Section 2.4 Formulas and Percents
23.
1 for
2
A
ha b b
1
22
2
2
Ahab
Ahab





24.
1 for
2
A
ha b a
A
1
22
2
2
Ahab
Ahab





25. for
A
xByC x
A
xByByCBy
A
xCBy
A
xCBy
A
A
CBy
x
A


This is the standard form of the equation of a line.
26. for
A
xByC y
27.
; 3% 0.03, 200APBP B 
0.03 200
6
APB
A
A

3% of 200 is 6.
18% of 40 is 7.2.
30.
; 16% 0.16, 90APBP B 
0.16 90 14.4
APB
A

80.4
80.4
0.4 0.4
20
A
PB
B
B
B

8 is 40% of 20.
33.
; 40.8, 24% 0.24APBA P 
40.8 0.24
APB
B

Chapter 2 Linear Equations and Inequalities in One Variable
34.
; 51.2, 32% 0.32APBA P 
A
35.
; 3, 15APBA B
A
PB
36.
;18;90APBA B
A
A
PB
37.
; 0.3, 2.5APBA B
APB
38.
;0.6,7.5APBA B
0.6 7.5
APB
P

39. The increase is
85 3.
40. The increase is
95 4.
A
PB
34
A
PB
P

28
28
P
P

or
ab ab
yy
xx
ab ab



Section 2.4 Formulas and Percents
45.
A
5
555
yabx
yabx
 
 
46.
A
8
888
8
yabx
yabx
yabx
 
 
 
47.
ycxdx
ycdx


48.
A
ycxdx
ycdx


49.
yAxBxC
yABxC
yC ABxCC
yC ABx

 
 
 
50.
yAxBxC
yC AxBxCC

  
51. a.
for
3
xyz
A
z

b. 90, 86, 88Axy
52. a.
for
4
xyzw
A
w

xyzw


b. 4;76,78,79wAxyzx y z 
4
480 76 78 79
87
wAxyz
w
w


Chapter 2 Linear Equations and Inequalities in One Variable
53. a.
for drt t
drt
rr
54. a.
932 for
5
FC C
9
55 32
5
FC




b.
5160
;59
9
5160
9
F
CF
F
C

55.
0.29 1800 522
522 workers stated that religion is the most taboo
topic to discuss at work.
56.
0.14 1800 252
252 workers stated that politics is the most taboo
57. a. This is the equivalent of asking: 5.85 is 5% of
what number?
A
PB

332,960 1.8
1.8 1.8
184,978
B
B
The average income in 1975, for the richest 5%
117
B
117 million households in the United States.
b. This is the equivalent of asking: $16,095 is
$15,042.
59. This is the equivalent of asking: 540 is what% of
1500?
540 1500
APB
P


Section 2.4 Formulas and Percents
60. This is the equivalent of asking: 105 is what% of
1500?
105 1500
APB
P


61. ; 7500, 60,000APBA B
7500 60,000
APB
P

62. This question is equivalent to, “225,000 is what
percent of $500,000?”
225,000 500,000
225,000 500,000 0.45
500,000 500,000
APB
P
PP


The charity has raised 45% of the goal.
A
65. a. The sales tax is 6% of $16,800.
0.06 16,800 1008
The sales tax due on the car is $1008.
66. a. The sales tax is 7% of $96.
0.07 96 6.72
The sales tax due on the graphing calculator is
$6.72.
0.12 860 103.20
The discount amount is $103.20.
b. The sale price is the regular price minus the
discount amount:
b. The sale price is the regular price minus the
discount amount.
$16.50 $6.60 $9.90
The sale price is $9.90.
69. The decrease is $840 $714 = $126.
126 840
A
PB
P


This is a 0.30 = 30% decrease.
Chapter 2 Linear Equations and Inequalities in One Variable
71. Investment dollars decreased in year 1 are
0.30 $10,000 $3000. This means that $10,000
$3000 = $7000 remains. Investment dollars
72. No; the first sale price is 70% of the original amount
and the second sale price is 80% of the first sale
price. The second sale price would be obtained by
the following computation:
73. – 74. Answers will vary.
75. makes sense
76. does not make sense; Explanations will vary.
Sample explanation: Sometimes you will solve for
one variable in terms of other variables.
79. false; Changes to make the statement true will vary.
A sample change is: If 0,
ax b
 then
ax b

b
83. 100 for
QC
C
100
100
M
CQ C C
CQ M
84. 520816
5 208 8 168
320 16
3 2020 1620
336
336
xx
xxxx
x
x
x
x


 
  


Mid-Chapter Check Point
85.
52 3 1 46 2
yy
 
10 15 1 24 8
10 16 24 8
10 16 8 24 8 8
yy
yy
yy yy
 

   
86.
0.3 1 0.3 1 0.3 0.7
xxxx x x
 
87.
13 7
x
Mid-Chapter Check Point – Chapter 2
1. Begin by multiplying both sides of the equation by
4, the least common denominator.
12
24
xx

2.
542 57
5 4242 5742
515
515
55
x
x
x
x



4.
0.06 140
8.4
APB
A
A


8.4 is 6% of 140.
1130
30
x
x
 
The solution set is
30 .
6.
13 5 423
yy

13 15 812
yy

Chapter 2 Linear Equations and Inequalities in One Variable
7.
2
Srh
π
9.
35
3
52 4
yy y
 
To clear fractions, multiply both sides by
the LCD, 20.
35
20 20 20 20 3
52 4
43 10 55 60
yy y
yy y
  
 
  
  
 
10.
2.4 6 1.4 0.5(6 9)
2.4 6 1.4 3 4.5
2.4 6 4.4 4.5
xxx
xxx
xx
 
  
 
To clear decimals, multiply both sides by 10.
10(2.4 6) 10(4.4 4.5)
xx
 
11.
576 2423
zz z
  
or
Ax By By C By
Ax C By
Ax C By
AA
CBy ByC
x
A
A



13.
673331
yyy
 
13
10 10 3 10 10 1
25
530610
55306510
30 10
xx
xx
xx xx
x
 
 
 
 



Section 2.5 An Introduction to Problem Solving
16.
32
23
4
mm

32
4423
4
5
mm

The solution set is
6.
5



17. The increase is 50 40 = 10.
A
PB

18.
12 4 8 4 4 5 2
20 8 20 8
ww w
ww
 
 
19. a.
582
2
5(14) 82
2
Ba
B
 
 
b.
582
2
5
22 82
2
Ba
a
 
 
2.5 Check Points
1. Let x = the number.
6468
644684
672
x
x
x

 
Let
18x
the median starting salary, in
thousands of dollars, for computer science majors.
(18)100
18 59
xx
x
 

The average salary for English majors is $41
thousand and the average salary for computer
science majors is
$41 $18 $59.
3. Let x = the page number of the first facing page.
Chapter 2 Linear Equations and Inequalities in One Variable
4. Let x = the number of eighths of a mile traveled.
20.25 10
2 2 0.25 10 2
x
x

  
5. Let x = the width of the swimming pool.
Let
3x
the length of the swimming pool.
22
320 2 3 2
Plw
xx

 
6. Let x = the original price.
the reduction
(40% of
Original the reduced
original price)
price price, $564
minus is
0.4 564xx
0.4 564
xx

2.5 Concept and Vocabulary Check
1.
46x
2.
215x
2.5 Exercise Set
1.
60 410
x

The number is 64.
3.
23 214
23 23 214 23
237
x
x
x

 
77
18
x
The number is 18.
6.
8272
x
19
19 19 5
19
95
x
x



The number is 95.
x
Section 2.5 An Introduction to Problem Solving
10.
53 59
354
18
x
x
x

The number is 18.
11.
57178
5771787
x
x

 
12.
68298
6306
51
x
x
x

The number is 51.
15.
2436
2836
228
14
x
x
x
x


The number is 14.
330
10
x
x
The number is 10.
19. 3434
5
330
x
x

4
312
4
348
16
x
x
x
Americans will spend 9 years watching TV and 28
years sleeping.
22. Let xthe number of years spent eating.
Let 24xthe number of years spent sleeping.
(24)32
xx
 
Chapter 2 Linear Equations and Inequalities in One Variable
23. Let xthe average salary, in thousands, for an
American whose final degree is a bachelor’s.
Let 270x the average salary, in thousands, for
an American whose final degree is a master’s.
(2 70) 173
xx

24. Let xthe average salary, in thousands, for an
American whose final degree is a bachelor’s.
Let 245x the average salary, in thousands, for
an American whose final degree is a doctorate.
(2 45) 198
245198
xx
xx


25. Let x = the number of the left-hand page.
Let x + 1 = the number of the right-hand page.
1629
1629
xx
xx


26. Let x = the number of the left-hand page.
Let x + 1 = the number of the right-hand page.
1 525
xx

27. Let xthe first consecutive odd integer (Babe
Ruth).
Let 2x the second consecutive odd integer
(Roger Maris).
(2)120
xx

Let 2x the second consecutive even integer
(Babe Ruth).
(2)118
2118
22118
xx
xx
x



200 0.15 200 320 200
0.15 120
0.15 120
0.15 0.15
x
x
x

180 0.25 180 395 180
0.25 215
0.25 215
x
x
x

Section 2.5 An Introduction to Problem Solving
31. Let xthe number of years after 2014.
37,600 1250 46,350
1250 8750
x
x

32. Let xthe number of years after 2014.
11.3 0.2 12.3
0.2 1
x
x

33. Let x = the width of the field.
Let 4x the length of the field.
22
500 2 4 2
Plw
xx

 
34. Let x = the width of the field.
Let 5x the length of the field.
22
288 2 5 2
Plw
xx

 
35. Let x = the width of a football field.
Let 200x the length of a football field.
22
160
200 360
Plw
x
x


A football field is 160 feet wide and 360 feet long.
86 2 26 2
86 4 26
60 4
15
xx
x
x
x


34(3)60
312 60
15 60
4
xx
xx
x
x


Chapter 2 Linear Equations and Inequalities in One Variable
38. As shown in the diagram,
let x = the length of a shelf and x + 3 = the height of
the bookcase,
4 shelves and 2 heights are needed.
39. Let x = the price before the reduction.
0.20 320
xx

40. Let x = the price before the reduction.
0.30 98
xx

41. Let x = the last year’s salary.
0.08 50,220
1.08 50,220
xx
x

42. Let x = the last year’s salary.
0.09 42, 074
xx

43. Let x = the price of the car without tax.
0.06 23,850
1.06 23,850
xx
x

160
x
The nightly cost without tax is $160.
35 35
11
x
63 1071
63 63
17
x
x
It took 17 hours of labor to repair the sailboat.
52. makes sense
53. makes sense
Section 2.5 An Introduction to Problem Solving
55. false; Changes to make the statement true will vary.
A sample change is: This should be modeled by
10 160.x
59. Let x = the number of inches over 5 feet.
100 5
135 100 5
135 100 100 100 5
Wx
x
x



60. Let x = the number of minutes.
Note that $0.55 is the cost of the first minute and
$0.40( 1)x is the cost of the remaining minutes.
0.55 0.40 1 6.95
x

61. Let x = the woman’s age.
Let 3x = the “uncle’s” age.
3202 20
xx
 
62. Let x = weight of unpeeled bananas.
The information in the cartoon translates into the
equation.
77
xx
877
87 777
7
xx
xx x x
x


The unpeeled banana weighs 7 ounces.
20
x

Check:
420 16
5
420 16

6181
611811
68
yy
yy
yy
 
 
The solution set is
0.
Chapter 2 Linear Equations and Inequalities in One Variable
65. 1 for
3
Vlwhw
1
3
Vlwh
66.
1
2
1
2
30 12
30 6
30 6
66
5
A
bh
h
h
h
h

2.6 Check Points
1.
24, 4Ab
1
2
A
bh
2. Use the formulas for the area and circumference of
a circle. The radius is 20 ft.
2
2
(20)
Ar
A
π
π
126
The circumference is 40
π
ft or approximately 126
ft.
3. The radius of the large pizza is 9 inches, and the
radius of the medium pizza is 7 inches.
large pizza:
2222
(9 in.) 81 in. 254 in.Ar
ππ π

2
(3) 5
45
V
π
π

The volume of the smaller cylinder is
3
45 in. .
π
Larger cylinder: r = 3 in., h = 10 in.
2
2
(3) 10
Vrh
V
π
π

Section 2.6 Problem Solving in Geometry
5. Use the formula for the volume of a sphere. The
radius is 4.5 in.
A
3
4
Vr
π
6. Let 3xthe measure of the first angle.
Let xthe measure of the second angle.
7. Step 1 Let x = the measure of the angle.
Step 2 Let 90 – x = the measure of its complement.
2.6 Concept and Vocabulary Check
1. 1
A
bh
6.
2
Vrh
π
7. 180°
2.6 Exercise Set
1. Use the formulas for the perimeter and area of a
rectangle. The length is 6 m and the width is 3 m.
22
Plw
3. Use the formula for the area of a triangle. The base
is 14 in and the height is 8 in.
11
(14)(8) 56
22
Abh 
The area is 56 square inches.
Chapter 2 Linear Equations and Inequalities in One Variable
4. Use the formula for the area of a triangle. The base
is 30 m and the height is 33 m.
A
5. Use the formula for the area of a trapezoid. The
bases are 16 m and 10 m and the height is 7 m.
6. Use the formula for the area of a trapezoid. The
bases are 37 meters and 26 meters and the height is
18 meters.
7. 1250, 25Aw
1250 25
50
A
lw
l
l

The length of the swimming pool is 50 feet.
A
9. 20, 5Ab
1
2
1
20 5
Abh
h

10. 30, 6Ab
1
10
h
The height is 10 ft.
The length of the rectangle is 50 cm.
12. 208, 46Pw
22
Plw

circle. The radius is 4 cm.
2
2
(4)
16
50
A
r
A
π
π
π
2
2
Section 2.6 Problem Solving in Geometry
14. Use the formula for the area and circumference of a
circle. The radius is 9m.
2
2
9
Ar
A
π
π
15. Since the diameter is 12 yd, the radius is
12 6 yd.
2
2
2
(6)
A
r
A
π
π
16. Since the diameter is 40 ft, the radius is
40 20 ft.
2
2
2
20
400
Ar
A
π
π
π
17.
2
14 2
14 2
22
Cr
r
r
π
ππ
ππ
ππ
18.
2
16 2
16 2
22
Cr
r
r
π
ππ
ππ
ππ
20. Use the formula for the volume of a rectangular
solid. The length is 5 cm and width and height are
each 3 cm.
Vlwh
150
471
π
The volume of the cylinder is
3
150 cm
π
or
approximately
3
471 cm
.
Chapter 2 Linear Equations and Inequalities in One Variable
23. Use the formula for the volume of a sphere. The
diameter is 18 cm, so the radius is 9 cm.
24. Use the formula for the volume of a sphere. The
diameter is 24 in., so the radius is 12 in.
3
4
3
Vr
π
25. Use the formula for the volume of a cone. The
radius is 4 m and the height is 9 m.
2
1
3
Vrh
π
26. Use the formula for the volume of a cone. The
radius is 5 m and the height is 16 m.
2
2
1
3
1516
3
Vrh
V
π
π

28.
2
22
1
3
or
Vrh
hh
rr
π
ππ

29. Smaller cylinder: r = 3 in, h = 4 in.
22
(3) 4 36Vrh
ππ π
 
3
smaller
36 1
V
π
So, the volume of the larger cylinder is 9 times the
volume of the smaller cylinder.
30. Smaller cylinder; r = 2 in., h = 3 in.
2
Large cylinder: r = 4(2 in.) = 8 in., h = 3 in.
2
2
83
192
Vrh
V
V
π
π
π

The volume of the larger cylinder is
3
192 in. .
π
The ratio of the volumes of the two cylinders is