( 5)

5

4 20 3

4 23

23

4

x

x x

x

x

x

x

=−

− = −−

− ==

=

Check.?

234

34

5

4 4

=−

=

The value of x is23

.4

47.22 12

2 5

x

x x

−=− +

2 2

2

2

2

2 1( 2 5)2 ( 2 5)

2 5

2 4 10 2 1

2 6 11 0

x

x x x x

x x

x x x

x x

−− + = − +

− +− + = −

− + =


19/39


254

4

6 36 4(2)(11)

4

6 36 88

6 52

4

x

±=

± −=

−

± −=

No real solutions. There is no value of . x .

49.5 3

11 1 x x

− = +− +

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

1 1 x x x x

x x

− + ⋅ − = − + + − +

2

2

2

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

1 5 5 3 3

1 8 2

x x x

x x x

x x

− − = + + −

− + = + + −

− + = +

20 8 1 x x = + +

8 64 4(1)(1)

2 x

− ± −=

8 60

2

− ±=

8 4 15

2

− ± ⋅=

8 2 15

2

4 15

− ±=

= − ±

Check 4 15− ± using a calculator. The valuesof x are 4 15− ± .

51.2 5

04 3

x x

x x

− += −

+ −

2 5( 4)( 3) 0 ( 4)( 3)

4 3

x x

x x x x

x x

− + + − ⋅ = + − − + −

2 2

2 2

0 ( 3)( 2) ( 4)( 5)

0 5 6 ( 9 20)

0 5 6 9 20

0 14 1414 14

1

x x x x

x x x x

x x x x

x

x

x

= − − − + +

= − + − + +

= − + − − −

= − −= −

= −

Check.? ( 1) 2 ( 1) 5

0( 1) 4 ( 1) 3

0 0

− − − += −

− + − −=

The value of x is 1.−

53.1 2

05 3

x x

x x

− += −

− +

1 2( 5)( 3) 0 ( 5)( 3)

5 3

x x

x x x x

x x

− +− + ⋅ = − + −

− +

2 2

2 2

0 ( 3)( 1) ( 5)( 2)

0

0

2 3 ( 3 10)

2 3 3 10

0 5 7

7 5

7

5

x x x x

x

x

x x x

x x x

x

x

x

= + − − − +=

=

+ − − − −

+ − − + += +

− =

− =

The x -intercept is7

,0.5

−

55. The student did not simplify the expression

correctly. The error was that the student treatedthe “expression” as an “equation” and multiplied

by the LCD.

57.6 4

1 x x

− =

6 41

6 4

2

x x

x x

x

x

− = ⋅ − ==

Check.?6 4

1(2) (2)

1 1

− ==

The solution is 2.

59. 6 4 6 4 6 4 2

1 1 x x x

x x x x x x x x x

−− − = − − ⋅ = − − =

61. 5 4 3

1

5 1 4 3 1

1 1 1

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

x x x

x x x

x x x x x x

x x x x x

x x x x x x

+ −+

+ + = ⋅ + ⋅ − ⋅ + + +

+ + − − + += = =+ + +

63. 5 4 3

1 x x x + =

+

5 4 3( 1) ( 1)

1 x x x x

x x x

+ + = + +


20/39


255

5( 1) 4 3( 1)

5 5 4 3 3

9 5 3 3

6 2

2 1

6 3

x x x

x x x

x x

x

x

+ + = ++ + = +

+ = += −

= − = −

Check.

( ) ( ) ( )

?

1 1 13 3 3

5 4 3

1

9 9

− − −+ =

+

− = −

The solution is1

.3

−

65. 2 2 2

2 1 4

5 6 4 6

x x

x x x x x

+ +− =

− + − − −

2 1 4

( 3)( 2) ( 2)( 2) ( 3)( 2)

x x

x x x x x x+ +− =

− − − + − +

2 1( 3)( 2)( 2)

( 3)( 2) ( 2)( 2)

4 ( 3)( 2)( 2)

( 3)( 2)

x x x x x

x x x x

x x x x x

+ +− − + −

− − − +

= − − +− +

2 2

2 2

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

4 4 ( 2 3) 4 8

4 4 2 3 4 8

6 7 4 8

2 15

15

2

x x x x x

x x x x x

x x x x x

x x

x

x

+ + − − + = −

+ + − − − = −

+ + − + + = −+ = −

= −

= −

Check.

( )

( ) ( )

( )

( ) ( ) ( )

15 15

2 2

2 2 215 15 15 15 15

2 2 2 2 2

2 1 4

5 6 4 6

16 16

231 231

− −

− − − − −

+ +− =

− + − − −

=

The solution is15

.2

−

67.2 2 2

2 1 4

5 6 4 6

x x

x x x x x+ +− +

− + − − −

2 1 4

( 2)( 3) ( 2)( 2) ( 3)( 2)

x x

x x x x x x

+ += − +

− − − + − +

2 2

2 2

2 2 1 3

( 2)( 3) 2 ( 2)( 2) 3

4 2

( 3)( 2) 2

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

( 2)( 3)( 2)

4 4 ( 2 3) 4 8

( 2)( 3)( 2)

4 4 2 3 4 8

( 2)( 3)( 2)

1

x x x x

x x x x x x

x

x x x

x x x x x

x x x

x x x x x

x x x

x x x x x

x x x

+ + + −−

− − + − + −

−+

− + −

=

+ + − + − + −=− − +

+ − − − + −=

− − +

+ + − + + + −=

− − +

=0 1

( 2)( 3)( 2)

x

x x x

−− − +

69. a. L

d

f is decreasing. This means that the sound

is lower the farther you are away from the

stereo speaker.

b.2

905

90

252250

k

k

k

=

=

=

c.2

2250( ) f d

d =

d.

e.2

2250(8) 35.16 decibels

8 f = ≈


21/39


256

f.2

2

2

225070

70 2250

2250

70

2250

70

5.67 feet

d

d

d

d

d

=

=

=

= ±

≈

71.0 5 1

2 and0 2 1

a a

b b

+ += =

+ +

5 12 2(1 ) 2(1 )

2 1

2 5(1 ) 2(1 )

By substituting 2 for in the second equationwe get the following:

a ab b

b b

b a b a

b a

+= + = +

+

= + = +

5(1 ) 2(1 2 )

5 5 2 4

3

b b

b b

b

+ = +

+ = +

= −

By substituting 3 in the first equation for

we get:b

−

23

6

a

a

=−

− =

The value for a is 6− and the value for b is 3.−


Homework 8.6

1. a. ( ) 1250 350C n n= +

b.1250 350

( ) n

M nn

+=

c.1250 350(30)

(30) $391.6730

M +

= ≈

d.1250 350

400

400 1250 350

50 1250

25

n

n

n n

n

n

+=

= +

=

=

The minimum number of students needed togo on the trip is 25.

3. a. ( ) 500 50T n n= +

b.500 50

( ) n

M nn

+=

c.500 50(270)

(270) $51.85270

M += ≈

If 270 people attend the reunion, the meancost per person is $51.85.

d.500 50

60

60 500 50

10 500

50

n

n

n n

n

n

+=

= +

=

=

For the mean cost per person to be $60, 50

people would have to attend the reunion.

e.

f. As n gets very large, ( ) M n decreases but

never drops below 50. This makes sense interms of the restaurant fees because eachperson pays $50 plus an equal share of $500– the cost for the band. So the more people

who attend (i.e., asn

gets larger), the closerthe mean cost per person gets to $50.

5. a. ( ) 90000 7000C n n= +

b.90000 7000

( ) n

B nn

+=

c.90000 7000

( ) 2000

90000 7000 2000

1

90000 7000 2000

90000 9000

nP n

n

n n

n n

n n

n n

n

n

+= +

+= + ⋅

+= +

+=


22/39


257

d.90000 9000(40)

(40) 1125040

P +

= =

If the manufacturer produces and sells 40cars, it would have to charge $11250 foreach car in order to make a profit of $2000per car.

e. As the values of n get very large, ( )P n gets

close to 9000. If a very large number of carsare produced and sold, the price per car canbe just a little more than $9000 to insure aprofit of $2000 per car.

7. a. The graph below represents a scatter plot ofthe data in the table.

The data appear to follow a linear model.Using linear regression we get:

( ) 221.52 1063.14 B t t = −

b. ( ) ( ) ( )

0.14 4.52 0.083 4.60

0.223 9.12

E t W t M t

t t

t

= += + + += +

c.( ) 221.52 1063.14

( )( ) 0.223 9.12

B t t A t

E t t

−= =

+

d.( )221.52 38 1063.14

(38) 418.020.223(38) 9.12

A−

= ≈+

The mean amount of money that will bespent per student in 2008 is about $418.02.

e.221.52 1063.14

4300.223 9.12

(0.223 9.12) 430

221.52 1063.14 ( 0.223 9.12)

0.223 9.12

95.89 3921.6 221.52 1063.144984.74 125.63

39.68

t

t

t

t t

t

t t t

t

−=

++ ⋅

−= +

+

+ = −=≈

About 40 years from 1970, or 2010, themean amount of money spent per studentwill equal $430.

9. Usingd

t s

= gives the following:

851.4

60t = ≈ hours

11. a.295

4.565

t = ≈ hours

b.295

( )65

T aa

=+

c.295 295

(5) 4.265 5 70

T = = ≈+

This means that if the student drives 70mph, which is 5 mph over the speed limit,the driving time is about 4.2 hours.

d.295

465

295(65 ) 4 (65 )

65

260 4 295

4 35

358.75

4

a

a aa

a

a

a

=+

+ ⋅ = ++

+ ==

= =

This result means that the student must drive8.75 mph over the speed limit or 73.75 mphin order to drive from Chicago to St. Louis

in 4 hours.

13. a.114 144

3.870 65

t = + ≈ hours

b.114 144

( )70 65

114 65 144 70

70 65 65 70

114 7410 144 10080

( 70)( 65) ( 70)( 65)

258 17490

( 70)( 65)

T aa a

a a

a a a a

a a

a a a a

a

a a

= ++ +

+ + = + + + + + + +

= ++ + + +

+= + +

c.114 144

(0) 3.80 70 0 65

T = + ≈+ +

hours

It is the same as the result from part a.


23/39


258

d.114 144

(10)10 70 10 65

114 1443.35

80 75

T = ++ +

= + ≈

This result means that if the student drives10 mph over the speed limit (i.e. 80 mph in

Michigan and 75 mph in Indiana) the trip

will take about 3.35 hours.

e. (0) (10) 3.8 3.35 0.45T T − = − =

This means the student will only save about

a half an hour in driving time by driving 10

mph over the speed limit.

15. a.253 410

( )75 70

T aa a

= ++ +

b. 2

2

2

8( 145 5250) 253( 70) 410( 75)

8 1160 42000 253 17710 410 30750

8 1160 42000 663 48460

a a a a

a a a a

a a a

+ + = + + +

+ + = + + +

+ + = +

28 497 6460 0a a+ − =

497 247009 4(8)( 6460)

16

497 453729

16

x− ± − −

=

− ±=

The only answer that makes sense in this

case is497 453729

11.0416

− +

≈ mph

The student would have to drive about 11

mph over the speed limit to make the trip in

8 hours.

In the table the value 11 for x produces the

closest value for y of 8.

Homework 8.7

1. w kt =

3.k

yt

=

5.3

k B

t =

7. w varies inversely at t

9. A varies directly as the cube of x

11. y varies inversely as 2u

13. y kx=

To find k solve the following equation:

6 (2)

3

k

k

=

=

3 y x=

15.k

y x

=


3 5

15

k

k

=

=

15 y

x=

17. y kx=


5.4 (2.7)

2

k

k

=

=

2 y x=

19. k pt

=


3.95.8

22.62

k

k

=

=

22.62 p

t =

21.2

A kr =

To find k solve the following equation:2

2

9 3

9 9

k

k

k

A r

π

π

π

π

= ⋅

=

=

=


24/39


259

23. 2log ( ) H k t =


2

2

12 log 8

12 3

4

4log ( )

k

k

k

H t

=

= ⋅

=

=

25.3

k w

t =


3

3

78

72

14

14

k

k

k

wt

=

=

=

=

27.2r k

A =


3

7

8 2

7

8 8

7

72r

k

k

k

A

=

=

=

=

29. Let c be the cost of tuition and h be the number

of credit hours a student takes.

c kh=

Let 1185c = and 15h = .

1185 (15)

79

k

k

=

=

Therefore, 79c h=

79(12) $948c = =

31. Let t be the tension in the string and r be theradius of the circle.

k t

r =

Let 80t = and 60r = .

8060

4800

k

k

=

=

Therefore,4800

t r

=

480096

50t = = newtons

33. Let d be the distance an object falls and t be the

time in motion.2

d kt =

Let 144.9d = and 3t = .2144.9 3

144.9 9

16.1

k

k

k

= ⋅

=

=

Therefore, 216.1d t = 216.1(3.4) 186.116d = = feet

35. Let i be the intensity of radiation and d be thedistance from the machine.

2

k i

d =

Let 90i = and 2.5d = .

290

2.5

906.25

562.5

k

k

k

=

=

=

Therefore,2

562.5i

d =

2

2

2

562.545

45 562.5

12.5

3.53 meters

d

d

d

d

=

=

=

≈

37. a.2

2

2

( )

302.6

30 6.76

202.8

202.8( )

k I f d

d

k

k

k

I f d d

= =

=

=

=

= =


25/39


260

b.2

2

2

2

202.8(1) 202.8

1

202.8 202.8(2) 50.7

42

202.8 202.8(3) 22.5393

202.8 202.8(4) 12.675

164

f

f

f

f

= =

= = =

= = ≈

= = =

The values are the intensities in2

/ W m of

1,2,3, and 4 km, respectively.

c.

The values of I decrease as d increases. In

terms of the situation, it means the intensity

decreases the farther the television signal is

from the transmitter.

d. The values of ( ) f d get very close to 0 for

extremely large values of .d This means

that the signal will be almost non-existent if

the television signal is too far away from the

transmitter.

39. a.2

( ) k

w f d d

= =

2200

4

20016

3200

k

k

k

=

=

=

Therefore,2

3200( )w f d

d = = .

b. If sea level is about 4 thousand miles from

the center of the Earth, then 1 thousand

miles above the surface would be a total of 5

thousand miles from the center of the Earth.

2

3200128 pounds

5w = =

c.2

2

32001

3200

d

d

=

=

6.573200 5 thousand

or 56570 miles from the center of the Earth

d = ≈

d.2

3200(239) 0.056 pounds

239 f = ≈

Written response

e. Written response

41, a. T kd =

Let 3T = and 3313d = .

3 (3313)

0.000906

k

k

=

=

Therefore, 0.000906T d = .

b. 4 0.000906

4415 feet

d

d

=

≈

c. For every additional foot away the lightning

strike, it takes another 0.000906 seconds to

hear the thunder.

d. Written response

43. a.

b. The model should be of the form ( ) k

f d d

= .

To find k take the average of all products

dh from the table.

10 16 20 7.3 30 4.8 40 3.8 50 3 60 2.5 70 2

7

148.86

k ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅

≈

=

So148.86

( ) f d d

=

c. This makes sense in this case because the

farther you are from the garage (i.e. the

bigger d is) the smaller the apparent height

( ) f d is.


26/39


261

d.148.86

(100) 1.4886 inches100

f = =

e.148.86

(1) 148.86 inches

1

f = =

44. a.

b. The model should be of the form ( ) k

F L L

= .

To find k take the average of all products

LF from the table.The sum of the products is 36467.715.

Divide this sum by 13 (the total number of

notes in the table) to get the average.

36467.7152805.21

13≈

So2805.21

( )F L L

=

The graph below shows the equation

graphed along with the scattergram of the

data. By inspection, it appears the model

fits the data extremely well.

c. F varies inversely as L

d. 2805.21

(7.58) 370.17.58

F = ≈ hertz

e.

121

2

2

2805.21( 1)

2805.21( 2) 2805.21

2805.212805.21 2

a

eq F a

eq F aa

a

=

= = ÷

= ⋅ =

Equation 2 is 2 times equation 1. So when

we halved the effective length the frequency

doubled.

47. a.2

k I

d =

b.2

Id k =

c. In list 3 are the values of2

d I .

A reasonable value for k is the average of

all values in list 3. Add all values in list 3

and divide this sum by 8.

sum of list 3 34276.44284.55

8 8k = = ≈

d.2

4284.55 I

d =

e.

The model fits the data very well.

f.

2

2

4284.55

0.167 mW/cm160 I = ≈

g. Written response

49. ( ) 5

5

f L L

k

==

51. 2

( )

2

f nn

k

=

=

53. ( ) 2

2

f r r

k

π

π

=

=

55. a. f is an increasing function

The number of CD’s sold increases as more

money is spent on advertising.


27/39

Chapter 8 Review Exercises SSM : Intermediate Algebra

262

b. The number of CD’s sold does not vary

directly as the amount of money spent on

advertising. “Varies directly” usually

implies a linear relationship. A linear model

would not fit this data well.

c. Written response

57. false; As a person gets older, the person doesn’t

necessarily keep getting taller. At some point we

quit growing.

59. true; Coffee will get cooler the longer the time

passes since it has been poured.

61. y kx=

Solving this equation for x we get the following:

y x

k

=

Yes, it follows that x varies directly as y. The

variation constant is1

.k

Chapter 8 Review Exercises

1.2

4 49 0

(2 7)(2 7) 0

x

x x

− =

− + =

2 7 0 or 2 7 0

2 7 2 7

7 7 2 2

x x

x x

x x

− = + =

= = −

= = −


7 7 and .

2 2−

2.2

12 13 35 0 x x+ − =

13 169 4(12)( 35)

24

13 169 1680

24

13 1849

24

13 43

24

Documents

Lehmann IA SSM Ch8