Section 2.2 The Multiplication Property of Equality
72. a. The bar graph indicates the median weekly
earnings, in 2013, for women with some college
or an associate’s degree is $657. Since 2013 is
b. 13 231
777 13 231
Wn
n
73. – 75. Answers will vary.
76. does not make sense; Explanations will vary.
Sample explanation: The addition property of
equality is not necessary for this equation.
80. false; Changes to make the statement true will vary.
A sample change is: If 721,x then 721
3.
77
x
81. false; Changes to make the statement true will vary.
A sample change is: If 3416,x then 320.x
answer and set x equal to this value. Then, multiply
both sides of this equation by 60 (since we will
divide both sides of the equation by 60 to solve).
For example, suppose we want the solution to be 3.
85. Answers will vary. As an example, start with an
integer solution, such as 10, and set it equal to x.
That is, we have 10x. The solution was obtained
42
Therefore, an example equation would be 525
42
x.
86. 3.7 19.46 9.988
3.7 9.988 19.46
x
x
72.8 14.6 4.98
455.43 4.98 4.98
67.82 14.6 455.43
67.82 14.6 14.6 455.43 14.6
67.82 440.83
67.82 440.83
yy
yy
y
y
y
y
Chapter 2 Linear Equations and Inequalities in One Variable
90.
3
3
4141
14
3
xx
93.
39 39
10 10 10
55 5 5
278
xx
x
2.3 Check Points
1. Simplify the algebraic expression on each side.
72531623
425132
xx x
xx
Collect variable terms on one side and constant
terms on the other side.
2. Simplify the algebraic expression on each side.
82(6)
8212
xx
xx
Collect variable terms on one side and constant
3. Simplify the algebraic expression on each side.
4(2 1) 29 3(2 5)
8429615
xx
xx
5
x
The solution set is
5.
4. Begin by multiplying both sides of the equation by
12, the least common denominator.
25
436
25
12 12
436
25
12 12 12
436
3810
xx
xx
xx
xx
0.48 3 0.2 1.2
100(0.48 3) 100(0.2 1.2)
48 300 20 120
48 300 300 20 120 300
xx
xx
xx
xx
6.
373(1)
3733
xx
xx
7.
3( 1) 9 8 6 5
33936
3636
xxx
xx
xx
8. 10 53
99
Dx
10 53
10 99
x
2.3 Concept and Vocabulary Check
1. simplify each side; combine like terms
2. 30
5. identity
1.
534 102
8412
412
412
xxx
xx
x
x
51
10 2
x
The solution set is
1.
2
12 64 40
12 24
2
x
x
x
The solution set is
2.
Chapter 2 Linear Equations and Inequalities in One Variable
5.
36 836
2623
xx x
xx
6.
32 638
2232
xx x
xx
7.
4120
4420
x
x
8.
326
36 6
30
0
x
x
x
x
The solution set is
0.
10.
42 3 32
81232
x
x
11.
38 30 2 1
38 30 2 2
x
x
The solution set is
3.
12.
20 44 8 2
x
13.
24 3 8 46
86846
8246
822462
848
848
38
z
z
z
z
z
z
Section 2.3 Solving Linear Equations
15.
631014
6 3 10 14
31014
xx
xx
x
16.
521410
5 2 14 10
31410
324
8
xx
xx
x
x
x
The solution set is
8.
17.
52 1 12 3
xx
18.
32 30
36 30
2630
224
12
xx
xx
x
x
x
The solution set is
12 .
19.
35 42 1
15 3 8 4
15 3 8 8 4 8
xx
xx
xx x x
20.
33 1 43 3
9 3 12 12
3312
315
5
xx
xx
x
x
The solution set is
5.
22.
833212
824636
22436
212
6
yy
yy
y
y
y
The solution set is
6.
23. 337143
xx
Chapter 2 Linear Equations and Inequalities in One Variable
24.
54 923
543623
xx x
xx x
25.
52 8 2 5 3 3
xx
10 40 2 5 15 3
10 42 5 12
xx
xx
26.
73 2 5 62 1 24
21 14 5 12 6 24
xx
xx
27.
641 3 1
64433
617
xx
xx
x
28.
100 1 4 6
100 1 4 24
100 3 23
xx
xx
x
29.
10 4 4 2 3 1 2 3
10 40 4 8 3 3 2 6
zzzz
zzzz
30.
2432262
2832 262
510 6
zz z
zz z
zz
5456
5
x
32.
13 22
2
x
To clear the equation of fractions, multiply both
26 44
26 26 44 26
70
x
x
x
Section 2.3 Solving Linear Equations
33.
257
3
x
To clear the equation of fractions, multiply both sides
by the least common denominator (LCD), which is 3.
2
3
3537
x
34.
396
4
x
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
35.
235
3412
y
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
which is 12.
23 5
12 12
34 12
y
4
36.
327
4312
y
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
which is 12.
The solution set is
3
37.
5
32 6
xx
To clear the equation of fractions, multiply both
38.
1
45
xx
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
which is 20.
xx
Chapter 2 Linear Equations and Inequalities in One Variable
39.
20 32
zz
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
which is 6.
620 6
zz
40.
1
52 6
zz
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
which is 30.
41.
22
35 55
yy
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
which is 15.
22
15 15
35 55
yy
42.
11
12 6 2 4
yy
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
which is 12.
11
12 12
yy
42
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
which is 8.
3
8382
42
240
240
22
20
xx
x
x
x
The solution set is
20 .
32 2
xx
Section 2.3 Solving Linear Equations
45.
35
1
54
xx
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
which is 20.
35
20 1 20
54
432055
xx
xx
46.
21
4
34
xx
To clear the equation of fractions, multiply both
sides by the least common denominator (LCD),
which is 12.
47.
3.6 2.9 6.3xx
To clear the equation of decimals, multiply both
sides by 10.
48.
1.2 3.6 2.4 0.3xx
To clear the equation of decimals, multiply both
sides by 10.
10(1.2 3.6) 10(2.4 0.3 )
12 36 24 3
12 60 3
15 60
4
xx
xx
xx
x
x
30
y
The solution set is
30 .
50. 0.15 0.1 2.5 1.04yy
To clear the equation of decimals, multiply both
0.3 4 0.1 1
xx
To clear the equation of decimals, multiply both
sides by 10.
10(0.3 4) 10(0.1 1)
xx
Chapter 2 Linear Equations and Inequalities in One Variable
52.
0.1( 80) 14 0.2
0.1 8 14 0.2
xx
xx
To clear the equation of decimals, multiply both
sides by 10.
53.
0.4(2 6) 0.1 0.5(2 3)
0.8 2.4 0.1 1.5
0.8 2.5 1.5
zz
zz
zz
To clear the equation of decimals, multiply both
sides by 10.
54.
1.4( 5) 0.2 0.5(6 8)
1.4 7 0.2 3 4
1.4 7.2 3 4
zz
zz
zz
55.
0.01( 4) 0.04 0.01(5 4)
0.01 0.4 0.04 0.05 0.4
0.01 0.36 0.05 0.4
xx
xx
xx
To clear the equation of decimals, multiply both
56.
0.02( 2) 0.06 0.01( 1)
0.02 0.04 0.06 0.01 0.01
0.02 0.04 0.01 0.05
xx
xx
xx
The solution set is
3.
57.
0.6( 300) 0.65 205
0.6 180 0.65 205
xx
xx
To clear the equation of decimals, multiply both
sides by 100.
58.
0.05(7 36) 0.4 1.2
0.35 1.8 0.4 1.2
xx
xx
To clear the equation of decimals, multiply both
59.
373 1
3733
373 333
73
xx
xx
xxxx
Section 2.3 Solving Linear Equations
60.
25210
210210
xx
xx
61.
244523
2828
xxx
xx
62.
3 18659
3333
xxx
xx
63.
723 5 832 1
xx
76 1086 3
6356
xx
xx
64.
232 7 943 1
26 21912 4
xx
xx
The solution set is
4.
3
65.
415 5 4
xx x
statement
11,
so the equation is an identity and
the solution set is all real numbers
is a real number .xx
The original equation is equivalent to the true
statement
55,
so the equation is an identity
and the solution set is all real numbers
is a real number .xx
Chapter 2 Linear Equations and Inequalities in One Variable
68.
53 12 35
xx x
69.
323
323
33 3
xx
xx xx
x
70.
545
54454
xx
xx x x
71.
2
33
xx
Multiply by the LCD, which is 3.
323
33
xx
xx
Since
12 0
is a false statement, the original
equation has no solution. The solution set is
.
73.
44
xx x
216416
16 4 16
16 4 16
xx x
xx
xx xx
74.
233
23
xx x
Multiply both sides by the LCD which is 6.
2
6363
xx x
Section 2.3 Solving Linear Equations
75.
25
2
36
xx
Multiply both sides by the LCD which is 6.
25
6626
36
22 12 5
xx
xx
76. 21
8
34
xx
Multiply both sides by the LCD which is 12.
21
12 12 8
34
xx
77. 0.06( 5) 0.03(2 7) 0.09
0.06 0.3 0.06 0.21 0.09
0.06 0.3 0.06 0.3
xx
xx
xx
To clear the equation of decimals, multiply both
78. 0.04( 2) 0.02(6 3) 0.02
0.04 0.08 0.12 0.06 0.02
0.04 0.08 0.12 0.08
xx
xx
xx
To clear the equation of decimals, multiply both
sides by 100.
100(0.04 0.08) 100(0.12 0.08)
xx
$
$
$
x
x
x
$
$
$
x
x
x
Δ
Δ
Δ
Chapter 2 Linear Equations and Inequalities in One Variable
81. First solve the equation for x.
2
53
2
55 35
xx
xx xx
82. First solve the equation for x.
33 4
244
33
444
24 4
xxx
xx x
83.
1116
35
LCD = 15
11
15 15 15 16
35
xx
xx
84.
2113
54
21
20 20 13
54
xx
xx
12
x
The number is 12.
86. 71
30
82
71
xx
250 10 65 50
250 50 10 65 50 50
200 10 650
200 650 10 650 650
850 10
x
x
x
x
x
Section 2.3 Solving Linear Equations
88.
10 65 50
400 10 650 50
Fx
x
89. 353
2
WH
3(6) 53
2
18 53
2
W
W
90. 353
2
WH
3(12) 53
2
36 53
2
W
W
91.
5
15 11
d
p
2046 5
2046
5
409.2
d
d
d
He descended to a depth of 409.2 feet below the
surface.
of 11 feet.
93. – 97. Answers will vary.
98. makes sense
99. makes sense
Chapter 2 Linear Equations and Inequalities in One Variable
105. false; Changes to make the statement true will vary.
A sample change is: The equation
11
32
x
is
equivalent to
11
66 6
32
x
or
623.x
106.
0.432 10.44
16 0.432 10.44
fh
h
107.
23 3 5
1
926
xx x
23 3 5
18 18 18 18 1
23 3 5
18 18 1
92 6
xx x
xx x
108.
23 4 3 23 1 2xxx
6832332
683231
68362
6892
689 929
xxx
xxx
xxx
xx
xxxx
The solution set is 10 .
3
109. 24 20 because
24
lies further to the left on a
number line.
p
113.
40.25
40.25
0.25 0.25
16
B
B
B
The solution set is
16 .
Section 2.4 Formulas and Percents
114.
1.3 26
A
P
2.4 Check Points
1.
A
lw
A
lw
ww
Al
w
A
3.
TDpm
TDpm
TD pm
pp
TDm
p
TD
mp
5. Use the formula : is percent of .
A
PB A P B
90.60
B
90.60
0.60 0.60
15
B
B
7. Use the formula : is percent of .
A
PB A P B
is
18 of
what percent 50?
18 50P
Find the price decrease:
$940 $611 $329
The price what the original
is
decrease percent of price?
329 940P
329 940
329 940
940 940
0.35
P
P
P
To change 0.35 to a percent, move the decimal point
two places to the right and add a percent sign.
Chapter 2 Linear Equations and Inequalities in One Variable
9. a.
Tax Paid Taxes Paid
Year increase/decrease
the Year Before This Year
The taxes for year 2 will be $1152.
b. The taxes for year 2 are less than those originally paid.
Find the tax decrease:
$1200 $1152 $48
2.4 Concept and Vocabulary Check
1. isolated on one side
2.
A
lw
2.4 Exercise Set
1. for
drt r
2. for drt t
or
drt
rr
dd
tt
rr
Section 2.4 Formulas and Percents
3. I = Prt for P
Pr
It
4. for IPrt r
IPrt
Pt PT
5.
2 for Crr
π
2
22
Cr
π
ππ
6.
for Cd d
π
Cd
π
ππ
7.
2
Emc
2
22
A
Emc
8.
2
for Vrhh
π
2
22
Vrh
rr
π
ππ
9. for ymxb m
ybmx
10. for ymxb x
ybmx
of a line.
11. for TDpm D
PC CMCC
PC MC
PC MC
13.
1for
2
A
bh b
1
22
or
Abh
bb
hh
This is the formula for the area of a triangle: area =
1
base height.
Chapter 2 Linear Equations and Inequalities in One Variable
14. 1 for
2
A
bh h
1
22
A
A
A
Abh
15. for
5
n
Mn
A
16. for
740
A
MA
17. 80 2 for
2
cFc
80 80 2 80
2
cF
18. 5
15 for
11
d
pd
19.
1for
2
A
ab a
1
22
Aab
1
22
2
2
Aab
Aab
21. for SPPrt r
or
SPPPrt P
SP SP
rr
Pt Pt
This is the formula for finding the sum of principle
and interest for simple interest problems.
22.
for SPPrt t
This is the formula for finding the sum of principle
and interest for simple interest problems.