Appendix A — Answer Key, Chapter 1: Translating and Evaluating Algebraic Expressions

SOL A.EO.1 · Covers textbook Chapter 1 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 122 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used throughout. Translations use $n$ for "a number" unless the item names a letter; any letter is acceptable if the student defines it. A translation is correct if it is equivalent as written — $8 + 3n$ and $3n + 8$ are both right for a phrase naming the sum, but $n - 8$ and $8 - n$ are never interchangeable. No answer in this chapter rationalizes a denominator, because A.EO.1b excludes it; $\dfrac{6}{\sqrt{2}}$ is a complete answer here and is not rewritten as $3\sqrt{2}$.

Lesson 1.1 — The Language of Algebra

Guided practice

1. $n + 15$
2. $n - 9$
3. $6n$
4. $\dfrac{n}{4}$
5. $2n - 7$
6. $3n + 5$
7. "ten more than a number," or "the sum of a number and $10$."
8. "one less than four times a number," or "the product of $4$ and a number, decreased by $1$."

Independent practice

9. a) $n - 12$ b) $n - 12$ c) $12 - n$ d) $12 - n$ e) $n - 12$. Only two expressions appear: parts a, b, and e all give $n - 12$, and parts c and d both give $12 - n$. "Less than" and "subtracted from" reverse the order of the quantities they name; "the difference of $12$ and a number" does not.
10. a) $\dfrac{n}{3}$ b) $\dfrac{20}{n}$ c) $\dfrac{n}{20}$ d) $n + n^2$ e) $(n + 3)^2$
11. a) $3(n + 8)$ b) $3n + 8$ c) $8 + 3n$ (equivalently $3n + 8$) d) $3n - 8$
12. a) "seven times a number" b) "eleven less than a number" c) "the sum of a number and $5$, divided by $2$" d) "nine more than twice a number" e) "six times the difference of a number and $4$"
13. a) three terms b) $5$ c) $-8$ (the sign belongs to the term) d) $3$
14. Completed table.

15. $10 - 4n$. The phrase starts at $10$ and takes away a product, so $10$ must come first: that eliminates $4n - 10$. The word decreased is subtraction, not multiplication of a difference, so $4(10 - n)$ is out. And nothing in the phrase divides, so $\dfrac{10}{4n}$ is out.
16. $n - 12$. The garage starts with $n$ cars and loses $12$, so the count you start from is $n$. The expression $12 - n$ would say you start with $12$ cars and remove $n$ of them, which describes a different situation and goes negative as soon as $n$ exceeds $12$.
17. "Less than" names the amount removed before it names what it is removed from, so the two quantities trade places when you write the expression. Replacing "a number" with $20$: $n - 7 = 20 - 7 = 13$, which is what "$7$ less than $20$" means, while $7 - n = 7 - 20 = -13$, which is not. The two expressions are opposites, so only one can be right.
18. The student translated in reading order and put the $5$ first. "Less than" reverses that order, so the product $3n$ is what you start from. The correct expression is $3n - 5$. The one-line check: replace the number with something concrete, say $10$. The phrase says "$5$ less than $30$," which is $25$; $3n - 5 = 25$ and $5 - 3n = -25$.

Exit ticket 1.1

19. $2n - 10$
20. $\dfrac{n}{6} + 5$
21. "three times the difference of a number and $2$," or "three times the quantity a number minus two."
22. $4n - 9$ is nine less than four times the number: start at $4n$ and go down $9$. $9 - 4n$ starts at $9$ and takes away four times the number. They are opposites of each other, so unless the number happens to make both zero, they never have the same value. For $n = 5$ they are $11$ and $-11$.

Lesson 1.2 — Expressions from Contextual Situations

Any letter is acceptable for the variable as long as the student writes a sentence saying what number it stands for.

Guided practice

23. Let $t$ = the number of tickets. Cost: $12t$ dollars.
24. Let $p$ = the number of toppings. Cost: $14 + 2p$ dollars.
25. Let $x$ = the number of dollars Sara spends. Amount left: $50 - x$ dollars.
26. $\dfrac{s}{5}$ students in each group.
27. $w + 4$ centimeters.
28. $2w + 2(w + 4) = 2w + 2w + 8 = 4w + 8$ centimeters.
29. Let $m$ = the number of months. Total cost: $30 + 22m$ dollars.

Independent practice

30. a) $3.25n$ dollars b) $3.25n + 5$ dollars. The bag is bought once, so it is a constant.
31. a) $55h$ miles b) $55h + 20$ miles
32. a) $25q + 10d$ cents b) $0.25q + 0.10d$ dollars. Each coin type keeps its own rate; the two products are added.
33. $80 - 6m$ gallons. Draining removes gallons, so the accumulated amount $6m$ is subtracted from the starting $80$.
34. The $40$ is the fixed monthly charge in dollars, paid no matter how much data is used. The $0.25$ is the rate in dollars per gigabyte over the limit. The $g$ is the number of gigabytes used beyond the limit.
35. a) $8n$ b) $n - 8$ c) $\dfrac{n}{8}$ d) $8 + n$ e) $8 - n$
36. a) Let $w$ = the number of weeks; $7w$ days. b) Let $f$ = the number of feet; $12f$ inches. c) $a - 5$ years.
37. Let $r$ = the number of rows. Total seats: $18r + 12$. With the balcony closed the $12$ disappears and the expression becomes $18r$. The rows term is unaffected, because the balcony seats were never part of it.
38. Addition is commutative, so $5 + 3n$ and $3n + 5$ add the same two quantities and always give the same value; the order they are written in does not change the total. Subtraction is not commutative. $5 - 3n$ starts at $5$ and removes $3n$, while $3n - 5$ starts at $3n$ and removes $5$. The two results are opposites — for $n = 4$ they are $-7$ and $7$ — so they cannot model the same situation.
39. The student wrote the numbers in the order they were read. "Less than" reverses that order: the shirt's price starts at the jacket's price and goes down $\$20$, so it is $c - 20$ dollars. The expression $20 - c$ would describe something else, such as the change you get back after paying for a $c$-dollar item with $\$20$.

Exit ticket 1.2

40. Let $m$ = the number of miles driven. Fare: $3.50 + 2m$ dollars.
41. $\dfrac{s}{6}$ marbles each.
42. One acceptable answer: a booth charges $\$15$ to set up plus $\$8$ for each item made, so $n$ = the number of items, $8$ = the cost in dollars per item, and $15$ = the one-time setup charge. Any situation with a per-unit rate of $8$ and a one-time amount of $15$ is correct.
43. The variable goes to the quantity that changes and that the other quantities are described in terms of. Pick the one whose value everything else depends on — the count of hours, items, or miles — and then express each remaining quantity using it. Choosing the other quantity is not wrong, but it usually forces a more awkward description.

Lesson 1.3 — Evaluating Expressions

Guided practice

44. $3(4) + 5 = 12 + 5 = 17$
45. $3(-4) + 5 = -12 + 5 = -7$
46. $7 - 2(-3) = 7 + 6 = 13$
47. $(-6)^2 = 36$
48. $-(-6)^2 = -(36) = -36$. The minus sign in front is applied after the squaring.
49. $|{-9}| = 9$
50. $-|{-9}| = -9$
51. $|{-9} - 4| = |{-13}| = 13$
52. $4\left(\tfrac{1}{2}\right) - 1 = 2 - 1 = 1$

Independent practice

53. a) $-2 + 5 = 3$ b) $-2 - 5 = -7$ c) $(-2)(5) = -10$ d) $5 - (-2) = 7$ e) $3(-2) + 2(5) = -6 + 10 = 4$
54. a) $(-3)^2 + 4^2 = 9 + 16 = 25$ b) $(-3 + 4)^2 = 1^2 = 1$ c) $2(-3)^2 - 4 = 18 - 4 = 14$ d) $-2(-3)(4) = 24$

Parts a and b are worth comparing: squaring first and then adding is not the same as adding first and then squaring.

55. a) $6\left(\tfrac{2}{3}\right) = 4$ b) $9\left(\tfrac{2}{3}\right) - 1 = 6 - 1 = 5$ c) $\tfrac{2}{3} + \tfrac{1}{6} = \tfrac{4}{6} + \tfrac{1}{6} = \tfrac{5}{6}$ d) $\dfrac{1}{2/3} = \tfrac{3}{2}$
56. a) $4(-1.5) = -6$ b) $(-1.5)^2 = 2.25$ c) $10 - 8(-1.5) = 10 + 12 = 22$ d) $|{-1.5}| = 1.5$
57. a) $12$ b) $-12$ c) $|7 - 15| = |{-8}| = 8$ d) $7 - 15 = -8$ e) $3 \cdot 4 = 12$

Parts c and d differ only in where the bars fall, and they come out opposite. In c the subtraction happens inside the bars, so the absolute value is taken last; in d each number's absolute value is taken first and the subtraction happens afterward, so the result can be negative.

58. a) $|2(5) - 7| = |3| = 3$ b) $|2(1) - 7| = |{-5}| = 5$ c) $|2(3.5) - 7| = |0| = 0$ d) $|2(-2) - 7| = |{-11}| = 11$
59. a) $\dfrac{5 + 3}{5 - 3} = \dfrac{8}{2} = 4$ b) $\dfrac{\tfrac{1}{2} + \tfrac{1}{4}}{\tfrac{1}{2} - \tfrac{1}{4}} = \dfrac{\tfrac{3}{4}}{\tfrac{1}{4}} = 3$
60. $5 - 2(1 - 3)^2 = 5 - 2(-2)^2 = 5 - 2(4) = 5 - 8 = -3$
61. $|11 - (-4)| = |15| = 15$ degrees Fahrenheit. Absolute value is right because the question asks how far apart the readings are, and a distance has no direction. Without the bars the answer would depend on which reading you subtracted from which; with them, both orders give $15$.
62. $25 + 15(3.5) = 25 + 52.5 = 77.5$, so the charge is $\$77.50$.
63. $(-6)^2 = 36$ and $-6^2 = -36$. In $(-6)^2$ the parentheses make $-6$ the base, so the negative is squared away. In $-6^2$ the base is just $6$; the expression means "the opposite of $6$ squared," so you square first and take the opposite second. The exponent applies only to what it is written on.
64. The error is in the substitution: the student multiplied $2$ by $3$ instead of by $-3$, dropping the negative sign. Correctly, $5 - 2(-3) = 5 - (-6) = 5 + 6 = 11$. The rule missed is that the replacement value goes in with its sign, inside parentheses, and subtracting a negative adds.

Exit ticket 1.3

65. a) $2(-5) - 9 = -10 - 9 = -19$ b) $(-5)^2 + (-5) = 25 - 5 = 20$
66. $6\left(\tfrac{1}{3}\right) + 1 = 2 + 1 = 3$
67. $|2.5 - 10| = |{-7.5}| = 7.5$ and $-|2.5 - 10| = -7.5$
68. Parentheses keep the sign attached to the value and keep the operation that was there before the substitution. Evaluating $x^2$ at $x = -4$ written properly is $(-4)^2 = 16$; written without parentheses it reads $-4^2 = -16$, which is a different number. The same protection matters for products: $3x$ at $x = -4$ is $3(-4) = -12$, while "$3-4$" is $-1$.

Lesson 1.4 — Absolute Value, Square Roots, and Cube Roots in Evaluation

Guided practice

69. $7$, since $7^2 = 49$.
70. $\tfrac{1}{2}$, since $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$.
71. $3$, since $3^3 = 27$.
72. $-2$, since $(-2)^3 = -8$.
73. $11$
74. $-5$, since $(-5)^3 = -125$.
75. $3\sqrt{16} - 2 = 3(4) - 2 = 12 - 2 = 10$
76. $\sqrt[3]{-27} + |{-27}| = -3 + 27 = 24$
77. $\dfrac{10}{\sqrt{5}}$. Since $5$ is not a perfect square, the division does not come out rational, and A.EO.1b does not ask for the denominator to be rationalized. This is the finished answer.

Independent practice

78. a) $8$ b) $12$ c) $\tfrac{3}{4}$, since $\left(\tfrac{3}{4}\right)^2 = \tfrac{9}{16}$ d) $0$ e) $1.2$, since $1.2^2 = 1.44$
79. a) $1$ b) $4$ c) $-3$ d) $-1$ e) $\tfrac{2}{3}$, since $\left(\tfrac{2}{3}\right)^3 = \tfrac{8}{27}$
80. a) $\sqrt{7 + 9} = \sqrt{16} = 4$ b) $\sqrt{-11 + 47} = \sqrt{36} = 6$ c) $\sqrt{\tfrac{1}{2} + \tfrac{1}{2}} = \sqrt{1} = 1$
81. a) $\sqrt[3]{9 - 1} = \sqrt[3]{8} = 2$ b) $\sqrt[3]{-7 - 1} = \sqrt[3]{-8} = -2$ c) $\sqrt[3]{1 - 1} = \sqrt[3]{0} = 0$
82. a) $\sqrt{7^2 - 4(2)(3)} = \sqrt{49 - 24} = \sqrt{25} = 5$ b) $\sqrt{6^2 - 4(1)(5)} = \sqrt{36 - 20} = \sqrt{16} = 4$
83. a) $|9| + \sqrt{9} = 9 + 3 = 12$ b) $\left|\tfrac{1}{4}\right| + \sqrt{\tfrac{1}{4}} = \tfrac{1}{4} + \tfrac{1}{2} = \tfrac{3}{4}$
84. a) $\dfrac{12}{\sqrt{3}}$ b) $\dfrac{12}{\sqrt{4}} = \dfrac{12}{2} = 6$. Part b finishes as a rational number because $4$ is a perfect square, so the radical becomes the whole number $2$ and the division goes through. Part a does not, because $3$ is not a perfect square; the answer keeps the radical in the denominator, which this chapter allows.
85. $2\sqrt{25} + \sqrt[3]{-8} = 2(5) + (-2) = 10 - 2 = 8$
86. Patio side: $\sqrt{169} = 13$ feet. Bin edge: $\sqrt[3]{216} = 6$ inches. The square root undoes squaring, which is how area was built from a side; the cube root undoes cubing, which is how volume was built from an edge.
87. $\sqrt{\dfrac{144}{16}} = \sqrt{9} = 3$ seconds. Simplify under the radical before taking the root.
88. Cubing preserves sign: a negative number cubed is negative, since $(-4)(-4)(-4) = -64$. So $-4$ is a real number whose cube is $-64$, and $\sqrt[3]{-64} = -4$. Squaring does not preserve sign — a positive squared is positive and a negative squared is also positive — so no real number squares to $-64$, and $\sqrt{-64}$ has no real value.
89. Correctly, $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$, not $7$. The shortcut assumes a square root distributes over addition, and it does not: the radical is a grouping symbol, so everything under it is combined before the root is taken. The student's rule would also fail the definition — if $\sqrt{x^2 + y^2}$ really were $x + y$, then squaring both sides would need $x^2 + y^2$ to equal $x^2 + 2xy + y^2$, which requires $2xy = 0$.

Exit ticket 1.4

90. a) $\sqrt{100} = 10$ b) $\sqrt[3]{-216} = -6$, since $(-6)^3 = -216$.
91. $5\sqrt[3]{8} - \sqrt{49} = 5(2) - 7 = 10 - 7 = 3$
92. $\dfrac{9}{\sqrt{6}}$. You are finished because $6$ is not a perfect square, so the denominator will not become rational by evaluating, and A.EO.1b explicitly excludes rationalizing the denominator.
93. Squaring any real number gives a result that is zero or positive, so nothing real squares to $-25$ and $\sqrt{-25}$ names no real number. Cubing preserves the sign of the number, so a negative number has a negative cube, and $\sqrt[3]{-25}$ is the real number whose cube is $-25$. That value is not a whole number — it is about $-2.92$ — but it is real.

Chapter 1 Review

Part A — Translating between verbal situations and algebraic expressions (A.EO.1a)

94. a) $n + 12$ b) $n - 8$ c) $\dfrac{n}{5}$ d) $3(n - 7)$ e) $2n + 9$
95. a) "two less than five times a number" b) "the sum of a number and $4$, divided by $3$" c) "four times the difference of a number and $6$"
96. "The difference of $6$ and a number" is $6 - n$: it names the two quantities in subtraction order, so it is written in the order you hear it. "$6$ less than a number" is $n - 6$: "less than" names the amount removed first, so the two quantities trade places. The expressions are opposites of each other; at $n = 10$ they are $-4$ and $4$.
97. Let $t$ = the number of text messages sent in the month. Bill: $30 + 0.10t$ dollars.
98. Let $w$ = the width. a) length $= 2w + 3$ b) perimeter $= 2w + 2(2w + 3) = 2w + 4w + 6 = 6w + 6$
99. $d - 18.75$ dollars.
100. $5n + 10d$ cents. A nickel is $5$ cents and a dime is $10$ cents, so each count is multiplied by its own value and the two products are added.
101. Let $r$ = the number of rows. Total seats: $24r + 8$.
102. a) three terms b) $-3$ c) $10$
103. Completed table.

104. "Less than" tells you how far down to go from a starting quantity, and the starting quantity is the one named after the phrase — here, $x$. So the expression is $x - 5$. Testing with $x = 12$: $x - 5 = 7$, which is what "$5$ less than $12$" means, while $5 - x = -7$, which is not. The two are opposites, so only one can match the phrase.
105. "The quotient of $8$ and a number" names $8$ as the dividend, so $8$ belongs on top: the correct expression is $\dfrac{8}{n}$. The student's $\dfrac{n}{8}$ translates "the quotient of a number and $8$," which is a different expression. Division, like subtraction, is not commutative, so the order matters.

Part B — Evaluating expressions, including absolute value and roots (A.EO.1b)

106. a) $3(-4) + 2 = -10$ b) $(-4)^2 - 2 = 16 - 2 = 14$ c) $(-4)(2)^2 = (-4)(4) = -16$ d) $\dfrac{-4 + 2}{2} = \dfrac{-2}{2} = -1$ e) $2(-4 - 3(2)) = 2(-4 - 6) = 2(-10) = -20$
107. a) $4\left(\tfrac{1}{2}\right) = 2$ b) $8\left(-\tfrac{3}{4}\right) = -6$ c) $\tfrac{1}{2} - \tfrac{3}{4} = -\tfrac{1}{4}$ d) $\tfrac{1}{2} - \left(-\tfrac{3}{4}\right) = \tfrac{5}{4}$ e) $\left(\tfrac{1}{2}\right)\left(-\tfrac{3}{4}\right) = -\tfrac{3}{8}$
108. a) $-10$ b) $(-2.5)^2 = 6.25$ c) $2.5$ d) $-2.5$ e) $6 - 2(-2.5) = 6 + 5 = 11$
109. a) $15$ b) $-15$ c) $|{-11}| = 11$ d) $15 + 4 = 19$ e) $-|{-11}| = -11$

Parts c, d, and e are the three arrangements students confuse. In c the addition happens inside the bars; in d each number is made nonnegative first; in e the bars do their work and the minus sign outside then negates the result.

110. a) $|3(4) - 5| = |7| = 7$ b) $|3(0) - 5| = |{-5}| = 5$ c) $\left|3\left(\tfrac{5}{3}\right) - 5\right| = |0| = 0$ d) $|3(-1) - 5| = |{-8}| = 8$
111. a) $14$ b) $\tfrac{5}{7}$, since $\left(\tfrac{5}{7}\right)^2 = \tfrac{25}{49}$ c) $1.5$, since $1.5^2 = 2.25$ d) $0$
112. a) $5$ b) $-5$ c) $\tfrac{1}{2}$ d) $-10$, since $(-10)^3 = -1000$.
113. a) $\sqrt{52 + 12} = \sqrt{64} = 8$ b) $\sqrt{-9 + 13} = \sqrt{4} = 2$ c) $\sqrt{0.75 + 0.25} = \sqrt{1} = 1$
114. $2\sqrt[3]{-27} + \sqrt{36} = 2(-3) + 6 = -6 + 6 = 0$
115. a) $\dfrac{14}{\sqrt{7}}$ — finished, since $7$ is not a perfect square and the denominator is not rationalized in this chapter. b) $\dfrac{14}{\sqrt{49}} = \dfrac{14}{7} = 2$
116. $\sqrt{11^2 - 4(3)(6)} = \sqrt{121 - 72} = \sqrt{49} = 7$
117. $25 + 15(2.5) = 25 + 37.5 = 62.5$, so the charge is $\$62.50$.
118. $\sqrt{6.25} = 2.5$ meters, since $2.5^2 = 6.25$.
119. $\sqrt[3]{343} = 7$ centimeters, since $7^3 = 343$.
120. $|9 - (-13)| = |22| = 22$ degrees Fahrenheit. Naming the readings in the other order gives $|-13 - 9| = |{-22}| = 22$, the same value, because the two differences are opposites and absolute value reports distance without direction.
121. $x^2 = (-5)^2 = 25$ and $-x^2 = -(-5)^2 = -25$. In $x^2$ the exponent is attached to $x$, and the replacement value $-5$ is substituted with its sign, so the negative is squared away. In $-x^2$ the exponent is still attached only to $x$; the minus sign sits outside and is applied after the squaring, so the result is the opposite of $25$. The minus sign in $-x^2$ is not part of the base.
122. This chapter's convention, taken straight from A.EO.1b, is that expressions are evaluated without rationalizing the denominator, so $\dfrac{20}{\sqrt{5}}$ is the finished answer and the student should stop there. Rationalizing is a legitimate technique and does give an equal value, $4\sqrt{5}$, but it is not part of A.EO.1. It is taught in Chapter 11, Radical Expressions, along with the rest of simplest radical form.

Workbook-only items

Page 2, fill in the blanks. An expression built from numbers, variables, and operations is an algebraic expression. Its pieces separated by $+$ and $-$ signs are its terms. The number multiplying a variable is the coefficient. A term with no variable is a constant. In $5x^2 - 8x + 3$ there are three terms, the coefficients are $5$ and $-8$, and the constant is $3$. The two order-reversing phrases are "less than" and "subtracted from." The frame completes as $\text{five less than twice a number} \longrightarrow \mathbf{2n - 5}$.

Page 7, fill in the blanks. Step one is always to define the variable in a full sentence. A quantity charged once is a constant in the expression. A quantity charged once per unit is multiplied by the variable. The frame completes as $\text{a } \$25 \text{ fee plus } \$15 \text{ per hour} \longrightarrow \mathbf{25 + 15h}$, where $h$ is the number of hours worked.

Page 12, fill in the blanks. Every replacement value is substituted inside parentheses. After substituting, follow the order of operations. $-6^2 = \mathbf{-36}$ but $(-6)^2 = \mathbf{36}$, because the exponent is attached to only the $6$ in the first expression and to the whole quantity $-6$ in the second. Absolute value reports a distance from zero, so it is never negative. $|-6| = \mathbf{6}$ and $-|-6| = \mathbf{-6}$.

Page 17, fill in the blanks. $\sqrt{a}$ asks for the nonnegative number whose square is $a$. $\sqrt[3]{a}$ asks for the number whose cube is $a$. $\sqrt[3]{-27} = \mathbf{-3}$, because $(\mathbf{-3})^3 = -27$. $\sqrt{-64}$ is not a real number, because no real number squares to a negative. $-\sqrt{16} = \mathbf{-4}$, and the minus sign is applied after the root is taken. A radical is a grouping symbol, so $\sqrt{9 + 16} = \mathbf{5}$ while $\sqrt{9} + \sqrt{16} = \mathbf{7}$.

Page 14, item 57 follow-up. The two parts showing that placement of the bars changes the answer are c and d: $|7 - 15| = 8$ but $|7| - |15| = -8$.

Page 8, item 33 follow-up. The rate is subtracted rather than added because the tank is losing water; each minute removes $6$ gallons from the starting $80$, so the accumulated loss $6m$ is taken away.

Page 19, item 84 follow-up. Part b finishes as a rational number, because $4$ is a perfect square and $\sqrt{4} = 2$ divides $12$ evenly. Part a keeps its radical denominator.

Page 20, item 88 frames. $(\mathbf{-4})^3 = -64$, so the cube root is $\mathbf{-4}$. Squaring a positive gives a positive and squaring a negative also gives a positive, so no real number squares to $-64$.

Page 22, item 96 frames. "the difference of $6$ and a number" is $\mathbf{6 - n}$; "$6$ less than a number" is $\mathbf{n - 6}$. See item 96 above for the explanation.

Page 27, item 117 frames. $25 + 15(2.5) = \mathbf{62.5}$, written as money as $\$\mathbf{62.50}$.

Page 28, item 122 frame. The technique the student is thinking of is taught in Chapter 11.