Chapter 2
Algebra Basics, Equations, and Inequalities
Section 2.1
1. A numerical expression is an algebraic expression that does not contain a variable.
3. 2x − 7 = 2⋅3− 7
=6 −7
= −1
Replace x with 3
5. −3(1− x) = −3[1− (−1)] Replace x with − 1
= −3(1+1)
= −3(2)
= −6
7. x − (1− 3x) = −2− [1− 3(−2)] Replace x with − 2
= −2− (1+ 6)
= −2− 7
=−9
9. 2x 2 + 3x + 2 = 2 ⋅32 + 3⋅ 3+ 2 Replace x with 3
= 18 + 9+ 2
= 29
11. 3x − y = 3⋅2 − (−4) Replace x with 2 and y with − 4
= 6+ 4
= 10
13. x 2 + xy + 3y2 = (−3)2 + (−3)⋅1+ 3 ⋅12 Replace x with − 3 and y with 1
= 9 − 3+ 3
=9
15.
17.
5− y 5− (−1)
=
Replace x with 2 and y with −1
3− x
3− 2
6
=
1
=6
a + b 3+ (−2)
=
Replace a with 3, b with − 2, and c with 4
c
4
1
=
4
19. (a + b)2 − 3c = [3+(−2)]2 − 3⋅4 Replace a with 3, b with − 2, and c with 4
=12 −12
=1−12
=−11
21. For x = –3.92, x – 2(x – 1) = 5.92.
23. For x = –1.2,
1
x+5
≈ −1.12 .
2x −1
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Chapter 2: Algebra Basics, Equations, and Inequalities
2
25. There are two terms in 2x + 1 but only
one term in 2(x + 1). In 2(x + 1) both 2
and x + 1 are factors.
SSM: Elementary Algebra
55. 2.1x – 4.2y + 0.3y + 1.2x = 3.3x –3.9y
57. Remove parentheses and combine like
terms.
27. Terms: 3y, x, –8
Coefficients: 3, 1, –8
59. −5(3x) = (−5 ⋅3)x = −15x
29. Terms: x2 , 3x, –6
Coefficients: 1, 3, –6
1
 1 
61. − (4y) =  − ⋅ 4 y = −y

4
4 
31. Terms: 7b, − c,
3a
7
Coefficients: 7, −1,
5  9   5 −9 
3
63. −  − x =  − ⋅  x = x
6  10   6 10 
4
3
7
65. 5(4 + 3b) = 5(4)+ 5(3b)
= 20 +15b
= 15b+ 20
33. 3x + 8x = (3 + 8)x = 11x
67. −3(5 − 2x) = −3(5) − 3(−2x)
=−15 + 6x
= 6x −15
35. –y + 3y = (–1 + 3)y = 2y
37. –3x – 5x = (–3 – 5)x = –8x
39. 4t 2 − t2 = 4t 2 −1⋅t 2
= (4 −1)t 2 = 3t 2
69. 2(3x + y) = 2(3x) + 2(y) = 6x + 2y
71. −3(−x + y − 5) = −3(−x) − 3(y)− 3(−5)
= 3x − 3y + 15
41. 6x – 7x – 3 = –x + 3
45. 2m – 7n – 3m + 6n = –m – n
73. − 3 (5x + 20) = − 3 (5x)− 3 (20)
5
5
5
= −3x −12
47. 10 – 2b + 3b – 9 = b + 1
75. – (x – 4) = –x + 4
49. 2a 2 + a 3 − a 2 − 5a3 =− 4a 3 + a 2
77. –(3 + n) = –3 – n
51. ab + 3ac – 2ab + 5ac = –ab + 8ac
79. – (2x + 5y – 3) = –2x – 5y + 3
43. 9x – 12 – 3x + 5 = 6x – 7
53.
1
1
2
2
7
a+ b − a + b = − a + b
3
2
3
3
6
81. 2(x −3) + 4 = 2 x − 6+ 4 Distributive Property
= 2x − 2
Combine like terms
83. 5− 3(x −1) = 5 − 3x + 3 Distributive Property
= −3x + 8 Combine like terms
85. −2(x − 3y) + y + 2x = −2x + 6y + y + 2x Distributive Property
= 7y
Combine like terms
87. (2a + 7) + (5 – a) = a + 12 Combine like terms
89. 2 + (2a + b) + a – b = 3a + 2 Combine like terms
Copyright © Houghton Mifflin Company. All rights reserved.
SSM: Elementary Algebra
Chapter 2: Algebra Basics, Equations, and Inequalities
91. 4(1−3x)− (3x + 4) = 4− 12x −3x − 4 Distributive Property
= −15x
Combine like terms
93. 2[5 − (2x − 3)] + 3x = 2[5 − 2x +3] + 3x
= 2[ −2x + 8] + 3x
= −4x +16 + 3x
= 16 − x
95. 4 + 5[x −3(x + 2) ] = 4 + 5[x −3x − 6 ]
= 4 + 5[ −2x − 6 ]
= 4 −10x − 30
= −10x − 26
Distributive Property
Simplify inside brackets
Distributive property
Combine like terms
Distributive Property
Simplify inside brackets
Distributive property
Combine like terms
97. The student will study for 8 hours, so t = 8.
8t + 20 = 8⋅8 + 20
= 64 + 20
= 84
The predicted grade is an 84.
99. (a) t = 1985 – 1980 = 5
t = 2005 – 1980 = 25
(b)
15 3
(5) = 937.5
2
15
(25) 3 = 117,187.5
2
The model is most accurate for year 2005.
101. 2(3n + 5) − 7 − [ 3(n+ 1)− 2(n − 4) ] = 2(3n + 5) − 7− [3n + 3− 2n + 8]
= 2(3n + 5) − 7 − [n+11]
= 6n +10 − 7 − n − 11
= 5n −8
103.
x +y
1.2 + (−3.1)
=
Replace x with 1.2 and y with − 3.1
2 x + 3y 2(1.2)+ 3(−3.1)
1.2− 3.1
=
2.4 − 9.3
−1.9
=
−6.9
≈ 0.28
105. 2a 2 + (c − 4)2 = 2(−5.3)2 + (2.15− 4)2 Replace a with − 5.3 and c with 2.15
= 2(−5.3)2 + (−1.85)2
= 2(28.09) + 3.4225
= 56.18 + 3.4225
= 59.6025
Section 2.2
1. The number lines are called the x-axis
and y-axis. Their point of intersection is
called the origin.
3. Quadrant I
5. Quadrant III
7. Quadrant II
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3
Chapter 2: Algebra Basics, Equations, and Inequalities
4
9. Quadrant IV
SSM: Elementary Algebra
27.
11. x-axis
13. y-axis
15. A(5, 0)
B(–7, 2)
C(2, 6)
D(0, –4)
E(–3, –5)
F(4, –4)
17. A(0, 6)
B(–5, 0)
C(7, 3)
D(–6, –3)
E(–2, 8)
F(3, –3)
The next ordered pair is (6, –12).
29.
19. The order of the coordinates is
different.
21.
The next ordered pair is (7, 10).
31.
23.
The next ordered pair is (10, –7).
33. (a) +, –
(b) +, +
(c) –, +
(d) –, –
25.
35. The point is either on the x-axis, or in
quadrant I or IV.
37. The point is either on the y-axis, or in
quadrant III or IV.
39. The point is in quadrant I or III.
41. The point is on the y-axis.
The next ordered pair is (2, 4).
43. The point is in quadrant II or IV.
45. (a) The point is on the x-axis.
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SSM: Elementary Algebra
Chapter 2: Algebra Basics, Equations, and Inequalities
(b) The point is on the y-axis.
5
65. (a) The second coordinates are:
–1, 1, 3, 5, 7
(c) The point is in quadrant IV.
(b) y = x + 2
(d) The point is in quadrant I or III.
47. P(0, 13)
67. (a) The second coordinates are:
3, 1, 0, –1, –3
49. P(9, 9)
51. P(–12, 6)
(b) y = –x
69. (a) The fourth vertex is (–40, –10).
53.
(b) The length is 50 feet, the width is
20
feet, so the area is 1000 square feet.
71. (a) Sales/marketing and Written
communications would be the same
distance above the x-axis because
they are both 35%
(b) Basic computer would be the
greatest distance above the x-axis
because 68% is the largest.
55.
73. They would all lie in the first quadrant.
75. (a) The years in which the number of
resolutions exceed the average for
the period
(b) The years in which the number of
resolutions was less than the
average for the period
57. B(–4, 6)
77. The midpoint is (2, 3).
59. B(17, 0)
79. The midpoint is (–4, 2).
61. The length is 5 and the width is 3. The
area is 15 and the perimeter is 16.
81. The other two vertices are 8 units above
or below the two given vertices. Since
one of these vertices must be in
quadrant I, it must be 8 units above the
vertex in quadrant IV, (7, –3). The
coordinates of P are (7, 5).
63. The length of each side is 3. The area is
9 and the perimeter is 12.
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Chapter 2: Algebra Basics, Equations, and Inequalities
6
SSM: Elementary Algebra
Section 2.3
1. It is not necessary to enter the expression each time it is to be evaluated.
3.
x
–1
x – 6 –7
5.
x
15.
–6
–2
5
6
12
23.8
0.7
–5.6
–18.2
x
–9
–4
3
5
x2 + 4x – 21
24
–21
0
24
–8
–1
2
10
0
14
20
36
x
–5
–1
–0.5
8
2x +1
9
1
0
17
2(x + 8)
13.
11
3
9. x
11.
4
–8
–2.1x + 7
7.
0
x
–5
1
25
50
x +1
2x
0.4
1
0.52
0.51
x
–2
1
4
7
10
x–3
–5
–2
1
4
7
(4, 1)
(7, 4)
(10, 7)
(x, x – 3) (–2, –5) (1, –2)
17.
x
3
7
0
–4
–5
2x + 5
11
19
5
–3
–5
(7, 19)
(0, 5)
(–4, –3) (–5, –5)
(x, 2x + 5) (3, 11)
19.
x
1
4
6
–3
–5
x2 – 3x – 4
–6
0
14
14
36
(4, 0)
(6, 14)
(–3, 14) (–5, 36)
(x, x2 – 3x – 4) (1, –6)
21. You obtain an error message for x = 1 because division by 0 is not defined.
23. When x = 3, y = 4.
25. When x = –1, y = 1.
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Chapter 2: Algebra Basics, Equations, and Inequalities
SSM: Elementary Algebra
7
27.
37.
x
2
5
7
2x –
–6
0
4
x
1
0
–32 –8
1
x + 3 –5
4
29.
39.
x
–30
–9
12
1
x+5
3
–5
2
9
(a) x = 3
(b) x = 9
(c) x = –5
(d) x = –10
31.
41.
(a) x = –1
x
–6
–(3x + 1) 17
0
5
–1
–16
(b) x = 2
(c) x = 5
33. The first coordinate, 3, is the value of
the variable, and the second coordinate,
7, is the value of the expression for
x = 3.
(d) x = 8
43.
35.
(a) x = –23
(b) x = 6
x
2
x – 4 –2
4
0
5
–1
–2
0
–4
1
–5
–6
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(c) x = 20
1
–12 16
6
0
4.5 10
7
28
Chapter 2: Algebra Basics, Equations, and Inequalities
8
(d) x = 33
SSM: Elementary Algebra
(a) x = –12, 22
45.
(b) x = –3, 13
(c) x = 5
(d) x = 1, 9
49. 2x + 1 = 4 – x when x = 1.
(a) x = –2, 11
51. 3x = 8 – x when x = 2.
(b) x = –1, 10
53.
(c) x = 1, 8
(d) x = 2, 7
47.
When x = 28, there are 3 pizzas left.
55.
Year
Reported
Number
Modeled
Number
Difference
1983
620
13(3) + 588 = 627
–7
1985
660
13(5) + 588 = 653
7
1987
680
13(7) + 588 = 679
1
1989
700
13(9) + 588 = 705
–5
1991
740
13(11) + 588 = 731
9
1993
750
13(13) + 588 = 757
The model is most accurate for 1987.
57. Year xValue of the expression
1965
1970
1975
1980
1985
1990
0
5
10
15
20
25
–7
61. The y-coordinate is one more than twice
the x -coordinate.
324
1787
3250
4714
6177
7640
63. The y-coordinate is twice the sum of the
x-coordinate and 1.
65. (a) For c = 0.5 and k = 6.1,
c 2 + 5k = 30.75.
59. For the year 2000, x = 35, and the value
of the expression is 10,567.
(b) For c = –0.5 and k = 6.1,
c2 + 5k = 30.75.
Copyright © Houghton Mifflin Company. All rights reserved.
SSM: Elementary Algebra
Chapter 2: Algebra Basics, Equations, and Inequalities
Section 2.4
31.
1. An equation has an = symbol, but an
expression does not.
3. Expression
5. Equation
7. Equation
x = 12
33.
9. Expression
11. A solution of an equation is a value of
the variable that makes the equation
true.
13.
15.
x−5 = 9
14 − 5 = 9 Replace x with 14
9 = 9 True
Yes
35.
5− 3x = 11
5 − 3⋅2 = 11 Replace x with 2
5− 6 = 11
−1 = 11 False
No
17. 10x − 3 = 5x +1
10 ⋅1− 3 = 5⋅1+1 Replace x with 1
10 − 3 = 5+1
7= 6
False
No
19.
x=4
x 2 + 3x + 2 = 0
(−1)2 + 3(−1)+ 2 = 0 Replace x
with −1
1 − 3+ 2 = 0
0 = 0 True
Yes
x = –14
37.
x = –32
39.
21. Yes
x = –20
23. Yes
25. Yes
27. No
41. Graph each side of the equation and
trace to the point of intersection. The
x-coordinate of that point is the
estimated solution.
29.
43.
x = 14
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x = –1.25
9
10
Chapter 2: Algebra Basics, Equations, and Inequalities
45.
SSM: Elementary Algebra
57.
x = 0.14
47.
x=8
59.
x = –1.71
49.
x = –9
61.
x = –7
51.
x = 15
63.
x = –12
53.
x = –9
65.
x = –12
55.
x = 15
67. (a) The graphs of the two sides of the
equation do not intersect. The
solution set is Ø.
(b) The graphs of the two sides of the
equation coincide. The solution set
is ℜ.
x = –20
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SSM: Elementary Algebra
69.
Chapter 2: Algebra Basics, Equations, and Inequalities
81.
x=0
71.
x = 24
83.
Ø
73.
Ø
85.
ℜ
75.
ℜ
87. (a) 2.1x + 22.8 = 37.50; 7 (1997)
(b) 2.1x + 20.8 = 39.70; 9 (1999)
89.
Ø
77.
x = 110
91.
ℜ
79.
x = –6, 30
93.
x=8
Ø
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11
Chapter 2: Algebra Basics, Equations, and Inequalities
12
Section 2.5
1. Equivalent equations are equations that
have exactly the same solutions.
3. 7x − 6x = 9
x = 9 Combine like terms
5. 1.8x − 0.8x = 3.5
x = 3.5 Combine like terms
7. (i)
9. (ii)
11.
x+5 = 9
x + 5 − 5 = 9 − 5 Subtract 5 from both sides
x=4
13.
−4 = x − 6
x − 6 = −4
Exchange sides
x − 6 + 6 = −4+ 6 Add 6 to both sides
x =2
15.
25 + x = 39
25 + x − 25 = 39 − 25 Subtract 25 from both sides
x = 14
17.
x + 27.6 = 8.9
x + 27.6 − 27.6 = 8.9− 27.6 Subtract 27.6 from both sides
x = −18.7
19.
12 = x +12
x + 12 = 12
Exchange sides
x +12 − 12 = 12 −12 Subtract 12 from both sides
x=0
21.
3x = 2x −1
3x − 2 x = 2x −1 − 2x Subtract 2 x from both sides
x =−1
23.
7x = 5 + 6x
7x − 6x = 5 + 6x − 6x Subtract 6 x from both sides
x =5
25.
−4x + 2 = −3x
−4x + 2 + 4x = −3x + 4x
2= x
x= 2
27.
9x − 6 = 10x
9x − 6− 9 x = 10x − 9x
−6 = x
x = −6
SSM: Elementary Algebra
Add 4 x to both sides
Exchange sides
Subtract 9x from both sides
Exchange sides
29. −3a + 5a = 9+ a + 6
2a = a + 15
Combine like terms
2a − a = a + 15− a Subtract a from both sides
a = 15
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SSM: Elementary Algebra
Chapter 2: Algebra Basics, Equations, and Inequalities
13
31. 3x + 3− 3x + 7= 6x − 6− 5x
10 = x − 6
Combine like terms
10 +6 = x − 6 +6
Add 6 to both sides
16 = x
x = 16
Exchange sides
33.
7x − 9 = 8 + 6x
7x − 9− 6 x = 8+ 6x − 6x Subtract 6x from both sides
x −9 = 8
x − 9+ 9 = 8 + 9
Add 9 to both sides
x = 17
35.
1− 3y = 4 − 4y
1 − 3y + 4y = 4− 4 y + 4y Add 4y to both sides
1+y = 4
1 + y − 1= 4 −1
Subtract 1 from both sides
y =3
37.
5x + 3= 6x +1
5x + 3 − 5x = 6x +1− 5x Subtract 5x from both sides
3 = x +1
3 −1= x +1 −1
Subtract 1 from both sides
2= x
x= 2
Exchange sides
39.
1− 5x = 1 − 4x
1 − 5x + 5x = 1− 4x + 5x Add 5x to both sides
1 = 1+ x
1 −1= 1 + x −1
Subtract 1 from both sides
0=x
x=0
Exchange sides
41. When we add 3 to both sides of x – 3 = 12, the left side becomes x – 3 + 3 = x + 0 = x. When
3x
we divide both sides of 3x = 12 by 3, the left side becomes
= x . In both cases, the variable
3
is isolated.
43. (ii)
45. (i)
47. All of the given steps are correct. In each case, the result is 1x = 20 because
(i)
4
3
and
are reciprocals whose product is 1.
3
4
(ii)
3
3
divided by is 1.
4
4
(iii) Multiplying by 4 and dividing by 3 is equivalent to multiplying by
49. 5x = 20
5x 20
=
Divide both sides by 5
5
5
x=4
Copyright © Houghton Mifflin Company. All rights reserved.
4
.
3
14
51.
Chapter 2: Algebra Basics, Equations, and Inequalities
SSM: Elementary Algebra
48 = −16x
−16x = 48
Exchange sides
−16x
48 Divide both sides by −16
=
−16
−16
x = −3
53. −y = −3
−y −3
=
Divide both sides by −1
−1 −1
y =3
55. −3x = 0
−3x
0
=
Divide both sides by − 3
−3 −3
x=0
57.
59.
61.
x
= −4
9
x
9 ⋅ = 9(−4) Multiply both sides by 9
9
x = −36
−t
= −5
4
−t
−4 ⋅ = −4(−5) Multiply both sides by − 4
4
t = 20
3
x =9
4
4 3
4
4
⋅ x = ⋅9 Multiply both sides by
3 4
3
3
x = 12
63. 2x − 7x = 0
−5x = 0
Combine like terms
−5x
0
=
Divide both sides by −5
−5
−5
x=0
65. y − 6y =−25
−5y =−25 Combine like terms
−5y −25
=
−5
−5 Divide both sides by − 5
y =5
67.
8 = 3t − 4t
8 =− t
Combine like terms
−t = 8
Exchange sides
−t
8
Divide both sides by −1
=
−1 −1
t =− 8
69.
9x = 8x
9x − 8x = 8x − 8x Subtract 8x from both sides
x= 0
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SSM: Elementary Algebra
Chapter 2: Algebra Basics, Equations, and Inequalities
71. −4x =−28
−4 x −28
=
Divide both sides by − 4
−4
−4
x =7
73.
5− 7x = −6x
5− 7x + 7x = −6x + 7x Add 7 x to both sides
5=x
x=5
Exchange sides
75.
−3+ x = 0
−3 + x + 3= 0+ 3 Add 3 to both sides
x=3
77.
1
=−2y
4
1
−2y =
4 Exchange sides
1
−2y 4
Divide both sides by − 2
=
−2 −2
1
y=−
8
79. Distributive Property
81. Multiplication Property of Equations
83. Addition Property of Equations
85. 0.7x = 3.5
0.7x 3.5
=
Divide both sides by 0.7
0.7 0.7
x= 5
The sales in 1991 were at $3.5 billion.
87. 0.7x = 5.6
0.7x 5.6
=
Divide both sides by 0.7
0.7 0.7
x= 8
The solution is 8, which means that sales for 1994 were projected to be $5.6 billion.
89.
x+a = b
x + a − a = b − a Subtract a from both sides
x = b −a
91.
5x − a = b
5x − a + a = b + a Add a to both sides
5x = a + b
5x a + b Divide both sides by 5
=
5
5
a +b
x=
5
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15
Chapter 2: Algebra Basics, Equations, and Inequalities
16
93.
x −k = 0
−4 − k = 0
Replace x with − 4
−4 − k + k = 0+ k Add k to both sides
−4 = k
k = −4
Exchange sides
95.
kx + 1 = 7
k ⋅ 2+ 1= 7
Replace x with 2
2k + 1 = 7
2k +1− 1 = 7−1 Subtract 1 from both sides
2k = 6
2k 6
=
Divide both sides by 2
2 2
k =3
97.
2x −1= 5
2x −1 +1= 5 +1
2x = 6
2x 6
=
2 2
x=3
3x − 4 = 3⋅3 − 4
= 9− 4
=5
SSM: Elementary Algebra
Solve for x
Add 1 to both sides
Divide both sides by 2
Replace x with 3
Section 2.6
1. Dividing both sides by 2 would create fractions and make the equation harder to solve.
The best first step is to add 5 to both sides.
3.
5x +14 = 4
5x + 14 −14 = 4 −14 Subtract 14 from both sides
5x =−10
5x −10 Divide both sides by 5
=
5
5
x =−2
The solution is –2.
5.
2 −3x = 0
2 − 3x − 2 = 0 − 2 Subtract 2 from both sides
−3x = −2
−3x −2
=
Divide both sides by −3
−3
−3
2
x=
3
2
The solution is .
3
7.
0.2 − 0.37t = −0.91
0.2 − 0.37t − 0.2 = −0.91 − 0.2 Subtract 0.2 from both sides
−0.37t = −1.11
−0.37t −1.11
=
Divide both sides by − 0.37
−0.37 −0.37
t=3
The solution is 3.
Copyright © Houghton Mifflin Company. All rights reserved.
SSM: Elementary Algebra
Chapter 2: Algebra Basics, Equations, and Inequalities
9.
7x = 2x − 5
7x − 2x = 2x − 5 − 2x Subtract 2 x from both sides
5x = −5
5x −5
=
Divide both sides by 5
5
5
x = −1
The solution is –1.
11.
3x = 5x +14
3x − 5x = 5x +14 − 5x Subtract 5x from both sides
−2x = 14
−2x 14
=
Divide both sides by −2
−2 −2
x = −7
The solution is –7.
13.
4x − 7 = 6 x + 3
4 x − 7 − 6x = 6x + 3 − 6x Subtract 6x from both sides
−2x −7 = 3
−2 x − 7 + 7 = 3+ 7
Add 7 to both sides
−2 x =10
−2x 10
=
Divide both sides by − 2
−2
−2
x = −5
The solution is –5.
15.
5x + 2 = 2 − 3x
5x + 2 + 3x = 2 − 3x + 3x Add 3x to both sides
8x + 2 = 2
8x + 2 − 2 = 2 − 2
Subtract 2 from both sides
8x =0
8x 0
=
Divide both sides by 8
8 8
x =0
The solution is 0.
17.
x + 6 = 5x −10
x + 6 − 5x = 5x −10 − 5x Subtract 5 x from both sides
−4x + 6 = −10
−4 x + 6 − 6 = −10 − 6
Subtract 6 from both sides
−4x = −16
−4x −16
=
Divide both sides by –4
−4
−4
x=4
The solution is 4.
19.
5 − 9x = 8x + 5
5 − 9x − 8x = 8x + 5 −8x Subtract 8x from both sides
5 −17x = 5
5 −17x − 5 = 5 − 5
Subtract 5 from both sides
−17x =0
−17x
0
Divide both sides by −17
=
−17 −17
x =0
The solution is 0.
Copyright © Houghton Mifflin Company. All rights reserved.
17
18
21.
Chapter 2: Algebra Basics, Equations, and Inequalities
9 − x = x +15
9 − x − x = x +15 − x Subtract x from both sides
9 − 2x = 15
9 − 2x −9 = 15 − 9
Subtract 9 from both sides
−2 x = 6
−2 x 6
Divide both sides by − 2
=
−2 −2
x = −3
The solution is –3.
23. 3x + 8− 5x = 2− x + 2x − 3
−2x +8 = x −1
−3x +8 = −1
−3x = −9
x=3
The solution is 3.
25. 3+ 5y − 24 + 3y = y −1+ y
8y − 21 = 2y −1
6y − 21 =−1
6y = 20
20 10
y=
=
6
3
10
The solution is
.
3
27.
SSM: Elementary Algebra
5− 5x +8 − 4 = −9+ x
9− 5x = −9+ x
9− 6x = −9
−6x = −18
x =3
The solution is 3.
Combine like terms
Subtract x from both sides
Subtract 8 from both sides
Divide by − 3
Combine like terms
Subtract 2y from both sides
Add 21 to both sides
Divide by 6
Combine like terms
Subtract x from both sides
Subtract 9 from both sides
Divide by − 6
29.
3(x +1) = 1
3x + 3 = 1
Remove grouping symbols
3x = −2 Subtract 3 from both sides
2
x=−
3 Divide by 3
2
The solution is − .
3
31.
5(2x −1) = 3(3x −1)
10x − 5 = 9 x − 3
Remove grouping symbols
x − 5 = −3
Subtract 9 x from both sides
x =2
Add 5 to both sides
The solution is 2.
33.
−10 = 3x + 8+ 4(x − 1)
−10 = 3x + 8 + 4x − 4 Remove grouping symbols
−10 = 7x + 4
Combine like terms
−14 = 7x
Subtract 4 from both sides
−2 = x
Divide by 7
The solution is –2.
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 2: Algebra Basics, Equations, and Inequalities
SSM: Elementary Algebra
35.
4(x −1) = 5 +3(x − 6)
4x − 4 = 5 +3x −18
4x − 4 = 3x − 13
x − 4 = −13
x = −9
The solution is –9.
Remove grouping symbols
Combine like terms
Subtract 3 x from both sides
Add 4 to both sides
37.
6x −(5x −3) = 0
6x − 5x + 3= 0 Remove grouping symbols
x + 3 = 0 Combine like terms
x = −3 Subtract 3 from both sides
The solution is –3.
39.
2 − 2(3x − 4) = 8− 2(5 + x )
2 − 6x + 8= 8 −10 − 2x
10 − 6x = −2 − 2x
12 − 6x = −2x
12 = 4x
3= x
The solution is 3.
Remove grouping symbols
Combine like terms
Add 2 to both sides
Add 6 x to both sides
Divide by 4
41. The number n is the LCD of all fractions in the equation.
43.
45.
47.
2
2 5
x− =
The LCD is 3
3
3 3
2
2
5
3 ⋅ x − 3 ⋅ = 3⋅
Multiply by 3 to clear denominators
3
3
3
2x − 2 = 5
2x = 7
Add 2 to both sides
7
x=
Divide by 2
2
7
The solution is .
2
3t 1
=
The LCD is 4
4 4
3t
1
4 ⋅1 + 4⋅ = 4⋅ Multiply by 4 to clear denominators
4
4
4+ 3t = 1
3t = −3 Subtract 4 from both sides
t = −1 Divide by 3
The solution is –1.
1+
1
2
1
x − 2 = 1+ x +
2
3
6
1
2
1
6⋅ x − 6 ⋅2 = 6 ⋅1 + 6⋅ x + 6 ⋅
2
3
6
3x −12 = 6 + 4 x +1
3x −12 = 7+ 4 x
−12 = 7+ x
−19 = x
The solution is –19.
The LCD is 6
Multiply by 6 to clear denominators
Combine like terms
Subtract 3x from both sides
Subtract 7 from both sides
Copyright © Houghton Mifflin Company. All rights reserved.
19
20
49.
Chapter 2: Algebra Basics, Equations, and Inequalities
x+5
x
= − +5
The LCD is 6
3
2
x+5
−x
6⋅
= 6⋅
+ 6⋅ 5 Multiply by 6 to clear denominators
3
2
2(x + 5)= −3x + 30
2x +10 = −3x + 30
Remove grouping symbols
5x +10 = 30
5x = 20
x=4
The solution is 4.
51.
53.
55.
57.
SSM: Elementary Algebra
Add 3 x to both sides
Subtract 10 from both sides
Divide by 5
3
(x −1) = 9
The LCD is 5
5
5 3
5
5
⋅ (x −1) = ⋅9 Multiply by
to clear denominators and isolate x
3 5
3
3
x −1 =15
x =16
Add 1 to both sides
The solution is 16.
x
3
+ 2 = (x + 3) − 6
3
2
x
3
6⋅ + 6⋅ 2 = 6⋅ (x + 3) − 6 ⋅6
3
2
2x +12 = 9(x + 3) −36
2x +12 = 9x + 27 − 36
2x +12 = 9x − 9
12 = 7x − 9
21 = 7x
3= x
The solution is 3.
The LCD is 6
Multiply by 6 to clear denominators
Remove grouping symbols
Combine like terms
Subtract 2 x from both sides
Add 9 to both sides
Divide by 7
3
5
1
x = (x +1)+ (8 − x)
7
14
7
3
5
1
14 ⋅ x = 14⋅ (x +1) +14⋅ (8 − x)
7
14
7
6x = 5(x + 1)+2(8 − x)
6x = 5x + 5 +16 − 2x
6x =3x + 21
3x = 21
x=7
The solution is 7.
The LCD is 14
Multiply by 14 to clear denominators
Remove grouping symbols
Combine like terms
Subtract 3x from both sides
Divide by 3
7.35x + 5.1 = 2.6x − 0.6
7.35x = 2.6 x − 5.7 Subtract 5.1 from both sides
4.75x = −5.7
Subtract 2.6 x from both sides
x = −1.2
Divide by 4.75
The solution is –1.2.
Copyright © Houghton Mifflin Company. All rights reserved.
SSM: Elementary Algebra
59.
Chapter 2: Algebra Basics, Equations, and Inequalities
2.7(3x −1.5) + 6.3 = 9.1x + 6.55
8.1x − 4.05 + 6.3 = 9.1x + 6.55
8.1x + 2.25 = 9.1x + 6.55
8.1x = 9.1x + 4.3
−x = 4.3
x = −4.3
The solution is –4.3.
Remove grouping symbols
Combine like terms
Subtract 2.25 from both sides
Subtract 9.1x from both sides
Multiply by −1
61.
x + 7 = 7+ x
x + 7 − x = 7+ x − x Subtract x from both sides
7 = 7 True
Identity
Solution set: ℜ
63.
x = x+3
x − x = x + 3− x Subtract x from both sides
0 = 3 False
Contradiction
Solution Set: Ø
65.
2x = 3x
2x − 2x = 3x − 2x Subtract 2 x from both sides
0= x
Conditional
The solution is 0.
67.
6t + 7 = 7
6t = 0 Subtract 7 from both sides
t = 0 Divide by 6
Conditional
The solution is 0.
69.
0x = 3
0 = 3 False
Contradiction
The solution set is Ø.
71.
6x − 3− x = 6x + 5
5x − 3= 6x + 5 Combine like terms
−3 = x + 5 Subtract 5x from both sides
−8 = x
Subtract 5 from both sides
The solution is –8.
73.
2 + 3x + 5 = 6x + 7 − 6x
3x + 7 = 7
Combine like terms
3x = 0
Subtract 7 from both sides
x =0
Divide by 3
The solution is 0.
Copyright © Houghton Mifflin Company. All rights reserved.
21
22
75.
Chapter 2: Algebra Basics, Equations, and Inequalities
SSM: Elementary Algebra
3
5 1
1
x + = (5 + x)+ x
The LCD is 4
4
2 2
4
3
5
1
1
4⋅ x + 4 ⋅ = 4⋅ (5 + x) + 4 ⋅ x Multiply by 4 to clear denominators
4
2
2
4
3x +10 = 2(5 + x )+ x
3x +10 =10 + 2x + x
Remove grouping symbols
3x +10 = 3x +10
10 =10 True
Solution set: ℜ
Combine like terms
Subtract 3x from both sides
77.
4(x − 2) +8 = 6(x +1)− 2x
4x −8 +8 = 6x + 6 − 2x Remove grouping symbols
4x = 4x + 6
Combine like terms
0 = 6 False
Subtract 4x from both sides
Solution set: Ø
79.
2(3x − 4) = 2 − 6(1− x)
6x − 8 = 2 − 6 + 6x Remove grouping symbols
6x − 8 = −4 + 6x
Combine like terms
−8 = −4 False Subtract 6x from both sides
Solution set: Ø
81.
3x −1− 2(x − 4) = 2x + 7 − x
3x −1− 2x + 8 = 2 x + 7− x Remove grouping symbols
x + 7 = x +7
Combine like terms
7 = 7 True
Subtract x from both sides
Solution set: ℜ
83. (a) (i) 5.88x + 40.58 = 150
The cost at a private college is $150,000.
(ii) 2.28x + 15.7 = 2(31.637)
2.28(7) + 15.7 = 31.66
The cost at a public college is double the 1997 cost.
(b) (i) 5.88x + 40.58 =150
5.88x =109.42
x =18.61
1990 + 19 = 2009
(ii) 2.28x +15.7 = 2(31.637)
2.28x = 47.574
x = 20.87
1990 + 21 = 2011
85.
For 1976, t = 20
Men: (−0.015)(20) +10.5 = −0.3+10.5
= 10.2 seconds
Women: ( −0.018)(20)+11.5 = −0.36 +11.5
= 11.14 seconds
87. The two times will be equal in the year 2289. For both men and women, the winning times are
5.51 seconds.
Copyright © Houghton Mifflin Company. All rights reserved.
SSM: Elementary Algebra
89.
Chapter 2: Algebra Basics, Equations, and Inequalities
x − 5{x − 5[ x − 5(x − 5)]} = 1
x − 5{ x − 5[ x − 5x + 25]} =1
x − 5{x − 5[−4x + 25]} = 1
x − 5{ x + 20x −125} = 1
x − 5{21x −125} = 1
x −105x + 625 = 1
−104x + 625 = 1
−104x = −624
x= 6
The solution is 6.
Remove parentheses
Combine like terms
Remove grouping symbols
Combine like terms
Remove grouping symbols
Combine like terms
Subtract 625 from both sides
Divide by − 104
91.
x − [2(x − 3) − (x + 1)] = −3(x −1)+ 4 + 3x
x − [ 2x − 6 − x −1] = −3x + 3 + 4+ 3x Remove parentheses
x − 2x + 6 + x +1 = −3x + 3 + 4+ 3x Remove brackets
7 = 7 True
Combine like terms
Solution set: ℜ
93.
ax + b = c
ax = c − b Subtract b from both sides
c−b
x=
Divide by a
a
95.
2t 2 −15 = t(t + 5)+ t 2
2t 2 −15 = t 2 + 5t + t 2
2t 2 −15 = 2t 2 + 5t
−15 = 5t
−3 = t
The solution is –3.
Remove grouping symbols
Combine like terms
Subtract 2 t2 from both sides
Divide by 5
Section 2.7
1. The value of the expression x + 3 is less than or equal to 7.
3.
x=3
x=5
x=9
x = 11
2+x>7
2+x>7
2+x>7
2+x>7
2+3>7
2+5>7
2+9>7
2 + 11 > 7
5>7
7>7
11 > 7
13 > 7
False
False
True
True
The solutions are x = 9 and x = 11.
Copyright © Houghton Mifflin Company. All rights reserved.
23
Chapter 2: Algebra Basics, Equations, and Inequalities
24
5.
SSM: Elementary Algebra
x = –4
x = –2
x = –1
x=0
4 ≤ 1 – 3x
4 ≤ 1 – 3x
4 ≤ 1 – 3x
4 ≤ 1 – 3x
4 ≤ 1 – 3(–4)
4 ≤ 1 – 3(–2)
4 ≤ 1 – 3(–1)
4≤1–3?0
4 ≤ 1 + 12
4≤1+6
4≤1+3
4≤1–0
4 ≤ 13
4≤7
4≤4
4≤1
True
True
True
False
The solutions are x = –4, –2, and –1.
7.
x = –1
x=2
x=5
x=6
x=9
–3 ≤ 2x – 7 < 5
–3 ≤ 2x – 7 < 5
–3 ≤ 2x – 7 < 5
–3 ≤ 2x – 7 < 5
–3 ≤ 2x – 7 < 5
–3 ≤ 2(–1) – 7 < 5
–3 ≤ 2 ? 2 – 7 < 5 –3 ≤ 2 ? 5 – 7 < 5 –3 ≤ 2 ? 6 – 7 < 5 –3 ≤ 2 ? 9 – 7 < 5
–3 ≤ –2 – 7 < 5
–3 ≤ 4 – 7 < 5
–3 ≤ 10 – 7 < 5
–3 ≤ 12 – 7 < 5
–3 ≤ 18 – 7 < 5
–3 ≤ –9 < 5
–3 ≤ –3 < 5
–3 ≤ 3 < 5
–3 ≤ 5 < 5
–3 ≤ 11 < 5
False
True
True
False
False
The solutions are x = 2 and x = 5.
9. The symbols [ and ] indicate that an
endpoint is included. The symbols ( and
) indicate that an endpoint is not
included.
11.
13.
15.
17.
31. x < –3
33. x ≥ 5
35. –4 < x ≤ 7
37. The point of intersection represents a
solution if the inequality symbol is ≤ or
≥.
39. x > –15
41. x ≤ 0
19.
43.
21.
23. n ≥ 0
25. x ≤ 3
x > –17
27. n < 0
29. x ≤ 6
Copyright © Houghton Mifflin Company. All rights reserved.
SSM: Elementary Algebra
45.
Chapter 2: Algebra Basics, Equations, and Inequalities
25
57.
x ≥ –5
x < 12
59. If the inequality is y1 > y2 or y1 ≥ y2
the solution set is ℜ. If the inequality is
y1 < y2 or y1 ≤ y2 , the solution set is Ø.
47.
61. The solution set is ℜ.
63. (a)
x < –3
49.
The solution set is ℜ.
(b)
x≤7
51.
The solution set is Ø.
65.
x<0
53.
–11 ≤ x ≤ 23
x<2
67.
55.
–19 < x ≤ 13
x ≥ –8
Copyright © Houghton Mifflin Company. All rights reserved.
26
Chapter 2: Algebra Basics, Equations, and Inequalities
69.
SSM: Elementary Algebra
81.
–13 < x ≤ 9
x<0
83.
71.
The solution set is Ø.
–6 < x < 8
85. x + 6 ≥ 3
73.
x ≥ –3
x≤9
87. 8 – x > –2
75.
x < 10
The solution set is Ø.
77.
x > –9
79.
The solution set is ℜ.
Copyright © Houghton Mifflin Company. All rights reserved.
SSM: Elementary Algebra
Chapter 2: Algebra Basics, Equations, and Inequalities
89. (a) –0.85x + 26.6 ≥ 20
27
97.
(b) –0.85x + 26.6 ≥ 20
The solution set is ℜ.
x ≤ 7.8; before 1988
99.
91. (a) –0.575x + 22.6 > –0.85x + 26.6
(b)
The solution set is Ø.
Section 2.8
x > 14.5; after 1995
93.
1. The properties allow us to add the same
number to both sides of an equation or
an inequality.
3. –4(–1) > 1(–1)
5. 3 + a > 3 + b
7. –4a > –4b
The solution set is Ø.
9.
95.
a
b
<
−5 −5
11.
x +5<1
x + 5 − 5 < 1− 5 Subtract 5 from
both sides
x <−4
13.
−3 ≤ 2 + x
−3 −2 ≤ 2 + x −2 Subtract 2
from both sides
−5 ≤ x
x ≥ −5
(a) The solution set is Ø.
(b) The solution set is ℜ.
Copyright © Houghton Mifflin Company. All rights reserved.
28
Chapter 2: Algebra Basics, Equations, and Inequalities
15.
5x −1≥ −1 +4x
5x −1− 4 x ≥ −1+ 4x − 4x Subtract 4 x from both sides
x −1≥ −1
x −1+1 ≥ −1+1
Add 1 to both sides
x≥ 0
17.
x − 2 > 2x + 6
x − 2 − x > 2x + 6 − x Subtract x from both sides
−2 > x + 6
−2 − 6 > x + 6 − 6 Subtract 6 from both sides
−8> x
x < −8
19.
3x > 12
3x 12
>
Divide both sides by 3
3
3
x> 4
21.
−5x <15
−5x 15
>
Divide both sides by − 5 and reverse the inequality
−5
−5
x > −3
23.
0 ≤ −x
−1(0)≥ −1(−x) Multiply by −1 and reverse the inequality
0 ≥x
x≤0
25.
27.
SSM: Elementary Algebra
3
x ≤12
4
4 3
4
4
⋅ x ≤ ⋅12 Multiply both sides by
3 4
3
3
x ≤16
x
≤1
−5
 x
−5   ≥ −5 ⋅1 Multiply by − 5 and reverse the inequality
 −5 
x ≥ −5
29. No. Because we multiplied both sides by a negative number, the inequality symbol must be
reversed: 32x > 89.
31.
5+ 3x >−4
3x >−9 Subtract 5 from both sides
3x −9
>
Divide both sides by 3
3
3
x >−3
33.
3− x ≤ 5
−x ≤ 2 Subtract 3 from both sides
−x
2
≥
Divide by −1 and reverse the inequality
−1 −1
x ≥ −2
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 2: Algebra Basics, Equations, and Inequalities
SSM: Elementary Algebra
35.
3− 4 x ≤ 19
−4 x ≤ 16
−4x 16 Subtract 3 from both sides
≥
Divide by − 4 and reverse the inequality
−4
−4
x ≥ −4
37.
5x + 5 ≤ 4x − 2
x + 5 ≤ −2
Subtract 4 x from both sides
x ≤ −7
Subtract 5 from both sides
39.
6− 3x > x − 2
6 − 4x >−2 Subtract x from both sides
−4x >−8 Subtract 6 from both sides
−4x −8
<
Divide by − 4 and reverse the inequality
−4 −4
x <2
41.
7+ x ≤ 3(1− x)
7 + x ≤ 3− 3x Remove grouping symbols
7 + 4x ≤ 3
4x ≤−4
4x −4
≤
4
4
x ≤−1
43.
45.
47.
Add 3x to both sides
Subtract 7 from both sides
Divide both sides by 4
6(2 − 3x) > 11(3− x)
12 −18x > 33 −11x
12 − 7x > 33
−7x > 21
−7x 21
<
−7
−7
x < −3
3y + 2(2y +1) >11+ y
3y + 4 y + 2 >11+ y
7y + 2 >11+ y
6y + 2 >11
6y > 9
6y 9
>
6 6
3
y>
2
Remove grouping symbols
Add 11x to both sides
Subtract 12 from both sides
Divide by − 7 and reverse the inequality
Remove grouping symbols
Combine like terms
Subtract y from both sides
Subtract 2 from both sides
Divide both sides by 6
6(t − 3) + 3≤ 2(4t + 3) + 5t
6t −18 + 3≤ 8t + 6 + 5t
6t −15 ≤13t + 6
−7t −15 ≤ 6
−7t ≤ 21
−7t 21
≥
−7 −7
t ≥ −3
Clear grouping symbols
Combine like terms
Subtract 13t from both sides
Add 15 to both sides
Divide by − 7 and reverse the inequality
Copyright © Houghton Mifflin Company. All rights reserved.
29
30
49.
51.
53.
Chapter 2: Algebra Basics, Equations, and Inequalities
SSM: Elementary Algebra
1 1
≥
The LCD is 6
2 3
1
1
6⋅2x + 6⋅ ≥ 6⋅
Clear denominators
2
3
12x + 3 ≥ 2
12x ≥ −1
Subtract 3 from both sides
12x −1
≥
Divide both sides by 12
12 12
1
x≥ −
12
2x +
1
3
1
t− <− t
The LCD is 12
2
4
3
1
3
1
12 ⋅ t −12 ⋅ <12 − t Clear denominators
 3 
2
4
6t − 9 < −4t
10t − 9 < 0
Add 4 t to both sides
10t < 9
Add 9 to both sides
10t 9
<
Divide both sides by 10
10 10
9
t<
10
−1.6t −1.4 ≥ 4.2 − 3t
1.4t −1.4 ≥ 4.2
Add 3t to both sides
1.4t ≥ 5.6
Add 1.4 to both sides
1.4t 5.6
≥
Divide both sides by 1.4
1.4 1.4
t≥4
55. If the resulting inequality is true, the solution set is ℜ. If the resulting inequality is false, the
solution set is Ø.
57.
3(x − 2) ≥ 2x + x
3x − 6 ≥ 2x + x Remove grouping symbols
3x − 6 ≥ 3x
Combine like terms
−6 ≥ 0
Subtract 3x from both sides
The solution set is Ø.
59.
2(3− x) + x ≤ 6− x
6 − 2x + x ≤ 6− x Remove grouping symbols
6 −x ≤ 6− x Combine like terms
6≤ 6
Add x to both sides
The solution set is ℜ.
61.
2x − 5+ x ≤−3(2 − x )
2x − 5 + x ≤−6 + 3x Remove grouping symbols
3x − 5 ≤ 3x − 6
Combine like terms
−5 ≤−6
Subtract 3x from both sides
The solution set is Ø.
63.
4 < x + 2 < 11
4 −2 < x + 2 −2 < 11− 2 Subtract 2 from all three parts
2 <x <9
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SSM: Elementary Algebra
Chapter 2: Algebra Basics, Equations, and Inequalities
65.
10 ≤ −5x < 35
10 −5x 35
≥
>
Divide by − 5 and reverse the inequalities
−5
−5 −5
−2 ≥ x >−7
−7 < x ≤−2
67.
y
≤2
The LCD is 3
3
y
3⋅ 0 ≤ 3⋅ ≤ 3⋅ 2 Multiply all three parts by 3
3
0 ≤ y≤6
69.
−5 < 2 x −1 <−1
−5 +1 < 2x −1 +1< −1+ 1 Add 1 to all three parts
−4 < 2x < 0
−4 2x 0
<
<
Divide all three parts by 2
2
2 2
−2 < x < 0
71.
−2 ≤ 4 − 3x ≤ 4
−2 − 4 ≤ 4− 3x − 4 ≤ 4 − 4 Subtract 4 from all three parts
−6 ≤ −3x ≤ 0
−6 −3x
0
≥
≥
Divide by − 3 and reverse the inequalities
−3 −3 −3
2≥ x ≥0
0 ≤ x ≤2
73.
−3x + 5≥ x −1
5≥ 4x −1 Add 3x to both sides
6 ≥ 4x
Add 1 to both sides
6 4x
≥
Divide both sides by 4
4
4
3
≥x
2
3
x≤
2
75.
5x − 6 ≥ 5x −1
−6 ≥ −1
Subtract 5 x from both sides
False, so the solution set is Ø.
77.
x+5 ≤ x+7
5 ≤7
Subtract x from both sides
True, so the solution set is ℜ.
79.
1− 3(x + 2) < 7
1− 3x − 6< 7
−3x − 5 < 7
−3x < 12
−3x 12
>
−3 −3
x > −4
0≤
Remove grouping symbols
Combine like terms
Add 5 to both sides
Divide by − 3 and reverse the inequality
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31
Chapter 2: Algebra Basics, Equations, and Inequalities
32
81.
83.
SSM: Elementary Algebra
3x −1
The LCD is 2
2
3x −1
2 ⋅2x + 2⋅1 > 2⋅
Clear denominator
2
4 x + 2 > 3x −1
x + 2 > −1
Subtract 3x from both sides
x > −3
Subtract 2 from both sides
2x +1>
2(x + 3) < 2x +1
2x + 6 < 2x +1 Remove grouping symbols
6<1
Subtract 2 x from both sides
False, so the solution set is Ø.
85.
2.3y + 0.75 ≥ 4.2
2.3y ≥ 3.45 Subtract 0.75 from both sides
2.3y 3.45
≥
2.3
2.3 Divide both sides by 2.3
y ≥1.5
87.
y2 =100 − (0.58x + 27.4)
y2 =100 − 0.58x − 27.4
y2 =−0.58x + 72.6
89. Until 1979 the number of employed women was less than the number of women who were not
employed.
91.
x+a < b
x < b − a Subtract a from both sides
93.
ax + b ≤ 0
ax ≤ −b
ax −b
≥
a
a
b
x≥ −
a
Subtract b from both sides
Divide by a and reverse the inequality since a < 0
95.
x(2x − 5) > 2x 2 + 5x + 20
2x 2 − 5x > 2x 2 + 5x + 20 Remove grouping symbols
−5x > 5x + 20
Subtract 2 x 2 from both sides
−10x > 20
Subtract 5x from both sides
−10x 20
<
Divide by −10 and reverse the inequality
−10 −10
x < −2
97.
3+ x < 2x − 4 < x + 6
3 + x − x < 2x − 4 − x < x + 6 − x Subtract x from all three parts
3< x−4 < 6
3+ 4< x−4 +4<6 +4
Add 4 to all three parts
7 < x < 10
Chapter 2 Review Exercises
1. (a) −4x + 7 = −4(−6) + 7 Replace x with − 6
= 24 + 7
= 31
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Chapter 2: Algebra Basics, Equations, and Inequalities
SSM: Elementary Algebra
(b) −x 2 +1= −(1)2 +1 Replace x with 1
= −1+1
=0
3. For x = –3, 7 + 2x − x 2 = −8.
5. (a) 3xy + x – 5xy + 4x = –2xy + 5x
(b) 8c 2 − 6c2 + 5c2 = 7c 2
7.
−3(2a − b+ 4) = −3(2a)− 3(−b) − 3(4) Distributive property
= −6a + 3b −12
9.
−(x + 2y) − 4(3x − y +1) = −x − 2y − 4(3x) − 4(−y)− 4(1) Distributive property
= −x − 2y − 12x + 4y − 4
= −13x + 2y − 4
Combine like terms
11. R(4, –1)
13. Point Q has coordinates (–3, 2).
15. The set of all points with a y-coordinate of 0 is the x-axis.
17. The point (0, 0) is called the origin.
19.
x
–4
2x – 3 –11
–1
3
7
–5
3
11
21.
x = 18
23. The y-coordinate will be –3 because the
y-coordinate is the value of the expression when x = 5.
25. Equation (b) is a conditional equation.
27. Yes, –3 is a solution.
29.
Yes, 4 is a solution.
31. The point is (–2, 20).
33. (a) The lines are parallel.
Copyright © Houghton Mifflin Company. All rights reserved.
33
34
Chapter 2: Algebra Basics, Equations, and Inequalities
SSM: Elementary Algebra
(b) The lines coincide.
35. Step (iii) would not produce an equivalent equation.
37. The next in solving equation (ii) would be to isolate the variable term.
39. 7 = 9x − 8x
7= x
Combine like terms
The solution is 7.
41.
43.
−5x + 8= −4x
−5x +8 + 5x = −4x + 5x Add 5x to both sides
8= x
The solution is 8.
x
= −6
3
 x
−3 −  = −3(−6) Multiply both sides by − 3
 3
x =18
The solution is 18.
−
45.
9 = 2x − 6
15 = 2x
Add 6 to both sides
15
=x
Divide both sides by 2
2
The solution is 15 .
2
47.
6 + 3x − 4 − 7x − 2 = 6− 5x − 10
−4x = −5x − 4
Combine like terms
x = −4
Add 5 x to both sides
The solution is –4.
49.
4
14 x 1
x+
= +
3
9
9 3
4
14
x
1
9 ⋅ x + 9 ⋅ = 9⋅ + 9⋅
3
9
9
3
12x +14 = x + 3
11x +14 =3
11x =−11
x =−1
The solution is –1.
The LCD is 9
Clear denominators
Subtract x from both sides
Subtract 14 from both sides
Divide both sides by 11
51.
2(x −3) − 3(x +1)= 0
2x − 6− 3x −3 = 0 Remove grouping symbols
−x − 9 = 0 Combine like terms
−9 = x Add x to both sides
The solution is –9.
53.
3x + 7 − x = 4 + 2x + 3
2x + 7 = 2x + 7
Combine like terms
7= 7
Subtract 2x from both sides
The solution set is ℜ.
Copyright © Houghton Mifflin Company. All rights reserved.
SSM: Elementary Algebra
55.
Chapter 2: Algebra Basics, Equations, and Inequalities
−3(x − 4) = 4x − (7x −13)
−3x + 12 = 4x − 7x + 13 Remove grouping symbols
−3x + 12 = −3x + 13
Combine like terms
12 = 13
Add 3 x to both sides
The solution set is Ø.
57. (iii) is a correct translation of x ≤ 5.
59. The point of intersection represents a solution for ≤ or ≥ inequalities.
61. (a) x ≤ 2
(b) –1 < x ≤ 3
63. (a) 8 + x > x – 6
(b) 8 + x < x – 6
65. Inequality (iii) is equivalent to x < 7.
67.
69.
−8 ≥ 3 − x
−11 ≥ −x Subtract 3 from both sides
−11 −x
≤
Divide by −1 and reverse the inequality
−1 −1
11 ≤ x
x ≥ 11
3
x ≤ 15
4
4  3 
4
4
− ⋅  − x ≥ − ⋅15 Multiply by − and reverse the inequality
3  4 
3
3
x ≥ −20
−
71.
−3(x + 1) < −(x −1)
−3x − 3 < −x +1 Remove grouping symbols
−3x < −x + 4
Add 3 to both sides
−2 x < 4
Add x to both sides
−2x 4
>
Divide by − 2 and reverse the inequality
−2 −2
x > −2
73.
−12 < −4x ≤16
−12 −4x 16
>
≥
Divide by − 4 and reverse all inequalities
−4
−4 −4
3> x ≥−4
−4 ≤ x < 3
75.
3(x −1)− (x +1) ≤ 2(x + 4)
3x − 3 − x −1≤ 2 x + 8 Remove grouping symbols
2x − 4 ≤ 2 x + 8 Combine like terms
−4 ≤ 8
Subtract 2 x from both sides
The solution set is ℜ.
Looking Ahead
1. 9 + 2(–6)
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35
Chapter 2: Algebra Basics, Equations, and Inequalities
36
3.
SSM: Elementary Algebra
3
(−28)
4
5. 2 – 6
7.
9
9
C + 32 = (30) + 32
5
5
= 54 + 32
= 86
9. Answers will vary.
11.
x − 0.32 x = 578
1x − 0.32 x = 578
(1− 0.32)x = 578
0.68x = 578
0.68x 578
=
0.68 0.68
x = 850
Chapter 2 Test
1.
3.
2x
2⋅ 2
− (x + y) =
− [2 + (−4)] Replace x with 2 and y with − 4
y
−4
4
=
− (−2)
−4
= −1 +2
=1
−(x − 1)+ 4(2 x + 3) =− x +1 +8x +12 Remove grouping symbols
= 7x +13
Combine like terms
5. Quadrant II
7. The y-coordinate of a point of the graph of an expression corresponds to the value of the
expression. If x is replaced with –4, the value of the expression 3x – 2 is –14. Thus if the
x-coordinate is –4, the y-coordinate is –14.
9.
x = 10
11. (a) The equation is an identity.
(b) The equation is a contradiction.
13.
14x − 20 = 1+ 13x
x − 20 = 1
Subtract 13x from both sides
x = 21
Add 20 to both sides
The solution is 21.
Copyright © Houghton Mifflin Company. All rights reserved.
SSM: Elementary Algebra
15.
17.
Chapter 2: Algebra Basics, Equations, and Inequalities
37
2
(x − 5) = x + 2
The LCD is 3
3
2
3 ⋅ (x − 5) = 3(x + 2) Clear denominator
3
2(x − 5) = 3(x + 2)
2x − 10 = 3x + 6 Remove grouping symbols
−10 = x + 6
Subtract 2 x from both sides
−16 = x
Subtract 6 from both sides
The solution is –16.
5− 5(1− x) = 3 + 5x
5 − 5+ 5x = 3+ 5x Remove grouping symbols
5x = 3 + 5x Combine like terms
0 =3
Subtract 5 x from both sides
The solution set is Ø.
19. x – 3 ≥ –2
21. (a) The solution set is Ø.
(b) The solution set is ℜ.
The graph of 9 – x is always above the graph of –11 – x. Thus the solution set for part (a) is Ø
and for part (b) is ℜ.
23.
25.
1
4x
≥−
−1.55
The LCD is 20
4
5
1
4x
20 ⋅x + 20⋅ ≥ 20 −  − 20(1.55) Clear denominators
 5
4
20x + 5 ≥ −16x − 31
36x + 5 ≥−31
Add 16 x to both sides
36x ≥ −36
Subtract 5 from both sides
36x −36
≥
Divide both sides by 36
36
36
x ≥ −1
x+
−4 ≤ 2(x − 7) < 0
−4 2(x − 7) 0
≤
<
Divide all three parts by 2
2
2
2
−2 ≤ x − 7 < 0
−2 + 7 ≤ x − 7 + 7 < 0 + 7 Add 7 to all three parts
5 ≤ x <7
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