Appendix A — Answer Key, Chapter 9: Equivalent Algebraic Expressions

SOL 7.PFA.2 (b–d) · Covers textbook Chapter 9 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 120 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Drawing convention used throughout: a long rectangle is an $x$-tile, a small square is a unit tile, a circle is a chip; blue pieces are positive and red pieces are negative.

Lesson 9.1 — Modeling Expressions with Algebra Tiles and Chips

Guided practice

1. $3x + 2$
2. Two $x$-tiles and five positive unit tiles.
3. One negative $x$-tile and four positive unit tiles.
4. $4x - 3$
5. A zero pair is worth $0$, since $x + (-x) = 0$ and $1 + (-1) = 0$. Removing it takes away a value of $0$, and subtracting $0$ leaves the total unchanged, so the model before and after represent equivalent expressions.
6. Two zero pairs cancel and one $x$-tile remains: $3x + (-2x) = x$.

Independent practice

7. a) $2x - 6$ b) $-3x + 1$ c) $x + 5$
8. Two negative $x$-tiles and three negative unit tiles.
9. $3$. Four zero pairs cancel and three positive chips remain.
10. $-3$. Two zero pairs cancel and three negative chips remain.
11. Four $x$-tiles and four negative $x$-tiles form four zero pairs, one for each tile, and nothing is left on the table. An empty model is worth $0$, so $4x + (-4x) = 0$ no matter what $x$ is.
12. Yes. Model A is $3x - 2$. Model B is $x - 5 + 2x + 3$: the one $x$-tile and two $x$-tiles make three $x$-tiles, and three of the five negative unit tiles pair off with the three positive unit tiles, leaving two negative unit tiles. So Model B is also $3x - 2$.
13. $3x + 2$. The model is three $x$-tiles and two positive unit tiles.
14. The student attached the subtraction sign to the wrong term. In $5 - 2x$, rewritten as $5 + (-2x)$, the $5$ is positive and the $2x$ is negative. The correct model is five positive unit tiles and two negative $x$-tiles.

Exit ticket 9.1

15. $2x - 4$
16. Three negative $x$-tiles and one positive unit tile.
17. Three zero pairs cancel and two $x$-tiles remain: $2x$.
18. The chips are opposites, so together they are worth $1 + (-1) = 0$. One counts a unit up and the other counts the same unit down, so they undo each other.

Lesson 9.2 — Like Terms and What Makes Them Alike

Guided practice

19. Coefficient $7$; constant $-4$.
20. Three terms: $3x$, $5$, and $-2x$.
21. $-1$, because $-x$ means $-1 \cdot x$.
22. No. $4x$ has the variable part $x$ and $4$ has no variable at all, so their variable parts are not the same.
23. $6x$ and $2x$ are like terms; $-3$ and $8$ are like terms.
24. Yes. Both have the variable part $x$; the coefficients being a decimal and a negative fraction does not matter.

Independent practice

25. $5x$, $-7$, and $x$
26. a) $9$ b) $-1$ c) $\tfrac{2}{3}$ d) $-2.5$
27. a) like b) like c) not like d) like
28. $x$-terms: $-2x$ and $5x$. Constants: $6$ and $-11$.
29. Three $x$-tiles are three long rectangles and $3$ is three small unit squares. They are different pieces, and since we do not know how many unit tiles fit in an $x$-tile, there is no way to merge the two piles into one count. So $3x + 3$ is already as short as it gets.
30. Yes, they are equivalent. At $x = 4$: $4 + 4 + 4 = 12$ and $3(4) = 12$. At $x = -2$: $-2 + (-2) + (-2) = -6$ and $3(-2) = -6$. Both models are three $x$-tiles.
31. $x$ and $x$ are like terms; $3$ and $3$ are like terms.
32. The sign in front of a term belongs to that term. Rewriting $8 - 3x$ as $8 + (-3x)$ shows the coefficient is $-3$, not $3$.

Exit ticket 9.2

33. Coefficient $-6$; constant $2.5$.
34. Yes. Both terms have the variable part $x$, and that is the only test for like terms.
35. Three terms: $4x$, $-9$, and $x$.
36. Two terms are like terms when their variable parts are exactly the same — both $x$, or both with no variable at all. The coefficients may be any rational numbers, positive or negative.

Lesson 9.3 — Combining Like Terms

Guided practice

37. $4x + 3x = (4 + 3)x = 7x$
38. $9x - 2x = (9 - 2)x = 7x$
39. $5x + 3x + 2 = (5 + 3)x + 2 = 8x + 2$
40. $-3x + 7x = (-3 + 7)x = 4x$
41. $6 + (-4x) + (-9) = -4x + (6 - 9) = -4x - 3$
42. $2x + (-5) + (-7x) + 1 = (2 - 7)x + (-5 + 1) = -5x - 4$
43. $0.5x + 1.5x = (0.5 + 1.5)x = 2x$
44. $\tfrac{1}{4}x + \tfrac{1}{2}x = \left(\tfrac{1}{4} + \tfrac{2}{4}\right)x = \tfrac{3}{4}x$

Independent practice

45. a) $(8 - 12)x = -4x$ b) $(-1 + 6)x = 5x$ c) $(3.2 - 1.2)x = 2x$ d) $\left(-\tfrac{2}{3} - \tfrac{1}{3}\right)x = -x$
46. $7x + 4 + (-2x) + (-9) = (7 - 2)x + (4 - 9) = 5x - 5$
47. $-5 + 3x + 8 + (-10x) = (3 - 10)x + (-5 + 8) = -7x + 3$
48. $\tfrac{1}{2}x + (-2x) + 3 + \tfrac{1}{2} = \left(\tfrac{1}{2} - 2\right)x + \tfrac{7}{2} = -\tfrac{3}{2}x + \tfrac{7}{2}$
49. $(-2.5 + 2.5)x + (-4 + 4) = 0x + 0 = 0$
50. $(6 - 1)x + (3 - 3) = 5x$
51. Both terms of $5x + 3x$ contain the factor $x$, so it can be factored out: $5x + 3x = (5 + 3)x = 8x$. In $5x + 3$ the second term has no factor of $x$, so there is nothing common to pull out front, and the expression stays as $5x + 3$.
52. $4x - x + 2 = (4 - 1)x + 2 = 3x + 2$. Check at $x = 5$: $4(5) - 5 + 2 = 20 - 5 + 2 = 17$ and $3(5) + 2 = 17$. Check at $x = -2$: $4(-2) - (-2) + 2 = -8 + 2 + 2 = -4$ and $3(-2) + 2 = -4$.
53. Perimeter $= x + (x + 4) + (2x - 1) = (1 + 1 + 2)x + (4 - 1) = 4x + 3$ inches.
54. $9$ and $-4x$ are not like terms — one has a variable and one does not — so they cannot be added, and the coefficient also lost its negative sign. The expression $9 - 4x$ is already simplified.

Exit ticket 9.3

55. $(3 + 8)x = 11x$
56. $-6x + x + 2 - 9 = (-6 + 1)x + (2 - 9) = -5x - 7$
57. $\left(\tfrac{3}{4} - \tfrac{1}{4}\right)x = \tfrac{2}{4}x = \tfrac{1}{2}x$
58. The two terms are not like terms. $7x$ has the variable part $x$ and $-7$ has no variable, so there is no common factor of $x$ to pull out and no way to add them into a single term.

Lesson 9.4 — The Distributive Property with Variables

Guided practice

59. $3(x) + 3(4) = 3x + 12$
60. $2(x) + 2(-5) = 2x - 10$
61. $(-4)(x) + (-4)(2) = -4x - 8$
62. $5(2x) + 5(-3) = 10x - 15$
63. $-1(x) + (-1)(-7) = -x + 7$
64. $\tfrac{1}{2}(6x) + \tfrac{1}{2}(8) = 3x + 4$
65. $3x + 6 + 4x = (3 + 4)x + 6 = 7x + 6$
66. $2x + 6 - x - 1 = (2 - 1)x + (6 - 1) = x + 5$

Independent practice

67. a) $6x + 6$ b) $(-2)(3x) + (-2)(-5) = -6x + 10$ c) $0.5(4x) + 0.5(2) = 2x + 1$ d) $-2x - 9$
68. $4x - 8 + 3x + 15 = (4 + 3)x + (-8 + 15) = 7x + 7$
69. $10x + 5 - 3x + 12 = (10 - 3)x + (5 + 12) = 7x + 17$
70. $-3x + 18 - 2x = (-3 - 2)x + 18 = -5x + 18$
71. $\tfrac{2}{3}(3x) + \tfrac{2}{3}(-9) + 4 = 2x - 6 + 4 = 2x - 2$
72. Draw a rectangle of height $3$ and width $x + 4$. Its area is $3(x + 4)$. Cutting it at the seam gives a left region of height $3$ and width $x$, with area $3x$, and a right region of height $3$ and width $4$, with area $12$. The two pieces together are the same rectangle, so $3(x + 4) = 3x + 12$.
73. $4(x + 3) = 4x + 12$ items.
74. The student multiplied $-3$ by $-2$ and wrote a negative result. A negative times a negative is positive, so $(-3)(-2) = +6$. The correct expansion is $-3(x - 2) = -3x + 6$.

Exit ticket 9.4

75. $7(x) + 7(-3) = 7x - 21$
76. $2x + 8 + 3x = (2 + 3)x + 8 = 5x + 8$
77. $-x + 5 + 2x = (-1 + 2)x + 5 = x + 5$
78. Multiplying by the outside factor makes that many copies of the whole quantity inside, and every term inside is part of each copy. The area model shows it: the factor is one side length of the full rectangle, so it multiplies every piece of the other side. Leaving a term out would account for only part of the rectangle.

Lesson 9.5 — Evaluating Expressions for Given Values

Guided practice

79. $3(4) + 5 = 12 + 5 = 17$
80. $3(-2) + 5 = -6 + 5 = -1$
81. $-2(3) + 7 = -6 + 7 = 1$
82. $4\left(-\tfrac{1}{2}\right) - 9 = -2 - 9 = -11$
83. $-5 + 2^3 = -5 + 8 = 3$
84. $5(2) - \sqrt{36} = 10 - 6 = 4$
85. $|-7| + 4 = 7 + 4 = 11$
86. $2[-4 + 3] - 1 = 2[-1] - 1 = -2 - 1 = -3$

Independent practice

87. a) $6(3) - 4 = 18 - 4 = 14$ b) $6(-3) - 4 = -18 - 4 = -22$ c) $6(0.5) - 4 = 3 - 4 = -1$
88. a) $-(12) + 10 = -12 + 10 = -2$ b) $-(-6) + 10 = 6 + 10 = 16$ c) $-\tfrac{1}{3} + 10 = 9\tfrac{2}{3}$
89. $3[2(4) - 5] + \sqrt{49} = 3[8 - 5] + 7 = 3[3] + 7 = 9 + 7 = 16$
90. $2^4 - 3(-2) = 16 + 6 = 22$
91. $|2(-3)| - 5 = |-6| - 5 = 6 - 5 = 1$
92. Simplified: $4(x - 1) + 2x = 4x - 4 + 2x = 6x - 4$. At $x = 3$: original $4(3 - 1) + 2(3) = 4(2) + 6 = 14$; simplified $6(3) - 4 = 14$. At $x = -\tfrac{1}{2}$: original $4\left(-\tfrac{1}{2} - 1\right) + 2\left(-\tfrac{1}{2}\right) = 4\left(-\tfrac{3}{2}\right) - 1 = -6 - 1 = -7$; simplified $6\left(-\tfrac{1}{2}\right) - 4 = -3 - 4 = -7$. The matching values are what equivalence means: simplifying did not change the value at either replacement value.
93. $12(3) + 5 = 36 + 5 = 41$, so the rental costs $41.
94. The student dropped the negative sign on the replacement value. Substituting correctly gives $5 - 2(-3) = 5 + 6 = 11$. Writing parentheses around $-3$ keeps the sign attached: $-2$ times $-3$ is $+6$, so the expression grows rather than shrinks.

Exit ticket 9.5

95. $7(-1) - 2 = -7 - 2 = -9$
96. $-4 + 3^2 = -4 + 9 = 5$
97. $|-10| - 6 = 10 - 6 = 4$
98. Without parentheses the substitution can be misread: $5 - 2x$ at $x = -3$ written as $5 - 2 - 3$ says something different from what was meant. Writing $5 - 2(-3)$ keeps the negative sign attached to the $3$ and keeps the multiplication visible, giving $5 + 6 = 11$.

Chapter 9 Review

Part A — Modeling with tiles and chips (7.PFA.2b)

99. $3x - 5$
100. Two negative $x$-tiles and four positive unit tiles.
101. Four zero pairs cancel and two $x$-tiles remain: $6x + (-4x) = 2x$.
102. $-3$. Eight zero pairs cancel and three negative chips remain.
103. Five $x$-tiles and five negative $x$-tiles pair off one to one, and each pair is worth $x + (-x) = 0$. All five pairs cancel, nothing is left on the table, and an empty model is worth $0$. So $5x + (-5x) = 0$ for every value of $x$.

Part B — Simplifying and generating equivalent expressions (7.PFA.2c)

104. a) $(9 + 5)x = 14x$ b) $(-7 + 3)x = -4x$ c) $(2.5 - 4.5)x = -2x$ d) $\left(\tfrac{5}{6} - \tfrac{2}{6}\right)x = \tfrac{3}{6}x = \tfrac{1}{2}x$
105. $8x + (-5x) + (-3) + 11 = (8 - 5)x + (11 - 3) = 3x + 8$
106. $5x - 10 + 3x = (5 + 3)x - 10 = 8x - 10$
107. $-8x - 2 + 6x = (-8 + 6)x - 2 = -2x - 2$
108. $\tfrac{3}{4}(8x) + \tfrac{3}{4}(-4) = 6x - 3$
109. $3x + 15 - 2x + 1 = (3 - 2)x + (15 + 1) = x + 16$
110. Two acceptable answers: $6(x + 2)$ and $4x + 12 + 2x$. The first is equivalent because distributing gives $6x + 12$ back. The second is equivalent because combining like terms gives $(4 + 2)x + 12 = 6x + 12$. Any expression that simplifies to $6x + 12$ is acceptable.

Part C — Evaluating expressions (7.PFA.2d)

111. a) $4(5) - 7 = 20 - 7 = 13$ b) $4(-2) - 7 = -8 - 7 = -15$ c) $4\left(\tfrac{1}{4}\right) - 7 = 1 - 7 = -6$
112. $-3(-1) + 2^3 = 3 + 8 = 11$
113. $|-6| + \sqrt{25} = 6 + 5 = 11$
114. $2[1.5 - 4] + 9 = 2[-2.5] + 9 = -5 + 9 = 4$
115. $10 - 5(-0.2) = 10 + 1 = 11$

Part D — Mixed application and reasoning

116. $3(2x - 1) + 4 = 6x - 3 + 4 = 6x + 1$, so the two expressions are equivalent. At $x = 2$: $3[2(2) - 1] + 4 = 3(3) + 4 = 13$ and $6(2) + 1 = 13$. At $x = -3$: $3[2(-3) - 1] + 4 = 3(-7) + 4 = -21 + 4 = -17$ and $6(-3) + 1 = -17$.
117. Total $= 2(2m + 3) = 4m + 6$ dollars. For $m = 7$: $4(7) + 6 = 28 + 6 = 34$, so the total is $34.
118. The student distributed $-2$ across the $x$ but not across the $3$, and also changed the sign of the second product. Distributing gives $-2(x + 3) = -2x - 6$, so $4 - 2(x + 3) = 4 - 2x - 6 = -2x - 2$.
119. Equivalence means the two expressions give the same value for every replacement value, and one test checks only one of infinitely many. For example, $x + 2$ and $3x$ both equal $3$ at $x = 1$, which looks like a match, but at $x = 2$ they give $4$ and $6$. One agreement proves nothing; one disagreement disproves equivalence. Only the algebra — distributing and combining like terms — proves two expressions are equivalent.
120. Model $2(x + 3)$ as two identical groups, each holding one $x$-tile and three positive unit tiles. Push the groups together and count: two $x$-tiles and six positive unit tiles. That is exactly the model for $2x + 6$. Since the same collection of tiles describes both, the expressions are equivalent.

Workbook-only items

Page 2, fill in the blanks. A long rectangle is an $x$-tile, worth $x$. A small square is a unit tile, worth 1. Blue pieces are positive and red pieces are negative. We never say how many unit tiles fit inside an $x$-tile because $x$ is unknown.

Page 6, table.

Page 10, why it works. $5x + 3x = (\mathbf{5} + \mathbf{3})x = \mathbf{8}x$. Pulling the common factor $x$ out front is the distributive property read backwards.

Page 10, item 41 frame. As a sum: $6 + (-4x) + (-9)$. Combine: $-4x - 3$.

Page 10, item 42 frame. As a sum: $2x + (-5) + (-7x) + 1$. $x$-terms: $2x$ and $-7x$, giving $-5x$. Constants: $-5$ and $1$, giving $-4$. Simplified: $-5x - 4$.

Page 14, watch the signs. $-3(x - 6) = (-3)(x) + (-3)(-6) = -3x + \mathbf{18}$.

Page 14, item 65 frame. Distribute: $3x + 6 + 4x$. Combine: $7x + 6$.

Page 14, item 66 frame. Distribute: $2x + 6 - x - 1$. Combine: $x + 5$.

Page 16, item 72 frame. The left region represents $3 \cdot x$, an area of $3x$. The right region represents $3 \cdot 4$, an area of $12$. Together they are the whole rectangle, whose area is $3(x + 4)$, so $3(x + 4) = 3x + 12$.

Page 18, substitution frame. $3x + 5$ at $x = -2$ becomes $3(\mathbf{-2}) + 5 = \mathbf{-1}$.

Page 20, item 92 table. At $x = 3$: both columns give $14$. At $x = -\tfrac{1}{2}$: both columns give $-7$.

Page 25, item 116 table. At $x = 2$: both columns give $13$. At $x = -3$: both columns give $-17$.