Exercise Set 6.9 153
26. V=k
P
28. P=k
W
440 = k
2.4
30. y=kx2
6=k·32
32
54 = k
34. y=kx
z
z
36. y=k·xz
w
38. y=k·xz
w2
b) 25 = k·53
1
5=k
200 = k·602
1
18 =k
44. V=kT
P
110 = k
46. D=kAv
222 = k·37.8·40
37
48. 125y3−27 = (5y−3)(25y2+15y+9)
154 Chapter 6: Rational Expressions and Equations
50. 2(3 −4t)=−66
3−4t=−33
−4t=−36
t=9
52. 2x−7=−17
kd2=πd
22
r=d
2
Chapter 6 Vocabulary Reinforcement
4. Expressions that have the same value for all allowable re-
placements are called equivalent expressions.
7. When a situation translates to an equation described by
y=k
8. When a situation translates to an equation described by
Chapter 6 Concept Reinforcement
3. True; −(2 −x)=−2+x=x−2.
Chapter 6 Study Guide
1. c2−11c+30=0
2. 2x2−2
4x2+24x+20 =2(x2−1)
4(x2+6x+5)
=x−1
2(x+5)
4. b2+3b−28
b2+5b−24 ÷b−4
b−3=b2+3b−28
b2+5b−24 ·b−3
b−4
b+8
5. 5
18 +7
60
5
=71
180
7. x
x−4+2x−4
4−x=x
x−4+2x−4
4−x·−1
−1
=x−2x+4
x−4
8. x
x2+x−2−5
x2−1
LCM is (x+2)(x−1)(x+1)
=x
(x+ 2)(x−1) ·x+1
x+1−5
(x+ 1)(x−1) ·x+2
x+2
9.
2
5−1
y
3
(The LCM of 5, y,and3is15y).
=
3
y−1
3
·15y
15y
10. 1
x=x(3 −x)·2
3−x
3−x=2x
11. y=kx
Substitute 2 for xand find y.
12. y=k
y=225
x=225
10 =22.5
Chapter 6 Review Exercises
solve:
x−6=0
3. x+5
x2−36
x=−6or x =6
4. x2−3x+2
(x+ 6)(x−5) = 0
5. −4
(x+2)
2
To determine the numbers for which the rational expres-
7. 4x2−8x
4x2+4x=4x(x−2)
4x(x+1)
8. 14x2−x−3
2x2−7x+3 =(2x−1)(7x+3)
(2x−1)(x−3)
9. (y−5)2
y2−25 =(y−5)(y−5)
(y+ 5)(y−5)
10. a2−36
10a·2a
a+6 =(a2−36)(2a)
10a(a+6)
11. 6t−6
2t2+t−1·t2−1
t2−2t+1
12. 10−5t
3÷t−2
12t=10 −5t
3·12t
t−2
3(t−2)
=5(−1)(t−2)·3·4t
3(t−2) 2−t=−1(t−2)
x2−1÷2x3
x2−2x+1 =4x4
x2−1·x2−2x+1
2x3
=4x4(x2−2x+1)
=2x(x−1)
x+1
4a−8=4(a−2)
LCM = 4(a−2)
16. y2−y−2=(y−2)(y+1)
Chapter 6 Summary and Review: Review Exercises 157
18. 3
3x−9+x−2
3−x=3
3(x−3) +x−2
3−x
=3
=3−3x+6
3(x−3)
19. 2a
a+1+4a
a2−1
=2a(a+1)
(a+ 1)(a−1)
20. d2
d−c+c2
c−d=d2
d−c+c2
c−d·−1
−1
21. 6x−3
x2−x−12 −2x−15
x2−x−12 =6x−3−(2x−15)
x2−x−12
=6x−3−2x+15
22. 3x−1
2x−x−3
x,LCM is 2x
=3x−1−2x+6
2x
24. 1
x2−25 −x−5
x2−4x−5
=1
(x+ 5)(x−5) ·x+1
x+1−x−5
(x−5)(x+1)·x+5
x+5
25. 3x
x+2−x
x−2+8
x2−4
26.
1
z+1
1
LCM of the denominators is z2.
z2−1z2
=
1
z·z2+1·z2
1
27.
c
d−d
c
1
c+1
d
28. 3
y−1
4=1
y,LCM = 4y
29. 15
x−15
x+2 =2,LCM = x(x+2)
15
Chapter 6 Summary and Review: Review Exercises 159
30. Familiarize. Let t= the time the job would take if the
crews worked together.
Translate. We use the work principle, substituting 9 for
4t+3t=36
7t=36
31. Familiarize. Let r= the speed of the slower plane, in
mph. Then r+ 80 = the speed of the faster plane. We
organize the information in a table.
r(r+ 80) ·950
r=r(r+ 80) ·1750
r+80
950(r+ 80) = 1750r
950r+76,000 = 1750r
32. Familiarize. Let r= the speed of the slower train, in
km/h. Then r+ 40 = the speed of the faster train. We
60 = rt,
70=(r+ 40)t
Since the times are the same, we solve each equation for t
and set the results equal to each other.
160 Chapter 6: Rational Expressions and Equations
train travels 70 km in 70/280, or 1/4 hr. Since the times
33. Familiarize. We can translate to a proportion, letting d=
the number of defective calculators that can be expected
in a sample of 5000.
34. a) Let x= the number of cups of onion that would be
used. Then we can write and solve a proportion.
6
13 =x
2
5
7=3
c
c=21
5,or 41
5
35. Familiarize. The ratio of blue whales tagged to the total
tagged to blue whales checked is 20
400.
Translate. Assuming the two ratios are the same, we can
translate to a proportion.
36. We write a proportion and solve it.
3.4
8.5=2.4
x
37. y=kx
38. y=k
Chapter 6 Summary and Review: Discussion and Writing Exercises 161
39. y=kxz
40. y=kx
41. t=k
rtvaries inversely as r.
of 1400 kL per minute.
42. N=ka N varies directly as a.
87 = k·28 Substituting
43. P=kC2Pvaries directly as the square of C.
P= 500
The circuit expends 500 watts of heat when the current is
44. 3x2−2x−1
x=0 or 3x+1=0
45. 1
x−5−1
x+5,LCM is (x−5)(x+5)
=1
x−5·x+5
x+5−1
x+5·x−5
x−5
46. 2a2+5a−3
a2·5a3+30a2
2a2+7a−4÷a2+6a
a2+7a+12
=2a2+5a−3
47. A+B
B=C+D
D
Chapter 6 Discussion and Writing Exercises
162 Chapter 6: Rational Expressions and Equations
2. Graph each side of the equation and determine the number
4. The larger the average gain per play, the smaller the num-
Chapter 6 Test
1. 8
2x
To determine the numbers for which the rational expres-
sion is not defined, we set the denominator equal to 0 and
2. 5
x+8
To determine the numbers for which the rational expres-
sion is not defined, we set the denominator equal to 0 and
3. x−7
x2−49
To determine the numbers for which the rational expres-
4. x2+x−30
x2−3x+2
To determine the numbers for which the rational expres-
sion is not defined, we set the denominator equal to 0 and
x−1=0 or x −2=0
5. 11
(x−1)2
1.
6. x+2
2
Since the denominator is the constant 2, there are no re-
placement numbers for which the expression is not defined.
7. 6x2+17x+7
x+3
8. a2−25
6a·3a
a−5=(a2−25)(3a)
6a(a−5)
9. 25x2−1
9x2−6x÷5x2+9x−2
3x2+x−2
=(5x+ 1)(x+1)
3x(x+2)
10. y2−9=(y+ 3)(y−3)
11. 16 + x
x3+7−4x
x3=16 + x+7−4x
x3=23 −3x
x3
12. 5−t
13. x−4
x−3+x−1
3−x=x−4
x−3+x−1
3−x·−1
−1
=−3
x−3
14. x−4
x−3−x−1
3−x=x−4
x−3−x−1
3−x·−1
−1
15. 5
t−1+3
t,LCD is t(t−1).
=8t−3
t(t−1)
16. 1
x2−16 −x+4
x2−3x−4
17. 1
x−1+4
x2−1−2
x2−2x+1
=1
x−1+4
(x+ 1)(x−1) −2
(x−1)(x−1),
=(x+ 1)(x−1)+4(x−1) −2(x+1)
(x+ 1)(x−1)(x−1)
18. We multiply the numerator and the denominator by the
LCM of the denominators, y2.
9−1
9−1
=
y2·y2
3·y2−1
y·y2
19. 7
y−1
3=1
4,LCM is 12y
12y7
y−1
3=12y·1
4
20. 15
x−15
x−2=−2,LCM is x(x−2)
15x−30 −15x=−2x2+4x
21. We substitute to find k.
y=kx
in the equation of variation.
y=2x
22. We substitute to find k.
y=kx
1.5=k·3
0.5=k
The equation of variation is y=0.5x.
23. We substitute to find k.
y=18
x
24. We substitute to find k.
y=22
x
25. Q=kxy
26. Familiarize and Translate. The problem states that we
have direct variation between the variables dand t. Thus,
an equation d=kt,k>0, applies.
Solve.
a) First find an equation of variation.
travel in 2 hr.
d= 120t
27. Familiarize and Translate. Let T= the time required
to do the job and let M= the number of concrete mixers
used. We have inverse variation between Tand Mso an
equation T=k
M,k>0, applies.
Solve.
a) First we find an equation of variation.
28. Familiarize. We can translate to a proportion, letting
d= the number of defective spark plugs that would be
The ratios are the same, so the answer checks.
29. Familiarize. The ratio of zebras tagged to the total zebra
Translate. Assuming the two ratios are the same, we can
translate to a proportion.
Zebras tagged
originally
Tagged zebras
caught later
−→ ←−
15
P=6
20
−→ ←−
Zebra
population
Zebras caught
later
Solve.
t
20 +t
30 =1,LCM is 60
31. Familiarize. Let r= Marilyn’s speed, in km/h. Then
Distance Speed Time
Marilyn 225 r t
166 Chapter 6: Rational Expressions and Equations
Since the times are the same, we solve each equation for t
32. We write a proportion and solve it.
12
9=20
x
33. 2
x−4+2x
x2−16=1
x+4,
34. Familiarize. Let r= the number of hours it would take
Solve.
40r+120=7r2+42r
0=7r2+2r−120
0=(7r+ 30)(r−4)
be a solution. If r= 4, then r+ 6 = 4 + 6 = 10. In 26
7hr,
or 20
7hr, then the portion of the job done is
Cumulative Review Chapters 1 – 6 167
35. 1+ 1
=1+ 1
a+1
2a+1
Cumulative Review Chapters 1 – 6
2. x3−2x2+x−1
3. If a free-range egg has 35% less cholesterol than a factory
farm egg, then it has 65% of the cholesterol of a factory
We convert 65% to decimal notation and multiply.
4. First we find the amount of decrease.
5. Let a= the amount originally borrowed. At a 6% rate, the
amount of simple interest owed after 1 yr is 6%a,or0.06a.
6. Familiarize. Let r= the speed of the faster car, in mph.
Then r−10 = the speed of the slower car. We organize
the information in a table.
Distance Speed Time
Slower car 75 r−10 t
and set the results equal to each other.
105 = rt,sot=105
168 Chapter 6: Rational Expressions and Equations
Then we have
105
r(r−10) ·105
r=r(r−10) ·75
r−10
If r= 35, then r−10 = 35 −10 = 25.
Check. If the speeds are 35 mph and 25 mph, then one
7. Familiarize. Let s= the length of a side of the original
enlarged square. Recall that the area of a square with side
original
square
plus
enlarged
square
is 452 ft2
↓↓ ↓↓↓
2s2+4s−448 = 0
2(s2+2s−224) = 0
s2+2s−224 = 0
and the area of a square with a side of 16 ft is (16 ft)2,or
256 ft2. The sum of the area is 196 ft2+256 ft2, or 452 ft2,
8. a) M=kB
b) We use the equation to find the value of Mwhen
B= 192 lb.
9. x2−3x3−4x2+5x3−2
10. 1
2x−3
8x−2
3+1
4x−1
3
11. 2x3
12. We begin by getting a common denominator in the numer-
ator and in the denominator.
=
10
2x+7
2x
4x−6
17x2
13. (5xy2−6x2y2−3xy3)−(−4xy3+7xy2−2x2y2)
Cumulative Review Chapters 1 – 6 169
14. (4x4+6x3−6x2−4) +(2x5+2x4−4x3−4x2+3x−5)
15. 2y+4
21 ·7
y2+4y+4 =(2y+ 4)(7)
21(y2+4y+4)
16. x2−9
x2+8x+15÷x−3
2x+10 =x2−9
x2+8x+15·2x+10
x−3
17. x2
x−4+16
4−x=x2
x−4+16
4−x·−1
−1
18. 5x
x2−4−−3
2−x=5x
(x+ 2)(x−2) −−3
2−x
=5x
(x+ 2)(x−2) −3
x−2
=5x−3(x+2)
(x+ 2)(x−2)
19. We use FOIL.
21. (2x3+ 1)(2x3−1) = (2x3)2−12=4x6−1
22. 9a2+52a−12
these factors to split the middle term and then factor by
grouping.
9a2+52a−12=9a2+54a−2a−12
25. x−[x−(x−1)]=2
x−[x−x+1]=2
26. 2x2+7x=4
2x=1 or x =−4
x=1
27. x2=10x
x2−10x=0
28. 3(x−2) ≤4(x+5)
29. 5x−2
4−4x−5
3=1,LCM is 12
−x+14=12
The solution is 2.
30. t=ax +ay
t=a(x+y)
t
34. m=y2−y1
x2−x1
=
3
4−3
4
−4−(−7) =0
−4+7 =0
3=0
36. y-intercept:
15 −40 ·0=−120y
x-intercept:
15 −40x=−120 ·0
37. The graph of y= 25 is a horizontal line with y-intercept
38. The graph of x=−1
4is a vertical line with x-intercept
−1
4,0. There is no y-intercept.
39. x=−3
40. y=−3
41. 3x−5y=15
First we will find the intercepts. To find the y-intercept
3x
5y 15
x
y
Cumulative Review Chapters 1 – 6 171
The x-intercept is (5,0).
Plot these points and draw the line containing them.
5
42. 2x−6y=12
First we will find the intercepts. To find the y-intercept
−6−6y=12
−6y=18
43. y=−1
3x−2
We find some ordered pairs that are on the graph.
We plot these points and draw the line.
x y
44. x−y=−5
we let x= 0 and solve for y.
0−y=−5
−y=−5
We can find a third point as a check. When x=−3, we
have
45.
1
x+x
172 Chapter 6: Rational Expressions and Equations