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Chapter 2 Algebra Basics, Equations, and...
Transcript of Chapter 2 Algebra Basics, Equations, and...
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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 Replace x with 3= 6 − 7= −1
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. 2x2 + 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. x2 + xy + 3y2 = (−3)2 + (−3) ⋅1+ 3 ⋅12 Replace x with − 3 and y with 1 = 9 − 3+ 3= 9
15.5− y3− x
= 5− (−1)3− 2
Replace x with 2 and y with −1
= 61
= 6
17.a + b
c= 3+ (−2)
4 Replace a with 3, b with − 2, and c with 4
= 14
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, x + 52x −1
≈ −1.12 .
2 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
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25. There are two terms in 2x + 1 but onlyone term in 2(x + 1). In 2(x + 1) both 2and x + 1 are factors.
27. Terms: 3y, x, –8Coefficients: 3, 1, –8
29. Terms: x2, 3x, –6Coefficients: 1, 3, –6
31. Terms: 7b, − c , 3a7
Coefficients: 7, −1, 37
33. 3x + 8x = (3 + 8)x = 11x
35. –y + 3y = (–1 + 3)y = 2y
37. –3x – 5x = (–3 – 5)x = –8x
39. 4t2 − t2 = 4 t2 −1 ⋅ t2
= (4 −1)t2 = 3t2
41. 6x – 7x – 3 = –x + 3
43. 9x – 12 – 3x + 5 = 6x – 7
45. 2m – 7n – 3m + 6n = –m – n
47. 10 – 2b + 3b – 9 = b + 1
49. 2a2 + a3 − a2 − 5a3 =− 4a3 + a2
51. ab + 3ac – 2ab + 5ac = –ab + 8ac
53.13
a+ 12
b − a + 23
b = − 23
a + 76
b
55. 2.1x – 4.2y + 0.3y + 1.2x = 3.3x –3.9y
57. Remove parentheses and combine liketerms.
59. −5(3x) = (−5 ⋅3)x = −15x
61. − 14
(4y) = − 14
⋅ 4
y = −y
63. − 56
− 910
x
= − 56
⋅ −910
x = 34
x
65. 5(4 + 3b) = 5(4)+ 5(3b)= 20 +15b= 15b+ 20
67. −3(5 − 2x) = −3(5) − 3(−2x)=−15 + 6x= 6x −15
69. 2(3x + y) = 2(3x) + 2(y) = 6x + 2y
71. −3(−x + y − 5) = −3(−x) − 3(y)− 3(−5)= 3x − 3y + 15
73. − 35
(5x + 20) = − 35
(5x)− 35
(20)
= −3x −12
75. – (x – 4) = –x + 4
77. –(3 + n) = –3 – n
79. – (2x + 5y – 3) = –2x – 5y + 3
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
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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
Distributive PropertySimplify inside bracketsDistributive propertyCombine like terms
95. 4 + 5 x −3(x + 2)[ ]= 4 + 5 x −3x − 6[ ] = 4 + 5 −2x − 6[ ] = 4 −10x − 30 = −10x − 26
Distributive PropertySimplify inside bracketsDistributive propertyCombine 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 = 5t = 2005 – 1980 = 25
(b)152
(5)3 = 937.5
152
(25)3 = 117,187.5
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
2 x + 3y= 1.2 + (−3.1)
2(1.2)+ 3(−3.1) Replace x with 1.2 and y with − 3.1
= 1.2− 3.12.4 − 9.3
= −1.9−6.9
≈ 0.28
105. 2a2 + (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-axisand y-axis. Their point of intersection iscalled the origin.
3. Quadrant I
5. Quadrant III
7. Quadrant II
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9. Quadrant IV
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)
19. The order of the coordinates isdifferent.
21.
23.
25.
The next ordered pair is (2, 4).
27.
The next ordered pair is (6, –12).
29.
The next ordered pair is (7, 10).
31.
The next ordered pair is (10, –7).
33. (a) +, –
(b) +, +
(c) –, +
(d) –, –
35. The point is either on the x-axis, or inquadrant I or IV.
37. The point is either on the y-axis, or inquadrant III or IV.
39. The point is in quadrant I or III.
41. The point is on the y-axis.
43. The point is in quadrant II or IV.
45. (a) The point is on the x-axis.
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 5
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(b) The point is on the y-axis.
(c) The point is in quadrant IV.
(d) The point is in quadrant I or III.
47. P(0, 13)
49. P(9, 9)
51. P(–12, 6)
53.
55.
57. B(–4, 6)
59. B(17, 0)
61. The length is 5 and the width is 3. Thearea is 15 and the perimeter is 16.
63. The length of each side is 3. The area is9 and the perimeter is 12.
65. (a) The second coordinates are:–1, 1, 3, 5, 7
(b) y = x + 2
67. (a) The second coordinates are:3, 1, 0, –1, –3
(b) y = –x
69. (a) The fourth vertex is (–40, –10).
(b) The length is 50 feet, the width is20feet, so the area is 1000 square feet.
71. (a) Sales/marketing and Writtencommunications would be the samedistance above the x-axis becausethey are both 35%
(b) Basic computer would be thegreatest distance above the x-axisbecause 68% is the largest.
73. They would all lie in the first quadrant.
75. (a) The years in which the number ofresolutions exceed the average forthe period
(b) The years in which the number ofresolutions was less than theaverage for the period
77. The midpoint is (2, 3).
79. The midpoint is (–4, 2).
81. The other two vertices are 8 units aboveor below the two given vertices. Sinceone of these vertices must be inquadrant I, it must be 8 units above thevertex in quadrant IV, (7, –3). Thecoordinates of P are (7, 5).
6 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
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Section 2.3
1. It is not necessary to enter the expression each time it is to be evaluated.
3. x –1 0 4 11
x – 6 –7 –6 –2 5
5. x –8 3 6 12
–2.1x + 7 23.8 0.7 –5.6 –18.2
7. x –9 –4 3 5
x2 + 4x – 21 24 –21 0 24
9. x –8 –1 2 10
2(x + 8) 0 14 20 36
11. x –5 –1 –0.5 8
2x +1 9 1 0 17
13. x –5 1 25 50
x +12x
0.4 1 0.52 0.51
15. x –2 1 4 7 10
x – 3 –5 –2 1 4 7
(x, x – 3) (–2, –5) (1, –2) (4, 1) (7, 4) (10, 7)
17. x 3 7 0 –4 –5
2x + 5 11 19 5 –3 –5
(x, 2x + 5) (3, 11) (7, 19) (0, 5) (–4, –3) (–5, –5)
19. x 1 4 6 –3 –5
x2 – 3x – 4 –6 0 14 14 36
(x, x2 – 3x – 4) (1, –6) (4, 0) (6, 14) (–3, 14) (–5, 36)
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.
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 7
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27.
x 2 5 7
2x –10
–6 0 4
29.
x –30 –9 12
13
x + 5 –5 2 9
31.
x –6 0 5
–(3x + 1) 17 –1 –16
33. The first coordinate, 3, is the value ofthe variable, and the second coordinate,7, is the value of the expression forx = 3.
35.
x 2 4 0 5 –1 –2
x – 4 –2 0 –4 1 –5 –6
37.
x –32 –8 –12 16 6 28
14
x + 3 –5 1 0 7 4.5 10
39.
(a) x = 3
(b) x = 9
(c) x = –5
(d) x = –10
41.
(a) x = –1
(b) x = 2
(c) x = 5
(d) x = 8
43.
(a) x = –23
(b) x = 6
(c) x = 20
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(d) x = 33
45.
(a) x = –2, 11
(b) x = –1, 10
(c) x = 1, 8
(d) x = 2, 7
47.
(a) x = –12, 22
(b) x = –3, 13
(c) x = 5
(d) x = 1, 9
49. 2x + 1 = 4 – x when x = 1.
51. 3x = 8 – x when x = 2.
53.
When x = 28, there are 3 pizzas left.
55. Year ReportedNumber
ModeledNumber
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 –7The model is most accurate for 1987.
57. Year xValue of the expression
1965 0 3241970 5 17871975 10 32501980 15 47141985 20 61771990 25 7640
59. For the year 2000, x = 35, and the valueof the expression is 10,567.
61. The y-coordinate is one more than twicethe x -coordinate.
63. The y-coordinate is twice the sum of thex-coordinate and 1.
65. (a) For c = 0.5 and k = 6.1,c2 + 5k = 30.75.
(b) For c = –0.5 and k = 6.1,c2 + 5k = 30.75.
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 9
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Section 2.4
1. An equation has an = symbol, but anexpression does not.
3. Expression
5. Equation
7. Equation
9. Expression
11. A solution of an equation is a value ofthe variable that makes the equationtrue.
13. x − 5 = 914 − 5 = 9 Replace x with 14
9 = 9 TrueYes
15. 5− 3x = 115 − 3 ⋅2 = 11 Replace x with 2
5− 6 = 11−1 = 11 False
No
17. 10x − 3 = 5x +110 ⋅1− 3 = 5⋅1+1 Replace x with 1
10 − 3 = 5+17 = 6 False
No
19. x2 + 3x + 2 = 0(−1)2 + 3(−1)+ 2 = 0 Replace x
with −11 − 3+ 2 = 0
0 = 0 TrueYes
21. Yes
23. Yes
25. Yes
27. No
29.
x = 14
31.
x = 12
33.
x = 4
35.
x = –14
37.
x = –32
39.
x = –20
41. Graph each side of the equation andtrace to the point of intersection. Thex-coordinate of that point is theestimated solution.
43.
x = –1.25
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45.
x = 0.14
47.
x = –1.71
49.
x = –7
51.
x = –12
53.
x = –12
55.
x = –20
57.
x = 8
59.
x = –9
61.
x = 15
63.
x = –9
65.
x = 15
67. (a) The graphs of the two sides of theequation do not intersect. Thesolution set is Ø.
(b) The graphs of the two sides of theequation coincide. The solution setis ℜ.
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 11
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69.
x = 0
71.
Ø
73.
ℜ
75.
Ø
77.
ℜ
79.
x = 8
81.
x = 24
83.
Ø
85.
ℜ
87. (a) 2.1x + 22.8 = 37.50; 7 (1997)
(b) 2.1x + 20.8 = 39.70; 9 (1999)
89.
x = 110
91.
x = –6, 30
93.
Ø
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Section 2.5
1. Equivalent equations are equations thathave exactly the same solutions.
3. 7x − 6x = 9x = 9 Combine like terms
5. 1.8x − 0.8x = 3.5x = 3.5 Combine like terms
7. (i)
9. (ii)
11. x + 5 = 9x + 5 − 5 = 9 − 5 Subtract 5 from both sides
x = 4
13. −4 = x − 6x − 6 = −4 Exchange sides
x − 6 + 6 = −4+ 6 Add 6 to both sidesx = 2
15. 25 + x = 3925 + 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 +12x + 12 = 12 Exchange sides
x +12 − 12 = 12 −12 Subtract 12 from both sidesx = 0
21. 3x = 2x −1 3x − 2 x = 2x −1 − 2x Subtract 2 x from both sides
x =−1
23. 7x = 5 + 6x7x − 6x = 5 + 6x − 6x Subtract 6x from both sides
x = 5
25. −4x + 2 = −3x−4x + 2 + 4x = −3x + 4x
2 = x x = 2
Add 4 x to both sides
Exchange sides
27. 9x − 6 = 10x9x − 6− 9 x = 10x − 9x
−6 = x x = −6
Subtract 9x from both sides
Exchange sides
29. −3a + 5a = 9+ a + 62a = a + 15
2a − a = a + 15− aa = 15
Combine like termsSubtract a from both sides
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31. 3x + 3− 3x + 7= 6x − 6− 5x10 = x − 6 Combine like terms
10 +6 = x − 6 +6 Add 6 to both sides16 = x
x = 16 Exchange sides
33. 7x − 9 = 8 + 6x7x − 9− 6 x = 8+ 6x − 6x
x −9 = 8x − 9+ 9 = 8 + 9
x = 17
Subtract 6x from both sides
Add 9 to both sides
35. 1− 3y = 4 − 4y1 − 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 +15x + 3 − 5x = 6x +1− 5x
3 = x +13 −1= x +1 −1
2 = xx = 2
Subtract 5x from both sides
Subtract 1 from both sides
Exchange sides
39. 1− 5x = 1 − 4x1 − 5x + 5x = 1− 4x + 5x
1 = 1+ x1 −1= 1 + x −1
0 = xx = 0
Add 5x to both sides
Subtract 1 from both sides
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
we divide both sides of 3x = 12 by 3, the left side becomes 3x3
= x . In both cases, the variable
is isolated.
43. (ii)
45. (i)
47. All of the given steps are correct. In each case, the result is 1x = 20 because
(i)43
and 34
are reciprocals whose product is 1.
(ii)34
divided by 34
is 1.
(iii) Multiplying by 4 and dividing by 3 is equivalent to multiplying by 43
.
49. 5x = 205x5
= 205
Divide both sides by 5
x = 4
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51. 48 = −16x−16x = 48−16x−16
= 48−16
x = −3
Exchange sidesDivide both sides by −16
53. −y = −3−y−1
= −3−1
Divide both sides by −1
y = 3
55. −3x = 0−3x−3
= 0−3
Divide both sides by − 3
x = 0
57.x9
= −4
9 ⋅ x9
= 9(−4) Multiply both sides by 9
x = −36
59.−t4
= −5
−4 ⋅ −t4
= −4(−5) Multiply both sides by − 4
t = 20
61.34
x = 9
43
⋅ 34
x = 43
⋅9 Multiply both sides by 43
x = 12
63. 2x − 7x = 0−5x = 0 Combine like terms−5x−5
= 0−5
Divide both sides by −5
x = 0
65. y − 6y =−25−5y =−25−5y−5
= −25−5
y = 5
Combine like terms
Divide both sides by − 5
67. 8 = 3t − 4t8 =− t
− t = 8−t−1
= 8−1
t =− 8
Combine like termsExchange sidesDivide both sides by −1
69. 9x = 8x9x − 8x = 8x − 8x Subtract 8x from both sides
x = 0
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71. −4x =−28−4 x−4
= −28−4
Divide both sides by − 4
x = 7
73. 5− 7x = −6x5− 7x + 7x = −6x + 7x
5 = xx = 5
Add 7 x to both sides
Exchange sides
75. −3+ x = 0−3 + x + 3= 0+ 3 Add 3 to both sides
x = 3
77. 1
4=−2y
−2y = 1
4−2y−2
=14−2
y = − 18
Exchange sides
Divide both sides by − 2
79. Distributive Property
81. Multiplication Property of Equations
83. Addition Property of Equations
85. 0.7x = 3.50.7x0.7
= 3.50.7
Divide both sides by 0.7
x = 5The sales in 1991 were at $3.5 billion.
87. 0.7x = 5.60.7x0.7
= 5.60.7
Divide both sides by 0.7
x = 8The solution is 8, which means that sales for 1994 were projected to be $5.6 billion.
89. x + a = bx + a − a = b − a Subtract a from both sides
x = b − a
91. 5x − a = b5x − a + a = b + a
5x = a + b5x5
= a + b5
x = a + b
5
Add a to both sides
Divide both sides by 5
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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 = 7k ⋅ 2+ 1= 7 Replace x with 22k + 1 = 7
2k +1− 1 = 7−1 Subtract 1 from both sides2k = 6 2k2
= 62
Divide both sides by 2
k = 3
97. 2x −1= 52x −1 +1= 5 +1
2x = 62x2
= 62
x = 3
Solve for xAdd 1 to both sides
Divide both sides by 2
3x − 4 = 3⋅3 − 4= 9 − 4= 5
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 = 45x + 14 −14 = 4 −14
5x =−105x5
= −105
x =−2
Subtract 14 from both sides
Divide both sides by 5
The solution is –2.
5. 2 −3x = 02 − 3x − 2 = 0 − 2 Subtract 2 from both sides
−3x = −2 −3x−3
= −2−3
Divide both sides by −3
x = 23
The solution is 23
.
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−0.37
= −1.11−0.37
Divide both sides by − 0.37
t = 3 The solution is 3.
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 17
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9. 7x = 2x − 57x − 2x = 2x − 5 − 2x Subtract 2x from both sides
5x = −5 5x5
= −55
Divide both sides by 5
x = −1 The solution is –1.
11. 3x = 5x +143x − 5x = 5x +14 − 5x Subtract 5x from both sides
−2x = 14−2x−2
= 14−2
Divide both sides by −2
x = −7 The solution is –7.
13. 4x − 7 = 6 x + 34 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−2
= 10−2
Divide both sides by − 2
x = −5The solution is –5.
15. 5x + 2 = 2 − 3x5x + 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 8x8
= 08
Divide both sides by 8
x = 0 The solution is 0.
17. x + 6 = 5x −10x + 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
−4= −16
−4Divide both sides by –4
x = 4The solution is 4.
19. 5 − 9x = 8x + 55 − 9x − 8x = 8x + 5 −8x
5 −17x = 55 −17x − 5 = 5 − 5
−17x =0−17x−17
= 0−17
x = 0
Subtract 8x from both sides
Subtract 5 from both sides
Divide both sides by −17
The solution is 0.
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21. 9 − x = x +159 − x − x = x +15 − x
9 − 2x = 159 − 2x −9 = 15 − 9
−2 x = 6−2 x−2
= 6−2
x = −3
Subtract x from both sides
Subtract 9 from both sides
Divide both sides by − 2
The solution is –3.
23. 3x + 8− 5x = 2− x + 2x − 3−2x +8 = x −1−3x +8 = −1
−3x = −9x = 3
Combine like termsSubtract x from both sidesSubtract 8 from both sidesDivide by − 3
The solution is 3.
25. 3+ 5y − 24 + 3y = y −1+ y8y − 21 = 2y −16y − 21 =−1
6y = 20
y = 206
= 103
Combine like terms Subtract 2y from both sides Add 21 to both sides
Divide by 6
The solution is 103
.
27. 5− 5x +8 − 4 = −9+ x9− 5x = −9+ x Combine like terms9− 6x = −9 Subtract x from both sides
−6x = −18 Subtract 9 from both sidesx = 3 Divide by − 6
The solution is 3.
29. 3(x +1) = 13x + 3 = 1
3x = −2
x = − 2
3
Remove grouping symbolsSubtract 3 from both sides
Divide by 3
The solution is − 23
.
31. 5(2x −1) = 3(3x −1)10x − 5 = 9 x − 3 Remove grouping symbols
x − 5 = −3 Subtract 9x from both sidesx = 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.
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 19
Copyright © Houghton Mifflin Company. All rights reserved.
35. 4(x −1) = 5 +3(x − 6)4x − 4 = 5 +3x −184x − 4 = 3x − 13
x − 4 = −13x = −9
Remove grouping symbolsCombine like termsSubtract 3 x from both sidesAdd 4 to both sides
The solution is –9.
37. 6x −(5x −3) = 06x − 5x + 3= 0 Remove grouping symbols
x + 3 = 0 Combine like termsx = −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 − 2x12 − 6x = −2x
12 = 4x3= x
Remove grouping symbolsCombine like termsAdd 2 to both sidesAdd 6 x to both sidesDivide by 4
The solution is 3.
41. The number n is the LCD of all fractions in the equation.
43.2
3x − 2
3= 5
3 The LCD is 3
3 ⋅ 2
3x − 3 ⋅ 2
3= 3 ⋅ 5
3 Multiply by 3 to clear denominators
2x − 2 = 5 2x = 7 Add 2 to both sides
x = 72
Divide by 2
The solution is 72
.
45. 1+ 3t4
= 14
The LCD is 4
4 ⋅1 + 4 ⋅ 3t4
= 4 ⋅ 14
Multiply by 4 to clear denominators
4+ 3t = 13t = −3 Subtract 4 from both sidest = −1 Divide by 3
The solution is –1.
47.12
x − 2 = 1+ 23
x + 16
6 ⋅ 12
x − 6 ⋅2 = 6 ⋅1 + 6 ⋅ 23
x + 6 ⋅ 16
3x −12 = 6 + 4 x +13x −12 = 7+ 4 x
−12 = 7+ x−19 = x
The LCD is 6
Multiply by 6 to clear denominators
Combine like termsSubtract 3x from both sidesSubtract 7 from both sides
The solution is –19.
20 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
Copyright © Houghton Mifflin Company. All rights reserved.
49.x + 5
3= − x
2+ 5 The LCD is 6
6 ⋅ x + 5
3= 6 ⋅ −x
2+ 6 ⋅ 5 Multiply by 6 to clear denominators
2(x + 5)= −3x + 30
2x +10 = −3x + 30 Remove grouping symbols
5x +10 = 30 Add 3 x to both sides
5x = 20 Subtract 10 from both sides
x = 4 Divide by 5The solution is 4.
51.35
(x −1) = 9 The LCD is 5
53
⋅ 35
(x −1) = 53
⋅9 Multiply by 53
to clear denominators and isolate x
x −1 =15x =16 Add 1 to both sides
The solution is 16.
53.x3
+ 2 = 32
(x + 3) − 6 The LCD is 6
6 ⋅ x3
+ 6 ⋅ 2 = 6 ⋅ 32
(x + 3) − 6 ⋅6 Multiply by 6 to clear denominators
2x +12 = 9(x + 3) −362x +12 = 9x + 27 − 36 Remove grouping symbols2x +12 = 9x − 9 Combine like terms
12 = 7x − 9 Subtract 2 x from both sides21 = 7x Add 9 to both sides3 = x Divide by 7
The solution is 3.
55.37
x = 514
(x +1)+ 17
(8 − x) The LCD is 14
14 ⋅ 37
x = 14 ⋅ 514
(x +1) +14 ⋅ 17
(8 − x) Multiply by 14 to clear denominators
6x = 5(x + 1)+2(8 − x)6x = 5x + 5 +16 − 2x Remove grouping symbols6x =3x + 21 Combine like terms3x = 21 Subtract 3x from both sides
x = 7 Divide by 3The solution is 7.
57. 7.35x + 5.1 = 2.6x − 0.67.35x = 2.6 x − 5.74.75x = −5.7
x = −1.2
Subtract 5.1 from both sidesSubtract 2.6 x from both sidesDivide by 4.75
The solution is –1.2.
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 21
Copyright © Houghton Mifflin Company. All rights reserved.
59. 2.7(3x −1.5) + 6.3 = 9.1x + 6.558.1x − 4.05 + 6.3 = 9.1x + 6.55
8.1x + 2.25 = 9.1x + 6.558.1x = 9.1x + 4.3
−x = 4.3x = −4.3
Remove grouping symbolsCombine like termsSubtract 2.25 from both sidesSubtract 9.1x from both sidesMultiply by −1
The solution is –4.3.
61. x + 7 = 7+ xx + 7 − x = 7+ x − x Subtract x from both sides
7 = 7 TrueIdentitySolution set: ℜ
63. x = x + 3x − x = x + 3− x Subtract x from both sides
0 = 3 FalseContradictionSolution Set: Ø
65. 2x = 3x2x − 2x = 3x − 2x Subtract 2x from both sides
0 = xConditionalThe solution is 0.
67. 6t + 7 = 76t = 0 Subtract 7 from both sidest = 0 Divide by 6
ConditionalThe solution is 0.
69. 0x = 30 = 3 False
ContradictionThe solution set is Ø.
71. 6x − 3− x = 6x + 55x − 3= 6x + 5
−3 = x + 5−8 = x
Combine like termsSubtract 5x from both sidesSubtract 5 from both sides
The solution is –8.
73. 2 + 3x + 5 = 6x + 7 − 6x3x + 7 = 7 Combine like terms
3x = 0 Subtract 7 from both sidesx = 0 Divide by 3
The solution is 0.
22 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
Copyright © Houghton Mifflin Company. All rights reserved.
75.3
4x + 5
2= 1
2(5 + x)+ 1
4x The LCD is 4
4 ⋅ 34
x + 4 ⋅ 52
= 4⋅ 12
(5 + x) + 4 ⋅ 14
x Multiply by 4 to clear denominators
3x +10 = 2(5 + x )+ x
3x +10 =10 + 2x + x Remove grouping symbols
3x +10 = 3x +10 Combine like terms
10 =10 True Subtract 3x from both sidesSolution set: ℜ
77. 4(x − 2) +8 = 6(x +1)− 2x4x −8 +8 = 6x + 6 − 2x Remove grouping symbols
4x = 4x + 6 Combine like terms0 = 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 symbols6x − 8 = −4 + 6x Combine like terms
−8 = −4 False Subtract 6x from both sidesSolution set: Ø
81. 3x −1− 2(x − 4) = 2x + 7 − x3x −1− 2x + 8 = 2 x + 7− x Remove grouping symbols
x + 7 = x +7 Combine like terms7 = 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 =1505.88x =109.42
x =18.61 1990 + 19 = 2009(ii) 2.28x +15.7 = 2(31.637)
2.28x = 47.574x = 20.87
1990 + 21 = 2011
85. For 1976, t = 20Men: (−0.015)(20) +10.5 = −0.3+10.5
= 10.2 secondsWomen: ( −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 are5.51 seconds.
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 23
Copyright © Houghton Mifflin Company. All rights reserved.
89. x − 5 x − 5 x − 5(x − 5)[ ]{ } = 1x − 5 x − 5 x − 5x + 25[ ]{ } =1 Remove parentheses
x − 5 x − 5 −4x + 25[ ]{ } = 1 Combine like termsx − 5 x + 20x −125{ } = 1 Remove grouping symbols
x − 5 21x −125{ }= 1 Combine like termsx −105x + 625 = 1 Remove grouping symbols
−104x + 625 = 1 Combine like terms−104x = −624 Subtract 625 from both sides
x = 6 Divide by − 104 The solution is 6.
91. x − 2(x − 3) − (x + 1)[ ] = −3(x −1)+ 4 + 3xx − 2x − 6 − x −1[ ] = −3x + 3 + 4+ 3x Remove parentheses
x − 2x + 6 + x +1 = −3x + 3 + 4+ 3x Remove brackets7 = 7 True Combine like terms
Solution set: ℜ
93. ax + b = cax = c − b Subtract b from both sides
x = c − ba
Divide by a
95. 2t2 −15 = t(t + 5)+ t2
2 t2 −15 = t2 + 5t + t2 Remove grouping symbols2 t2 −15 = 2t2 + 5t Combine like terms
−15 = 5t Subtract 2 t2 from both sides−3 = t Divide by 5
The solution is –3.
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.
24 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
Copyright © Houghton Mifflin Company. All rights reserved.
5. 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 anendpoint is included. The symbols ( and) indicate that an endpoint is notincluded.
11.
13.
15.
17.
19.
21.
23. n ≥ 0
25. x ≤ 3
27. n < 0
29. x ≤ 6
31. x < –3
33. x ≥ 5
35. –4 < x ≤ 7
37. The point of intersection represents asolution if the inequality symbol is ≤ or≥.
39. x > –15
41. x ≤ 0
43.
x > –17
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 25
Copyright © Houghton Mifflin Company. All rights reserved.
45.
x ≥ –5
47.
x < –3
49.
x ≤ 7
51.
x < 0
53.
x < 2
55.
x ≥ –8
57.
x < 12
59. If the inequality is y1 > y2 or y1 ≥ y2the solution set is ℜ. If the inequality isy1 < y2or y1 ≤ y2 , the solution set is Ø.
61. The solution set is ℜ.
63. (a)
The solution set is ℜ.
(b)
The solution set is Ø.65.
–11 ≤ x ≤ 23
67.
–19 < x ≤ 13
26 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
Copyright © Houghton Mifflin Company. All rights reserved.
69.
–13 < x ≤ 9
71.
–6 < x < 8
73.
x ≤ 9
75.
The solution set is Ø.
77.
x > –9
79.
The solution set is ℜ.
81.
x < 0
83.
The solution set is Ø.
85. x + 6 ≥ 3
x ≥ –3
87. 8 – x > –2
x < 10
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 27
Copyright © Houghton Mifflin Company. All rights reserved.
89. (a) –0.85x + 26.6 ≥ 20
(b) –0.85x + 26.6 ≥ 20
x ≤ 7.8; before 1988
91. (a) –0.575x + 22.6 > –0.85x + 26.6
(b)
x > 14.5; after 1995
93.
The solution set is Ø.
95.
(a) The solution set is Ø.
(b) The solution set is ℜ.
97.
The solution set is ℜ.
99.
The solution set is Ø.
Section 2.8
1. The properties allow us to add the samenumber to both sides of an equation oran inequality.
3. –4(–1) > 1(–1)
5. 3 + a > 3 + b
7. –4a > –4b
9.a
−5< b
−5
11. x + 5 < 1x + 5 − 5 < 1− 5 Subtract 5 from
both sidesx <−4
13. −3 ≤ 2 + x−3 −2 ≤ 2 + x −2 Subtract 2
from both sides−5 ≤ x
x ≥ −5
28 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
Copyright © Houghton Mifflin Company. All rights reserved.
15. 5x −1≥ −1 +4x5x −1− 4 x ≥ −1+ 4x − 4x Subtract 4 x from both sides
x −1≥ −1x −1+1 ≥ −1+1 Add 1 to both sides
x ≥ 0
17. x − 2 > 2x + 6x − 2 − x > 2x + 6 − x Subtract x from both sides
−2 > x + 6−2 − 6 > x + 6 − 6 Subtract 6 from both sides
−8> xx < −8
19. 3x > 123x3
> 123
Divide both sides by 3
x > 4
21. −5x <15−5x−5
> 15−5
Divide both sides by − 5 and reverse the inequality
x > −3
23. 0 ≤ −x−1(0)≥ −1(−x) Multiply by −1 and reverse the inequality
0 ≥ xx ≤ 0
25.34
x ≤12
43
⋅ 34
x ≤ 43
⋅12 Multiply both sides by 43
x ≤16
27.x
−5≤ 1
−5x
−5
≥ −5 ⋅1 Multiply by − 5 and reverse the inequality
x ≥ −5
29. No. Because we multiplied both sides by a negative number, the inequality symbol must bereversed: 32x > 89.
31. 5+ 3x >−43x >−9 Subtract 5 from both sides3x
3> −9
3 Divide both sides by 3
x >−3
33. 3− x ≤ 5−x ≤ 2 Subtract 3 from both sides−x
−1≥ 2
−1 Divide by −1 and reverse the inequality
x ≥ −2
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 29
Copyright © Houghton Mifflin Company. All rights reserved.
35. 3− 4 x ≤ 19−4 x ≤ 16−4x−4
≥ 16−4
x ≥ −4
Subtract 3 from both sidesDivide by − 4 and reverse the inequality
37. 5x + 5 ≤ 4x − 2x + 5 ≤ −2 Subtract 4x from both sides
x ≤ −7 Subtract 5 from both sides
39. 6− 3x > x − 26 − 4x >−2 Subtract x from both sides
−4x >−8 Subtract 6 from both sides−4x−4
< −8−4
Divide by − 4 and reverse the inequality
x < 2
41. 7+ x ≤ 3(1− x)
7 + x ≤ 3− 3x
7 + 4x ≤ 3
4x ≤−44x4
≤ −44
x ≤−1
Remove grouping symbols
Add 3x to both sides
Subtract 7 from both sides
Divide both sides by 4
43. 6(2 − 3x) > 11(3− x)12 −18x > 33 −11x Remove grouping symbols12 − 7x > 33 Add 11x to both sides
−7x > 21 Subtract 12 from both sides−7x−7
< 21−7
Divide by − 7 and reverse the inequality
x < −3
45. 3y + 2(2y +1) >11+ y3y + 4 y + 2 >11+ y
7y + 2 >11+ y6y + 2 >11
6y > 96y6
> 96
y > 32
Remove grouping symbolsCombine like termsSubtract y from both sidesSubtract 2 from both sidesDivide both sides by 6
47. 6(t − 3) + 3≤ 2(4 t + 3) + 5t6t −18 + 3≤ 8t + 6 + 5t
6t −15 ≤13t + 6−7t −15 ≤ 6
−7t ≤ 21−7t−7
≥ 21−7
t ≥ −3
Clear grouping symbolsCombine like termsSubtract 13t from both sidesAdd 15 to both sidesDivide by − 7 and reverse the inequality
30 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
Copyright © Houghton Mifflin Company. All rights reserved.
49. 2x + 12
≥ 13
The LCD is 6
6 ⋅2x + 6 ⋅ 12
≥ 6 ⋅ 13
Clear denominators
12x + 3 ≥ 212x ≥ −1 Subtract 3 from both sides12x12
≥ −112
Divide both sides by 12
x ≥ − 112
51.1
2t − 3
4< − 1
3t The LCD is 12
12 ⋅ 1
2t −12 ⋅ 3
4<12 − 1
3t
Clear denominators
6 t − 9 < −4t10t − 9 < 0 Add 4 t to both sides
10t < 9 Add 9 to both sides10t10
< 910
Divide both sides by 10
t < 910
53. −1.6t −1.4 ≥ 4.2 − 3t1.4t −1.4 ≥ 4.2 Add 3t to both sides
1.4t ≥ 5.6 Add 1.4 to both sides1.4t1.4
≥ 5.61.4
Divide both sides by 1.4
t ≥ 4
55. If the resulting inequality is true, the solution set is ℜ. If the resulting inequality is false, thesolution set is Ø.
57. 3(x − 2) ≥ 2x + x3x − 6 ≥ 2x + x Remove grouping symbols3x − 6 ≥ 3x Combine like terms
−6 ≥ 0 Subtract 3x from both sidesThe solution set is Ø.
59. 2(3− x) + x ≤ 6− x6 − 2x + x ≤ 6− x Remove grouping symbols
6 −x ≤ 6− x Combine like terms6 ≤ 6 Add x to both sides
The solution set is ℜ.
61. 2x − 5+ x ≤−3(2 − x )2x − 5 + x ≤−6 + 3x
3x − 5 ≤ 3x − 6−5 ≤−6
Remove grouping symbolsCombine like termsSubtract 3x from both sides
The solution set is Ø.
63. 4 < x + 2 < 114 −2 < x + 2 −2 < 11− 2 Subtract 2 from all three parts
2 < x < 9
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 31
Copyright © Houghton Mifflin Company. All rights reserved.
65. 10 ≤ −5x < 3510
−5≥ −5x
−5> 35
−5 Divide by − 5 and reverse the inequalities
−2 ≥ x >−7−7 < x ≤−2
67.0 ≤ y
3≤ 2 The LCD is 3
3 ⋅ 0 ≤ 3⋅ y3
≤ 3⋅ 2 Multiply all three parts by 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 −42
< 2x2
< 02
Divide all three parts by 2
−2 < x < 0
71. −2 ≤ 4 − 3x ≤ 4−2 − 4 ≤ 4− 3x − 4 ≤ 4 − 4
−6 ≤ −3x ≤ 0−6−3
≥ −3x−3
≥ 0−3
2 ≥ x ≥ 00 ≤ x ≤ 2
Subtract 4 from all three parts
Divide by − 3 and reverse the inequalities
73. −3x + 5≥ x −15≥ 4x −1 Add 3x to both sides6 ≥ 4x Add 1 to both sides6
4≥ 4x
4 Divide both sides by 4
3
2≥ x
x ≤ 3
2
75. 5x − 6 ≥ 5x −1−6 ≥ −1 Subtract 5x from both sides
False, so the solution set is Ø.
77. x + 5 ≤ x + 75 ≤ 7 Subtract x from both sides
True, so the solution set is ℜ.
79. 1− 3(x + 2) < 71− 3x − 6< 7
−3x − 5 < 7−3x < 12−3x−3
> 12−3
x > −4
Remove grouping symbolsCombine like termsAdd 5 to both sidesDivide by − 3 and reverse the inequality
32 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
Copyright © Houghton Mifflin Company. All rights reserved.
81. 2x +1> 3x −12
The LCD is 2
2 ⋅2x + 2 ⋅1 > 2 ⋅ 3x −12
Clear denominator
4 x + 2 > 3x −1x + 2 > −1 Subtract 3x from both sides
x > −3 Subtract 2 from both sides
83. 2(x + 3) < 2x +12x + 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 ≥ 4.22.3y ≥ 3.452.3y
2.3≥ 3.45
2.3y ≥1.5
Subtract 0.75 from both sides
Divide both sides by 2.3
87. y2 =100 − (0.58x + 27.4)y2 =100 − 0.58x − 27.4y2 =−0.58x + 72.6
89. Until 1979 the number of employed women was less than the number of women who were notemployed.
91. x + a < bx < b − a Subtract a from both sides
93. ax + b ≤ 0
ax ≤ −baxa
≥ −ba
x ≥ − ba
Subtract b from both sides
Divide by a and reverse the inequality since a < 0
95. x(2x − 5) > 2x2 + 5x + 202x2 − 5x > 2x2 + 5x + 20 Remove grouping symbols
−5x > 5x + 20 Subtract 2x2 from both sides−10x > 20 Subtract 5x from both sides−10x−10
< 20−10
Divide by −10 and reverse the inequality
x < −2
97. 3+ x < 2x − 4 < x + 63 + x − x < 2x − 4 − x < x + 6 − x Subtract x from all three parts
3 < x − 4 < 63 + 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
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 33
Copyright © Houghton Mifflin Company. All rights reserved.
(b) −x2 +1= −(1)2 +1 Replace x with 1= −1+1= 0
3. For x = –3, 7 + 2x − x2 = −8.
5. (a) 3xy + x – 5xy + 4x = –2xy + 5x
(b) 8c2 − 6c2 + 5c2 = 7c2
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)= −x − 2y − 12x + 4y − 4= −13x + 2y − 4
Distributive property
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 –1 3 7
2x – 3 –11 –5 3 11
21.
x = 18
23. The y-coordinate will be –3 because they-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.
34 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
Copyright © Houghton Mifflin Company. All rights reserved.
(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 − 8x7 = x Combine like terms
The solution is 7.
41. −5x + 8= −4x−5x +8 + 5x = −4x + 5x Add 5x to both sides
8 = xThe solution is 8.
43. − x3
= −6
−3 − x3
= −3(−6) Multiply both sides by − 3
x =18The solution is 18.
45. 9 = 2x − 615 = 2x Add 6 to both sides152
= x Divide both sides by 2
The solution is 152
.
47. 6 + 3x − 4 − 7x − 2 = 6− 5x − 10−4x = −5x − 4 Combine like terms
x = −4 Add 5 x to both sidesThe solution is –4.
49.43
x + 149
= x9
+ 13
The LCD is 9
9 ⋅ 43
x + 9 ⋅149
= 9 ⋅ x9
+ 9 ⋅ 13
Clear denominators
12x +14 = x + 311x +14 =3 Subtract x from both sides
11x =−11 Subtract 14 from both sidesx =−1 Divide both sides by 11
The solution is –1.
51. 2(x −3) − 3(x +1)= 02x − 6− 3x −3 = 0
−x − 9 = 0−9 = x
Remove grouping symbolsCombine like termsAdd x to both sides
The solution is –9.
53. 3x + 7 − x = 4 + 2x + 32x + 7 = 2x + 7 Combine like terms
7 = 7 Subtract 2x from both sidesThe solution set is ℜ.
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 35
Copyright © Houghton Mifflin Company. All rights reserved.
55. −3(x − 4) = 4x − (7x −13)−3x + 12 = 4x − 7x + 13−3x + 12 = −3x + 13
12 = 13
Remove grouping symbolsCombine like termsAdd 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. −8 ≥ 3 − x−11 ≥ −x Subtract 3 from both sides−11−1
≤ −x−1
Divide by −1 and reverse the inequality
11 ≤ xx ≥ 11
69. − 34
x ≤ 15
− 43
⋅ − 34
x
≥ − 43
⋅15 Multiply by − 43
and reverse the inequality
x ≥ −20
71. −3(x + 1) < −(x −1)−3x − 3 < −x +1
−3x < −x + 4−2 x < 4−2x−2
> 4−2
x > −2
Remove grouping symbolsAdd 3 to both sidesAdd x to both sidesDivide by − 2 and reverse the inequality
73. −12 < −4x ≤16−12
−4> −4x
−4≥ 16
−4 Divide by − 4 and reverse all inequalities
3> x ≥−4−4 ≤ x < 3
75. 3(x −1)− (x +1) ≤ 2(x + 4)3x − 3 − x −1≤ 2 x + 8
2x − 4 ≤ 2 x + 8−4 ≤ 8
Remove grouping symbolsCombine like termsSubtract 2 x from both sides
The solution set is ℜ.
Looking Ahead
1. 9 + 2(–6)
36 Chapter 2: Algebra Basics, Equations, and Inequalities SSM: Elementary Algebra
Copyright © Houghton Mifflin Company. All rights reserved.
3.34
(−28)
5. 2 – 6
7.95
C + 32 = 95
(30) + 32
= 54 + 32= 86
9. Answers will vary.
11. x − 0.32 x = 5781x − 0.32 x = 578(1− 0.32)x = 578
0.68x = 5780.68x0.68
= 5780.68
x = 850
Chapter 2 Test
1.2xy
− (x + y) = 2 ⋅ 2−4
− [2 + (−4)] Replace x with 2 and y with − 4
= 4−4
− (−2)
= −1 +2= 1
3. −(x − 1)+ 4(2 x + 3) =− x +1 +8x +12= 7x +13
Remove grouping symbolsCombine like terms
5. Quadrant II
7. The y-coordinate of a point of the graph of an expression corresponds to the value of theexpression. If x is replaced with –4, the value of the expression 3x – 2 is –14. Thus if thex-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+ 13xx − 20 = 1 Subtract 13x from both sides
x = 21 Add 20 to both sidesThe solution is 21.
SSM: Elementary Algebra Chapter 2: Algebra Basics, Equations, and Inequalities 37
Copyright © Houghton Mifflin Company. All rights reserved.
15.23
(x − 5) = x + 2 The LCD is 3
3 ⋅ 23
(x − 5) = 3(x + 2) Clear denominator
2(x − 5) = 3(x + 2)2x − 10 = 3x + 6 Remove grouping symbols
−10 = x + 6 Subtract 2x from both sides−16 = x Subtract 6 from both sides
The solution is –16.
17. 5− 5(1− x) = 3 + 5x5 − 5+ 5x = 3+ 5x Remove grouping symbols
5x = 3 + 5x Combine like terms0 = 3 Subtract 5x 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 Øand for part (b) is ℜ.
23. x + 14
≥ − 4x5
−1.55 The LCD is 20
20 ⋅x + 20 ⋅ 14
≥ 20 − 4x5
− 20(1.55) Clear denominators
20x + 5 ≥ −16x − 3136x + 5 ≥−31 Add 16x to both sides
36x ≥ −36 Subtract 5 from both sides36x36
≥ −3636
Divide both sides by 36
x ≥ −1
25. −4 ≤ 2(x − 7) < 0−42
≤ 2(x − 7)2
< 02
Divide all three parts by 2
−2 ≤ x − 7 < 0−2 + 7 ≤ x − 7 + 7 < 0 + 7 Add 7 to all three parts
5 ≤ x < 7