Appendix A — Answer Key, Chapter 4: Equivalent Algebraic Expressions

SOL 8.PFA.1 · Covers textbook Chapter 4 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 126 across the chapter. Reasoning answers show an acceptable response, not the only wording.

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

Where a check by substitution is shown, it is there because it is the fastest way to catch a sign error. Substitution is evidence, not proof; the properties of real numbers are the proof.

Lesson 4.1 — Representing Expressions with Tiles and Pictures

Guided practice

1. $2x - 5$
2. Four $x$-tiles and three positive unit tiles.
3. Three negative $x$-tiles and two positive unit tiles.
4. $3x - 4$
5. A zero pair is worth $0$, since $x + (-x) = 0$ and $1 + (-1) = 0$ by the additive inverse property. Removing it takes away a value of $0$, and taking away $0$ leaves the total unchanged, so the model before and after represent equivalent expressions.
6. Three zero pairs cancel and two $x$-tiles remain: $5x + (-3x) = 2x$.

Independent practice

7. a) $4x - 7$ b) $-2x + 3$ c) $x - 6$
8. Six positive unit tiles and two negative $x$-tiles. Rewriting $6 - 2x$ as $6 + (-2x)$ shows which term the subtraction sign belongs to.
9. Two identical rows, each holding one $x$-tile and four unit tiles. Counting the whole rectangle gives two $x$-tiles and eight unit tiles: $2(x + 4) = 2x + 8$.
10. Three identical rows, each holding two $x$-tiles and one unit tile. Counting the whole rectangle gives six $x$-tiles and three unit tiles: $3(2x + 1) = 6x + 3$.
11. Lay out six $x$-tiles and four unit tiles, then split the pile into two equal groups, because the factor $\tfrac{1}{2}$ asks for one of two equal parts. Each group holds three $x$-tiles and two unit tiles, so $\tfrac{1}{2}(6x + 4) = 3x + 2$.
12. Two zero pairs of $x$-tiles cancel, leaving one $x$-tile. Five of the eight negative unit tiles pair off with the five positive unit tiles, leaving three negative unit tiles. Simplified: $x - 3$.
13. $5(x + 3) = 5x + 15$ items. The arrangement is five identical rows, each holding one $x$-tile and three unit tiles; counting the whole rectangle gives five $x$-tiles and fifteen unit tiles.
14. The student attached the subtraction sign to the wrong term. In $2 - 5x$, rewritten as $2 + (-5x)$, the $2$ is positive and the $5x$ is negative. The correct model is two positive unit tiles and five negative $x$-tiles.

Exit ticket 4.1

15. $4x - 2$
16. One negative $x$-tile and five positive unit tiles.
17. Two identical rows, each holding three $x$-tiles and two unit tiles. Expanded: $2(3x + 2) = 6x + 4$.
18. Each row of the rectangle contains a full copy of everything inside the parentheses, so a rectangle with $n$ rows contains $n$ copies of every term, not just of the first one. In $3(x + 2)$ there are three $x$-tiles and three copies of the two unit tiles, which is why the $2$ gets multiplied too.

Lesson 4.2 — The Properties of Real Numbers

Guided practice

19. Commutative property of addition.
20. Associative property of multiplication.
21. Distributive property.
22. Additive identity property.
23. Additive inverse property.
24. $0.5(8) + 3 = 4 + 3 = 7$

Independent practice

25. a) additive inverse property b) associative property of addition c) multiplicative inverse property d) multiplicative identity property
26. a) distributive property b) commutative property of addition c) associative property of addition
27. $\tfrac{1}{2}(10) + 4 \cdot 3 - 6 = 5 + 12 - 6 = 11$
28. $2.5(4) - 3 + 0.5 = 10 - 3 + 0.5 = 7.5$
29. No. $4 - 9 = -5$ and $9 - 4 = 5$, and $-5 \neq 5$, so the order matters for subtraction. Rewriting $4 - 9$ as $4 + (-9)$ turns it into a sum, and the commutative property does apply to addition: $4 + (-9) = (-9) + 4$. That is why the first move in every simplification is to rewrite subtraction as adding the opposite.
30. Read backwards, the distributive property says $ab + ac = a(b + c)$ — you may only collect terms that share a factor. In $6x + 4$ the first term has a factor of $x$ and the second does not, so there is no common factor of $x$ to pull out and no way to write the sum as a single $x$-term. A substitution confirms it: at $x = 2$, $6(2) + 4 = 16$ while $10(2) = 20$.
31. $2(x + 2.5) + 2(1.5) = 2x + 5 + 3 = 2x + 8$. The distributive property expanded $2(x + 2.5)$, and the associative property of addition regrouped $5 + 3$ into $8$. (The commutative property of addition is also acceptable as the regrouping step.)
32. The commutative property lets you reorder terms of a sum; it never lets you merge two unlike terms into one. So $3 + 4x = 4x + 3$ is correct and is as far as it goes. Writing $7x$ claims the $3$ has a factor of $x$, which it does not. Correct simplification: $4x + 3$.

Exit ticket 4.2

33. Multiplicative identity property.
34. Distributive property, since $8(x) + 8\left(-\tfrac{1}{2}\right) = 8x - 4$.
35. $6 + 2 \cdot 9 - \tfrac{1}{3}(12) = 6 + 18 - 4 = 20$
36. The commutative property of addition. Rewriting $-7 + 2x$ as $2x + (-7)$ reorders the two terms of a sum, and $2x + (-7)$ is written $2x - 7$.

Lesson 4.3 — Combining Like Terms with Rational Coefficients

Guided practice

37. $5x + 8x = (5 + 8)x = 13x$
38. $4.5x - 1.5x = (4.5 - 1.5)x = 3x$
39. $\tfrac{2}{5}x + \tfrac{1}{5}x = \left(\tfrac{2}{5} + \tfrac{1}{5}\right)x = \tfrac{3}{5}x$
40. $3x + (-7) + (-8x) + 2 = (3 - 8)x + (-7 + 2) = -5x - 5$
41. $\tfrac{1}{2}x + \tfrac{1}{4}x + 4 = \left(\tfrac{2}{4} + \tfrac{1}{4}\right)x + 4 = \tfrac{3}{4}x + 4$
42. $0.75x + (-0.25x) + (-2.5) + 1 = (0.75 - 0.25)x + (-2.5 + 1) = 0.5x - 1.5$
43. $\left(-\tfrac{4}{6} + \tfrac{1}{6}\right)x + 5 = -\tfrac{3}{6}x + 5 = -\tfrac{1}{2}x + 5$
44. $(-1.2 + 0.2)x + (6 - 9) = -x - 3$

Independent practice

45. a) $\left(\tfrac{3}{4} + \tfrac{1}{4}\right)x = x$ b) $(-2.4 + 0.4)x = -2x$ c) $\left(\tfrac{5}{6} - \tfrac{3}{6}\right)x = \tfrac{2}{6}x = \tfrac{1}{3}x$ d) $\left(-\tfrac{3}{8} - \tfrac{1}{8}\right)x = -\tfrac{4}{8}x = -\tfrac{1}{2}x$
46. $(7 - 3)x + \left(\tfrac{1}{2} - \tfrac{3}{2}\right) = 4x + \left(-\tfrac{2}{2}\right) = 4x - 1$
47. $(-0.5 + 2.5)x + (6 - 10) = 2x - 4$
48. $\left(\tfrac{1}{3} + \tfrac{1}{3}\right)x + (-4 + 4) = \tfrac{2}{3}x$
49. $\left(-\tfrac{3}{5} + \tfrac{3}{5}\right)x + (9 - 9) = 0x + 0 = 0$, by the additive inverse property applied twice.
50. $(2.5 - 2.5)x + (1.5 + 3.5) = 0x + 5 = 5$
51. Both terms of $\tfrac{1}{2}x + \tfrac{1}{4}x$ contain the factor $x$, so the distributive property read backwards pulls it out: $\tfrac{1}{2}x + \tfrac{1}{4}x = \left(\tfrac{1}{2} + \tfrac{1}{4}\right)x = \tfrac{3}{4}x$. In $\tfrac{1}{2}x + \tfrac{1}{4}$ the second term has no factor of $x$, so there is nothing common to pull out and the expression stays as it is.
52. $1.5x + (-0.5x) + 2 = (1.5 - 0.5)x + 2 = x + 2$. Check at $x = 4$: $1.5(4) + 2 - 0.5(4) = 6 + 2 - 2 = 6$ and $4 + 2 = 6$. Check at $x = -2$: $1.5(-2) + 2 - 0.5(-2) = -3 + 2 + 1 = 0$ and $-2 + 2 = 0$.
53. Perimeter $= \tfrac{1}{2}x + \left(\tfrac{1}{2}x + 3\right) + (2x - 1.5) = \left(\tfrac{1}{2} + \tfrac{1}{2} + 2\right)x + (3 - 1.5) = 3x + 1.5$ inches.
54. The student added the numerators and the denominators instead of finding a common denominator. Correctly, $\tfrac{1}{2} = \tfrac{3}{6}$ and $\tfrac{1}{3} = \tfrac{2}{6}$, so $\tfrac{1}{2}x + \tfrac{1}{3}x = \left(\tfrac{3}{6} + \tfrac{2}{6}\right)x = \tfrac{5}{6}x$. A quick sanity check: $\tfrac{2}{5}$ is smaller than $\tfrac{1}{2}$, so it cannot be the sum of $\tfrac{1}{2}$ and something positive.

Exit ticket 4.3

55. $\left(\tfrac{7}{8} - \tfrac{3}{8}\right)x = \tfrac{4}{8}x = \tfrac{1}{2}x$
56. $(-1.4 + 0.4)x + (3 - 5) = -x - 2$
57. $\left(-\tfrac{4}{6} + \tfrac{1}{6}\right)x + 4 = -\tfrac{1}{2}x + 4$
58. The two terms are not like terms. $0.6x$ has the variable part $x$ and $-0.6$ has no variable at all, so there is no common factor of $x$ to pull out and no way to add them into one term. Having the same coefficient is irrelevant; only the variable part decides.

Lesson 4.4 — Expanding with the Distributive Property

Guided practice

59. $4(x) + 4(2.5) = 4x + 10$
60. $\tfrac{1}{2}(4x) + \tfrac{1}{2}(6) = 2x + 3$
61. $(-3)(x) + (-3)(-1.5) = -3x + 4.5$
62. $\tfrac{2}{3}(9x) + \tfrac{2}{3}(-6) = 6x - 4$
63. $-1(x) + (-1)(-2.5) = -x + 2.5$
64. $0.2(10x) + 0.2(5) = 2x + 1$
65. $5x + 7 - 2x = (5 - 2)x + 7 = 3x + 7$
66. $6x - 3 + 3 = 6x$

Independent practice

67. a) $6(x) + 6\left(\tfrac{1}{3}\right) = 6x + 2$ b) $(-4)(2x) + (-4)(-0.25) = -8x + 1$ c) $\tfrac{1}{5}(10x) + \tfrac{1}{5}(-15) = 2x - 3$ d) $-0.5x - 7$
68. $3x - 4.5 + 2x + 8 = (3 + 2)x + (-4.5 + 8) = 5x + 3.5$
69. $3x + 4 - 3x + 1 = (3 - 3)x + (4 + 1) = 5$. The variable terms are additive inverses, so no $x$ survives and the expression has the value $5$ for every $x$.
70. $-2x - 5 - 4x + 4 = (-2 - 4)x + (-5 + 4) = -6x - 1$
71. $2x - 4 + 3x = (2 + 3)x - 4 = 5x - 4$
72. $2x - 3 - x + 5 = (2 - 1)x + (-3 + 5) = x + 2$
73. Draw a strip of height $\tfrac{1}{2}$ and width $4x + 6$; its area is $\tfrac{1}{2}(4x + 6)$. Cutting at the seam gives a left region of height $\tfrac{1}{2}$ and width $4x$, with area $2x$, and a right region of height $\tfrac{1}{2}$ and width $6$, with area $3$. The two regions together are the same strip, so $\tfrac{1}{2}(4x + 6) = 2x + 3$.
74. $6(x + 1.5) = 6x + 9$ dollars.
75. The student multiplied only the first term by $\tfrac{1}{3}$ and left the $9$ untouched. The factor must multiply every term inside: $\tfrac{1}{3}(6x) = 2x$ and $\tfrac{1}{3}(9) = 3$, so $\tfrac{1}{3}(6x + 9) = 2x + 3$.
76. The student kept the second product negative. A negative times a negative is positive, so $(-5)(-0.4) = +2$ and $-5(x - 0.4) = -5x + 2$. A check at $x = 0$ confirms it: $-5(-0.4) = 2$, not $-2$.

Exit ticket 4.4

77. $\tfrac{2}{5}(10x) + \tfrac{2}{5}(-5) = 4x - 2$
78. $3x + 1.5 - x + 1.5 = (3 - 1)x + (1.5 + 1.5) = 2x + 3$
79. $(-6)\left(\tfrac{1}{2}x\right) + (-6)\left(-\tfrac{1}{3}\right) = -3x + 2$
80. Multiplying by a factor makes that many copies of the whole quantity inside — and a fractional factor takes that fraction of the whole quantity, which means that fraction of each part. The area model shows it: the factor is one side length of the full rectangle, so it multiplies every piece of the other side. Taking half of $4x + 6$ but only half of the $4x$ would leave part of the strip unaccounted for.

Lesson 4.5 — Generating Equivalent Expressions

Guided practice

81. Any two expressions that simplify to $4x + 10$. For example $4x + 6 + 4$ (splitting the numeric term) and $2(2x + 5)$ (factoring out $2$). Also acceptable: $10 + 4x$, $\tfrac{1}{2}(8x + 20)$.
82. $6x + 15 = 3(2x) + 3(5) = 3(2x + 5)$
83. $0.5x + 1.5 = 0.5(x) + 0.5(3) = 0.5(x + 3)$
84. $6x - 2 - 4x - 2 = (6 - 4)x + (-2 - 2) = 2x - 4$
85. $2x + 3 - 3 = 2x$
86. Yes. $3(x - 2) + 5 = 3x - 6 + 5 = 3x - 1$. Check at $x = 2$: $3(2 - 2) + 5 = 0 + 5 = 5$ and $3(2) - 1 = 5$.

Independent practice

87. a) $4(2x + 3)$ b) $-3(2x - 3)$ c) $\tfrac{1}{2}(x + 3)$ d) $0.3(x - 4)$ For (b), factoring out the negative flips the sign of every term inside: $-6x = -3(2x)$ and $9 = (-3)(-3)$.
88. Any three expressions that simplify to $2x + 6$. For example: $6 + 2x$ (commutative property of addition), $2(x + 3)$ (distributive property read backwards), and $2x + 4 + 2$ (splitting the numeric term, then the associative property of addition puts it back together). Also acceptable: $\tfrac{1}{2}(4x + 12)$.
89. $5 - 2x + 7 = -2x + (5 + 7) = -2x + 12$. Note that the $-2$ multiplied the $-3.5$ to give $+7$.
90. $4x - 6 + 2x + 1 = (4 + 2)x + (-6 + 1) = 6x - 5$
91. $4[x + 2x - 2] = 4[3x - 2] = 12x - 8$
92. $-3x + 2 + 3x - 3 = (-3 + 3)x + (2 - 3) = -1$
93. Not equivalent. Distributing gives $2(x) + 2(4.5) = 2x + 9$, so the student's version left the $4.5$ unmultiplied. One substitution proves they differ: at $x = 0$, $2(0 + 4.5) = 9$ but $2(0) + 4.5 = 4.5$. The expression $2(x + 4.5)$ is equivalent to $2x + 9$.
94. Distributing gives $6(x + 2) = 6(x) + 6(2) = 6x + 12$, and the distributive property is a statement about all real numbers, so the equality holds for every replacement value at once. A single test could not have established that, because agreement at one value can happen by accident: $x + 3$ and $4x$ both equal $4$ at $x = 1$, yet at $x = 2$ they give $5$ and $8$. One agreement proves nothing; one disagreement disproves equivalence.
95. $4(x + 4.5) = 4x + 18$ dollars. Another equivalent form: $4x + 10 + 8$, or the factored form $2(2x + 9)$.
96. The student subtracted before multiplying. The order of operations requires the multiplication $3(x + 2)$ to be done first, because the parentheses are a grouping symbol and the factor $3$ multiplies them. Correctly: $10 - 3(x + 2) = 10 - 3x - 6 = -3x + 4$.

Exit ticket 4.5

97. $0.4x + 1.6 = 0.4(x) + 0.4(4) = 0.4(x + 4)$
98. $3[2x - x - 1] = 3[x - 1] = 3x - 3$
99. $\tfrac{1}{2}(x + 10)$, since $\tfrac{1}{2}(x) + \tfrac{1}{2}(10) = \tfrac{1}{2}x + 5$. Also acceptable: $\tfrac{1}{2}(x + 4) + 3$.
100. Every step of a simplification replaces a quantity with something equal to it — a regrouping (associative property), a reordering (commutative property), an added $0$ or a multiplied $1$ (identity properties), a canceled pair of opposites (inverse properties), or a distribution. None of those change the amount, and each holds for all real numbers, so the value stays the same whatever $x$ is. Only the way the expression is written changes.

Chapter 4 Review

Part A — Representing expressions with manipulatives and pictures (8.PFA.1a)

101. $5x - 3$
102. Four negative $x$-tiles and two positive unit tiles.
103. Four identical rows, each holding one $x$-tile and three unit tiles. Counting the whole rectangle gives four $x$-tiles and twelve unit tiles: $4(x + 3) = 4x + 12$.
104. One zero pair of $x$-tiles cancels, leaving three $x$-tiles. Two of the six negative unit tiles pair off with the two positive unit tiles, leaving four negative unit tiles. Simplified: $3x - 4$.
105. The three $x$-tiles pair off one-to-one with the three negative $x$-tiles, and the five unit tiles pair off with the five negative unit tiles. Every piece finds a partner, nothing is left on the table, and an empty model is worth $0$. The property is the additive inverse property: $3x + (-3x) = 0$ and $5 + (-5) = 0$.
106. Lay out six $x$-tiles and nine unit tiles, then split the pile into three equal groups, since the factor $\tfrac{1}{3}$ asks for one of three equal parts. Each group holds two $x$-tiles and three unit tiles, so $\tfrac{1}{3}(6x + 9) = 2x + 3$.
107. Factored: $2(3x + 1)$. Expanded: $6x + 2$. Both readings describe the same six $x$-tiles and two unit tiles.
108. An $x$-tile is a long rectangle and a unit tile is a small square, and we never say how many unit tiles fit inside an $x$-tile, because $x$ is unknown. Without knowing that, there is no single count that describes both piles, so the shortest honest description keeps them separate — which in symbols is why $3x + 6$ cannot become $9x$.

Part B — Simplifying and generating equivalent expressions (8.PFA.1b)

109. a) $\left(\tfrac{5}{8} + \tfrac{1}{8}\right)x = \tfrac{6}{8}x = \tfrac{3}{4}x$ b) $(-3.6 + 1.6)x = -2x$ c) $\left(\tfrac{7}{10} - \tfrac{2}{10}\right)x = \tfrac{5}{10}x = \tfrac{1}{2}x$ d) $(0.9 - 0.9)x = 0$
110. $(6 - 4)x + (-2.5 + 4) = 2x + 1.5$
111. $\left(\tfrac{2}{6} - \tfrac{5}{6}\right)x + (7 - 2) = -\tfrac{3}{6}x + 5 = -\tfrac{1}{2}x + 5$
112. $5x - 6 + 2x = (5 + 2)x - 6 = 7x - 6$
113. $-\tfrac{3}{4}(8x) + \left(-\tfrac{3}{4}\right)(-12) = -6x + 9$
114. $2x + 4 - x + 3 = (2 - 1)x + (4 + 3) = x + 7$
115. $2[3x - x + 4] = 2[2x + 4] = 4x + 8$
116. $6x + 4 - 3x + 4 = (6 - 3)x + (4 + 4) = 3x + 8$
117. a) $5(2x + 5)$ b) $-4(2x - 1)$ c) $1.5(x + 3)$
118. Any two expressions that simplify to $5x - 20$, at least one factored. For example $5(x - 4)$ and $5x - 12 - 8$. The first is equivalent because distributing gives $5(x) + 5(-4) = 5x - 20$ back; the second is equivalent because $-12 + (-8) = -20$, so combining the numeric terms returns the original.
119. a) distributive property b) commutative property of addition c) additive identity property
120. $7 - 6x + 3 = -6x + (7 + 3) = -6x + 10$. The $-2$ multiplied the $-1.5$ to give $+3$.

Part C — Mixed application and reasoning

121. $4(x - 0.75) + 3 = 4x - 3 + 3 = 4x$, so the two expressions are equivalent. Check at $x = 2$: $4(2 - 0.75) + 3 = 4(1.25) + 3 = 5 + 3 = 8$ and $4(2) = 8$. Check at $x = -1$: $4(-1 - 0.75) + 3 = 4(-1.75) + 3 = -7 + 3 = -4$ and $4(-1) = -4$.
122. Total weight $= 3(x + 2.5) = 3x + 7.5$ kilograms.
123. The student kept the second product negative. Multiplying $-\tfrac{1}{2}$ by $-4$ gives $+2$, so $-\tfrac{1}{2}(6x - 4) = -3x + 2$. Checking at $x = 0$ settles it: $-\tfrac{1}{2}(-4) = 2$, not $-2$.
124. Equivalence means the two expressions give the same value for every replacement value, and one test checks only one of infinitely many. For example, $x + 3$ and $4x$ both equal $4$ at $x = 1$, which looks like a match, but at $x = 2$ they give $5$ and $8$. One agreement proves nothing; one disagreement disproves equivalence. Only the properties of real numbers — distributing, combining like terms, reordering and regrouping — prove that two expressions are equivalent.
125. Build three identical rows, each holding one $x$-tile and two unit tiles. Reading by rows gives three copies of $(x + 2)$, which is $3(x + 2)$. Reading the whole rectangle as one collection gives three $x$-tiles and six unit tiles, which is $3x + 6$. The same tiles describe both expressions, so they are equivalent — and the six unit tiles show that the factor $3$ multiplied the $2$ as well as the $x$.
126. Any expression with a fractional coefficient that simplifies to $2x - 1$. For example $\tfrac{1}{2}(4x - 6) + 2$: distributing gives $2x - 3 + 2$ (distributive property), and combining the numeric terms gives $2x - 1$ (associative property of addition, since $-3 + 2 = -1$). Another acceptable answer is $\tfrac{3}{2}x + \tfrac{1}{2}x - 1$, where the two $x$-terms combine because they share the factor $x$.

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. Two expressions are equivalent when they give the same value for every replacement value.

Page 4, read the rectangle two ways. By rows: 3 rows of $(x + 2)$, which is $3(x + 2)$. By counting the whole rectangle: 3 $x$-tiles and 6 unit tiles, which is $3x + 6$.

Page 7, properties table.

Page 7, careful. $4 - 9 = \mathbf{-5}$ but $9 - 4 = \mathbf{5}$.

Page 11, why it works. $5x + 8x = (\mathbf{5} + \mathbf{8})x = \mathbf{13}x$. Pulling the common factor $x$ out front is the distributive property read backwards. And $\tfrac{1}{2}x + \tfrac{1}{4}x = \left(\tfrac{2}{4} + \tfrac{1}{4}\right)x = \mathbf{\tfrac{3}{4}}x$.

Page 11, item 40 frame. As a sum: $3x + (-7) + (-8x) + 2$. $x$-terms: $3x$ and $-8x$, giving $-5x$. Numeric terms: $-7$ and $2$, giving $-5$. Simplified: $-5x - 5$.

Page 11, item 42 frame. As a sum: $0.75x + (-2.5) + (-0.25x) + 1$. Simplified: $0.5x - 1.5$.

Page 15, watch the signs. $-3(x - 1.5) = (-3)(x) + (-3)(-1.5) = -3x + \mathbf{4.5}$.

Page 15, item 65 frame. Distribute: $5x + 7 - 2x$. Combine: $3x + 7$.

Page 15, item 66 frame. Distribute: $6x - 3 + 3$. Combine: $6x$.

Page 17, item 73 frame. The left region represents half of $4x$, an area of $2x$. The right region represents half of $6$, an area of $3$. Together they are the whole strip, whose area is $\tfrac{1}{2}(4x + 6)$, so $\tfrac{1}{2}(4x + 6) = 2x + 3$.

Page 19, many forms table. $10 + 4x$ — commutative property of addition. $4x + 6 + 4$ — the numeric term was split, and the associative property of addition puts $6 + 4 = 10$ back together. $2(2x + 5)$ — distributive property, read backwards.

Page 19, factoring frame. $6x + 15 = 3(\mathbf{2x}) + 3(\mathbf{5}) = 3(\mathbf{2x + 5})$.

Page 19, nested grouping frame. $4[x + 2(x - 1)] = 4[\mathbf{3x - 2}] = \mathbf{12x - 8}$.

Page 19, item 86 frame. Simplified: $3x - 1$. At $x = 2$ both give 5. Equivalent: yes.

Page 13, item 52 table. At $x = 4$: both columns give $6$. At $x = -2$: both columns give $0$.

Page 26, item 121 table. At $x = 2$: both columns give $8$. At $x = -1$: both columns give $-4$.

Page 26, item 125 frame. Reading by rows: $3(x + 2)$. Reading the whole rectangle: $3x + 6$.