Appendix A — Answer Key, Chapter 2: Multistep and Literal Linear Equations

SOL A.EI.1 (a, b, d, e, f) · Covers textbook Chapter 2 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 124 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Every numeric solution below was substituted back into the original equation and confirmed, and every literal rearrangement was checked symbolically and again with numbers. Properties are named the way the standard asks: properties of equality change both sides, properties of real numbers rewrite one side.

Lesson 2.1 — Writing an Equation for a Situation

Workbook fill-ins. A known total gives $\text{rate} \cdot x + \text{fixed} = \text{total}$; two options compared give $\text{option A} = \text{option B}$. "Seven less than four times a number" is $4n - 7$. The rideshare row is $1.75m + 2.50 = 16.50$; the gym row is $12m + 75 = 27m$.

Guided practice

1. Let $t$ be the number of tickets. $9.50t + 4.25 = 61.25$, so $9.50t = 57.00$ and $t = 6$. Check: $9.50(6) + 4.25 = 57.00 + 4.25 = 61.25$. True.
2. Let $n$ be the number. $3n + 5 = 26$, so $3n = 21$ and $n = 7$. Check: $3(7) + 5 = 26$. True.
3. Let $m$ be the number of miles. $2.25m + 3.00 = 23.25$, so $2.25m = 20.25$ and $m = 9$. Check: $2.25(9) + 3 = 20.25 + 3 = 23.25$. True.
4. Let $w$ be the width in centimeters, so the length is $2w - 3$. $2(2w - 3) + 2w = 54$, which expands to $4w - 6 + 2w = 54$ and simplifies to $6w - 6 = 54$, so $6w = 60$ and $w = 10$. The width is 10 cm and the length is 17 cm. Check: $2(17) + 2(10) = 34 + 20 = 54$ cm. True.

Independent practice

5. a) Let $w$ be the number of weeks. $18w + 120 = 12w + 210$, so $6w = 90$ and $w = 15$. Check: $18(15) + 120 = 390$ and $12(15) + 210 = 390$. True. They each have $\$390$ after 15 weeks. b) Let $y$ be the number of years. $42 + 3y = 60 + 1.5y$, so $1.5y = 18$ and $y = 12$. Check: $42 + 3(12) = 78$ and $60 + 1.5(12) = 78$. True. Both are 78 inches tall after 12 years.
6. a) Let $g$ be the number of guests. $22g + 150 = 1030$, so $22g = 880$ and $g = 40$. Check: $22(40) + 150 = 880 + 150 = 1030$. True. b) Let $c$ be the mass of one crate in kilograms. $3c + 14 = 71$, so $3c = 57$ and $c = 19$. Check: $3(19) + 14 = 57 + 14 = 71$. True.
7. Let $h$ be the number of hours. $68 - 1.5h = 50$, so $-1.5h = -18$ and $h = 12$. Check: $68 - 1.5(12) = 68 - 18 = 50$. True.
8. Let $n$ be the number. $2(n + 6) = 5n - 3$, which expands to $2n + 12 = 5n - 3$, so $15 = 3n$ and $n = 5$. Check: $2(5 + 6) = 22$ and $5(5) - 3 = 22$. True.
9. Let $t$ be the number of minutes. $4500 - 150t = 1800$, so $-150t = -2700$ and $t = 18$. Check: $4500 - 150(18) = 4500 - 2700 = 1800$. True.
10. Let $s$ be the number of shirts. $8s + 60 = 12s$, so $60 = 4s$ and $s = 15$. Check: $8(15) + 60 = 180$ and $12(15) = 180$. True. The bills are equal at 15 shirts, $\$180$ each.
11. "Eight less than a number" means the number with 8 taken away from it, which is $n - 8$. The phrase less than names the amount being subtracted first and the thing it is subtracted from second, so the words arrive in the opposite order from the symbols. The test: substitute a number and retell the story. With $n = 20$, "eight less than 20" is plainly 12, and $n - 8 = 12$ while $8 - n = -12$.
12. Store B charges nothing for shipping, so no $\$6$ belongs on its side. Writing $+6$ on both sides describes a situation in which both stores charge shipping, and it makes the shipping cancel out entirely, which is why the student's equation would give the same answer no matter what the shipping cost. The correct equation is $14b + 6 = 17b$, so $6 = 3b$ and $b = 2$. Check: $14(2) + 6 = 34$ and $17(2) = 34$. True.

Exit ticket 2.1

13. Let $m$ be the number of months. $25m + 40 = 290$, so $25m = 250$ and $m = 10$. Check: $25(10) + 40 = 290$. True.
14. Let $n$ be the number. $6n - 4 = 2n + 20$, so $4n = 24$ and $n = 6$. Check: $6(6) - 4 = 32$ and $2(6) + 20 = 32$. True.
15. Let $w$ be the width in meters, so the length is $2w$. $2(2w) + 2w = 96$, which simplifies to $6w = 96$, so $w = 16$. The width is 16 m and the length is 32 m. Check: $2(32) + 2(16) = 64 + 32 = 96$ m. True.
16. Each side of the equation is one complete quantity. Build the total for one option on the left and the total for the other option on the right; the equal sign is the claim that the two totals match. When only one quantity is described and its total is known, the known number is the whole other side.

Lesson 2.2 — Solving Multistep Equations, One Named Property at a Time

Workbook fill-ins. The order of business: expand, combine, undo addition or subtraction, undo multiplication or division, verify. Property table for $5x + 3x - 7 = 41$: combining $5x + 3x$ into $8x$ is the distributive property; adding 7 to both sides is the addition property of equality; dividing both sides by 8 is the division property of equality. Properties of equality change both sides; properties of real numbers rewrite one side.

Guided practice

17. $9x - 5 = 40$ (distributive property, combining like terms), so $9x = 45$ (addition property of equality) and $x = 5$ (division property of equality). Check: $7(5) + 2(5) - 5 = 35 + 10 - 5 = 40$. True.
18. $3x + 12 - 5 = 22$ (distributive property), so $3x + 7 = 22$ (combining like terms), $3x = 15$ (subtraction property of equality), and $x = 5$ (division property of equality). Check: $3(5 + 4) - 5 = 27 - 5 = 22$. True.
19. $-6x + 8 = 26$ (distributive property), so $-6x = 18$ (subtraction property of equality) and $x = -3$ (division property of equality). The second product changes sign because $(-2)(-4) = +8$. Check: $-2(3(-3) - 4) = -2(-13) = 26$. True.
20. $\frac{3}{4}x = 12$ (addition property of equality), then multiply both sides by the reciprocal $\frac{4}{3}$ (multiplication property of equality): $x = 16$ (multiplicative inverse and identity properties). Check: $\frac{3}{4}(16) - 2 = 12 - 2 = 10$. True.

Independent practice

21. a) $11x - 8 = 47$, so $11x = 55$ and $x = 5$; check $6(5) + 5(5) - 8 = 30 + 25 - 8 = 47$. b) $5x - 10 = 35$, so $5x = 45$ and $x = 9$; check $5(9 - 2) = 5(7) = 35$. Dividing by 5 first gives $x - 2 = 7$ and the same answer. c) $-3x = 21$, so $x = -7$; check $8 - 3(-7) = 8 + 21 = 29$. d) $8x + 2 - 3x = 27$, so $5x + 2 = 27$, $5x = 25$, and $x = 5$; check $2(4(5) + 1) - 3(5) = 2(21) - 15 = 42 - 15 = 27$.
22. a) $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$, so $\frac{5}{6}x = 10$ and $x = 12$; check $\frac{1}{2}(12) + \frac{1}{3}(12) = 6 + 4 = 10$. b) $0.25 + 0.75 = 1$, so $x - 3 = 9$ and $x = 12$; check $0.25(12) + 0.75(12) - 3 = 3 + 9 - 3 = 9$.
23. $-4x - 8 + 9 = 1$ (distributive property), so $-4x + 1 = 1$, then $-4x = 0$ (subtraction property of equality) and $x = 0$ (division property of equality). Check: $-4(0 + 2) + 9 = -8 + 9 = 1$. True. Zero is an ordinary solution: it is a number, and it makes the sentence true. The variable did not disappear here — only its value turned out to be 0, which is a different thing from "no solution."
24. $4x - 6 = 26$ (distributive property), so $4x = 32$ and $x = 8$. Check: $\frac{2}{5}(10(8) - 15) = \frac{2}{5}(65) = 26$. True.
25. $6x - 3 + 4x + 8 = 55$ (distributive property twice), so $10x + 5 = 55$ (combining like terms), $10x = 50$, and $x = 5$. Check: $3(2(5) - 1) + 4(5 + 2) = 3(9) + 4(7) = 27 + 28 = 55$. True.
26. Let $h$ be the number of hours. $45h + 120 = 525$, so $45h = 405$ and $h = 9$. Check: $45(9) + 120 = 405 + 120 = 525$. True. The landscaper worked 9 hours; 9 is a sensible number of hours for one job, so the answer is usable.
27. $5(x - 3) + 2x = 34$ — given. $5x - 15 + 2x = 34$ — distributive property. $7x - 15 = 34$ — combining like terms, which is the distributive property read backwards, together with the commutative and associative properties of addition that bring $5x$ and $2x$ next to each other. $7x = 49$ — addition property of equality. $x = 7$ — division property of equality, with the multiplicative identity property leaving $x$ alone. Check: $5(7 - 3) + 2(7) = 20 + 14 = 34$. True. Combining like terms rewrites one side without changing what it is worth, so the other side is untouched. Adding 15 changes what the left side is worth, so the identical change must be made on the right or the two sides stop being equal.
28. The factor $-3$ must multiply the $-6$ as well, and $(-3)(-6) = +18$, not $-18$. Correctly: $-3x + 18 = 15$, so $-3x = -3$ and $x = 1$. Check: $-3(1 - 6) = -3(-5) = 15$. True. The student's answer fails the check: $-3(-11 - 6) = -3(-17) = 51 \neq 15$.

Exit ticket 2.2

29. $13x - 6 = 46$ (combining like terms), so $13x = 52$ (addition property of equality) and $x = 4$ (division property of equality). Check: $4(4) + 9(4) - 6 = 16 + 36 - 6 = 46$. True.
30. $6x + 12 - 5 = 25$ (distributive property), so $6x + 7 = 25$, $6x = 18$ (subtraction property of equality), and $x = 3$ (division property of equality). Check: $6(3 + 2) - 5 = 30 - 5 = 25$. True.
31. $\frac{5}{6}x = 15$ (subtraction property of equality), so $x = 18$ (multiplication by the reciprocal $\frac{6}{5}$). Check: $\frac{5}{6}(18) + 4 = 15 + 4 = 19$. True.
32. A property of equality authorizes a change to both sides, because it changes what each side is worth — for example the subtraction property of equality, used to take 6 from both sides of $6x + 12 - 5 = 25$. A property of real numbers authorizes rewriting one side into an equivalent form without changing its value, so nothing needs to be done to the other side — for example the distributive property, used to turn $6(x + 2)$ into $6x + 12$.

Lesson 2.3 — The Variable on Both Sides, and Clearing Fractions and Decimals

Workbook fill-ins. The extra step is to gather the variable terms on one side, using the addition or subtraction property of equality. Move the smaller variable term so the coefficient you divide by stays positive. The least common denominator for $\frac{1}{2}x + 7 = \frac{3}{4}x + 1$ is 4; the multiplier that clears $0.15x + 2.4 = 0.05x + 3.6$ is 100. Multiplying both sides is authorized by the multiplication property of equality, and every term must be multiplied, by the distributive property.

Guided practice

33. Subtract $3x$ from both sides (subtraction property of equality): $5x + 5 = 30$. Subtract 5 (subtraction property of equality): $5x = 25$. Divide by 5 (division property of equality): $x = 5$. Check: $8(5) + 5 = 45$ and $3(5) + 30 = 45$. True.
34. Expand: $2x + 12 = 5x - 9$. Subtract $2x$, the smaller variable term, so the coefficient stays positive: $12 = 3x - 9$. Add 9: $21 = 3x$. Divide by 3: $x = 7$. Check: $2(7 + 6) = 26$ and $5(7) - 9 = 26$. True.
35. Multiply both sides by 6: $2x + 12 = x + 30$. Subtract $x$: $x + 12 = 30$. Subtract 12: $x = 18$. Check in the original: $\frac{1}{3}(18) + 2 = 6 + 2 = 8$ and $\frac{1}{6}(18) + 5 = 3 + 5 = 8$. True.
36. Multiply both sides by 10: $2x + 14 = 5x - 4$. Subtract $2x$: $14 = 3x - 4$. Add 4: $18 = 3x$. Divide by 3: $x = 6$. Check in the original: $0.2(6) + 1.4 = 1.2 + 1.4 = 2.6$ and $0.5(6) - 0.4 = 3 - 0.4 = 2.6$. True.

Independent practice

37. a) Subtract $5x$: $4x - 7 = 13$, so $4x = 20$ and $x = 5$; check $9(5) - 7 = 38$ and $5(5) + 13 = 38$. b) Expand: $4x - 4 = 2x + 10$. Subtract $2x$: $2x - 4 = 10$, so $2x = 14$ and $x = 7$; check $4(7 - 1) = 24$ and $2(7) + 10 = 24$. c) Add $5x$: $12 = 2 + 2x$, so $10 = 2x$ and $x = 5$; check $12 - 5(5) = -13$ and $2 - 3(5) = -13$. d) Expand both sides: $5x + 15 = 3x + 21$. Subtract $3x$: $2x + 15 = 21$, so $2x = 6$ and $x = 3$; check $5(3 + 3) = 30$ and $3(3 + 7) = 30$.
38. a) Subtract $\frac{1}{3}x$: $\frac{1}{3}x + 1 = 6$, so $\frac{1}{3}x = 5$ and $x = 15$; check $\frac{2}{3}(15) + 1 = 11$ and $\frac{1}{3}(15) + 6 = 11$. Multiplying by 3 at the start gives $2x + 3 = x + 18$ and the same answer. b) Subtract $0.4x$: $0.2x - 1.2 = 2$, so $0.2x = 3.2$ and $x = 16$; check $0.6(16) - 1.2 = 9.6 - 1.2 = 8.4$ and $0.4(16) + 2 = 6.4 + 2 = 8.4$.
39. Combine on the left: $4x + 4 = x + 22$. Subtract $x$: $3x + 4 = 22$, so $3x = 18$ and $x = 6$. Check: $6(6) + 4 - 2(6) = 36 + 4 - 12 = 28$ and $6 + 22 = 28$. True.
40. Expand: $5 - 3x + 6 = 2x - 9$, so $11 - 3x = 2x - 9$. Add $3x$: $11 = 5x - 9$. Add 9: $20 = 5x$. Divide by 5: $x = 4$. Check: $5 - 3(4 - 2) = 5 - 6 = -1$ and $2(4) - 9 = -1$. True.
41. Multiply both sides by 3: $x + 5 = 3x - 15$. Subtract $x$: $5 = 2x - 15$. Add 15: $20 = 2x$. Divide by 2: $x = 10$. Check: $\frac{10 + 5}{3} = \frac{15}{3} = 5$ and $10 - 5 = 5$. True.
42. Let $m$ be the number of miles. $0.60m + 49 = 0.85m + 29$. Subtract $0.60m$: $49 = 0.25m + 29$. Subtract 29: $20 = 0.25m$. Divide by 0.25: $m = 80$. Check: $0.60(80) + 49 = 48 + 49 = 97$ and $0.85(80) + 29 = 68 + 29 = 97$. True. At 80 miles both rentals cost $\$97$.
43. Subtracting $4x$: $6 = 2x - 10$, so $16 = 2x$ and $x = 8$. Subtracting $6x$: $-2x + 6 = -10$, so $-2x = -16$ and $x = 8$. Check: $4(8) + 6 = 38$ and $6(8) - 10 = 38$. True. Both routes are legal applications of the subtraction property of equality, and a legal move never changes which numbers make the sentence true. Two correct routes therefore describe the same solution set, so they cannot disagree; the only difference is how many negative signs you handled along the way.
44. A property of equality must be applied to both sides. Subtracting $2x$ from the left only makes the left smaller than the right, so the line $3x + 2 = 17$ is not equivalent to the original — it is a different equation that happens to have the same solution here, and nothing guarantees that in general. Correctly: subtract $2x$ from both sides (subtraction property of equality) to get $3x + 2 = 17$, subtract 2 (subtraction property of equality) to get $3x = 15$, divide by 3 (division property of equality) to get $x = 5$. Check: $5(5) + 2 = 27$ and $2(5) + 17 = 27$. True.

Exit ticket 2.3

45. Subtract $6x$: $4x - 3 = 17$, so $4x = 20$ and $x = 5$. Check: $10(5) - 3 = 47$ and $6(5) + 17 = 47$. True.
46. Expand: $3x + 12 = 5x - 2$. Subtract $3x$: $12 = 2x - 2$. Add 2: $14 = 2x$, so $x = 7$. Check: $3(7 + 4) = 33$ and $5(7) - 2 = 33$. True.
47. Multiply both sides by 4: $x + 12 = 2x - 4$. Subtract $x$: $12 = x - 4$, so $x = 16$. Check in the original: $\frac{1}{4}(16) + 3 = 4 + 3 = 7$ and $\frac{1}{2}(16) - 1 = 8 - 1 = 7$. True.
48. Multiplying both sides by the same nonzero number is the multiplication property of equality, so the two sides stay equal to each other. The new equation is true for exactly the numbers that made the old one true — no value can be gained or lost — because multiplying by a nonzero number is reversible: dividing by the same number returns the original equation.

Lesson 2.4 — One Solution, No Solution, Infinitely Many

Workbook fill-ins. Table: $5x - 7 = 2x + 8$ ends at $x = 5$, one solution; $4x + 9 = 4x - 3$ ends at $9 = -3$, false, no solution; $3(x + 2) = 3x + 6$ ends at $6 = 6$, true, infinitely many. Graph blanks: one solution — lines cross once; no solution — lines are parallel, same slope and different $y$-intercepts; infinitely many — the lines coincide, same slope and same $y$-intercept. Number-line blanks: a single dot, nothing shaded, the entire line shaded. "The variable disappeared" is not the same as "the answer is zero."

Guided practice

49. Expanding gives $2x + 10 = 2x + 10$, and subtracting $2x$ from both sides leaves $10 = 10$. The last line is true and no longer contains the variable, so every real number makes the original sentence true: infinitely many solutions. The two sides were the same expression written two ways, which is what an identity is.
50. Subtracting $7x$ from both sides leaves $-4 = 1$, which is false and contains no variable, so no solution. Graphed, $y = 7x - 4$ and $y = 7x + 1$ have the same slope 7 and different $y$-intercepts: parallel lines that never meet.
51. Subtract $5x$: $3x + 3 = 18$. Subtract 3: $3x = 15$. Divide by 3: $x = 5$. One solution. Check: $8(5) + 3 = 43$ and $5(5) + 18 = 43$. True.
52. Expanding gives $-6x + 3 = -6x + 3$, and adding $6x$ to both sides leaves $3 = 3$, true: infinitely many solutions. The negative factor changes nothing about the reasoning, because distributing it correctly is just rewriting one side; what decides the case is whether the two sides turn out to be the same expression.

Independent practice

53. a) $4x - 8 = 4x - 8$, so $-8 = -8$: infinitely many solutions. b) Subtracting $9x$ leaves $5 = -5$, false: no solution. c) Subtract $2x$: $4x + 1 = 21$, so $4x = 20$ and $x = 5$: one solution; check $6(5) + 1 = 31$ and $2(5) + 21 = 31$. d) $5x + 15 = 5x + 3$, so $15 = 3$, false: no solution.
54. a) The left side becomes $2x + 3x + 3 = 5x + 3$, so the equation is $5x + 3 = 5x + 3$ and the last line is $3 = 3$: infinitely many solutions. b) $6x + 12 = 6x + 10$, so $12 = 10$, false: no solution.
55. $k = 4$. With $k = 4$ the equation reads $4x + 7 = 4x + 7$, which is an identity, so every real number is a solution. For any other value of $k$ the equation becomes $(4 - k)x = 0$ with $4 - k \neq 0$, which has the single solution $x = 0$. No value of $k$ gives no solution here, because the constants on the two sides already agree.
56. $c = 8$, which makes both sides read $3x + 8$. For any other value of $c$, subtracting $3x$ from both sides leaves $c = 8$, a statement with no variable in it that is simply false — and a false statement cannot be repaired by choosing $x$, since $x$ is no longer there to choose. So every other $c$ gives no solution at all, not even one.
57. Expanding gives $10 - 2x - 6 = -2x + 4$, so $4 - 2x = -2x + 4$. Adding $2x$ to both sides leaves $4 = 4$: infinitely many solutions.
58. Let $s$ be the number of shirts. $5s + 40 = 5s + 60$, and subtracting $5s$ leaves $40 = 60$, which is false: no solution. Interpretation: the two shops charge the same amount per shirt, so shop B's bill is always exactly $\$20$ more than shop A's, no matter how many shirts are ordered. The bills are never equal, and shop A is cheaper for every order.
59. Write each side in the form $mx + b$ and compare. If $a \neq c$ the slopes differ, so the lines cross exactly once: one solution — for example $6x + 1 = 2x + 21$, with solution $x = 5$. If $a = c$ but $b \neq d$ the sides differ by the same nonzero amount at every $x$: no solution — for example $9x + 5 = 9x - 5$. If $a = c$ and $b = d$ the two sides are the same expression: infinitely many solutions — for example $4(x - 2) = 4x - 8$.
60. The student read a true statement as a failure. Reaching $0 = 0$ means the variable dropped out and what remained is true, which says every value of $x$ works — infinitely many solutions. "No solution" is what a false leftover, such as $0 = 7$, would mean. Graphed, the two sides are the same line drawn twice, so they agree everywhere; two parallel lines that never meet would be the no-solution picture.

Exit ticket 2.4

61. $5x - 5 = 5x - 5$, so $-5 = -5$, true: infinitely many solutions.
62. Subtracting $8x$ leaves $2 = -6$, false: no solution.
63. Subtract $4x$: $3x - 3 = 12$, so $3x = 15$ and $x = 5$: one solution. Check: $7(5) - 3 = 32$ and $4(5) + 12 = 32$. True.
64. Graph each side as its own function. One solution: the slopes differ, so the lines cross at exactly one point, and its $x$-coordinate is the solution. No solution: the slopes match but the $y$-intercepts differ, so the lines are parallel and never meet. Infinitely many: the slopes and the $y$-intercepts match, so the two graphs are the same line and every point is a place where the sides agree. Slope alone separates one solution from the other two cases; the $y$-intercept then separates those two from each other.

Lesson 2.5 — Literal Equations and Formulas

Workbook fill-ins. A literal equation has more than one letter; a formula is a literal equation people use. Treat every other letter as a number. Undo in reverse order. Divide the whole side, not one term. The figure's two columns use the same properties: the subtraction property of equality, then the division property of equality. From $P = 2l + 2w$ the rearrangement is $w = \frac{P - 2l}{2}$, and with $P = 84$ and $l = 25$ it gives $w = 17$.

Guided practice

65. Divide both sides by $r$ (division property of equality): $t = \frac{d}{r}$, for $r \neq 0$. Check with $d = 120$ and $r = 40$: $t = \frac{120}{40} = 3$, and $120 = 40(3)$. Agreed.
66. Multiply both sides by 2 (multiplication property of equality): $2A = bh$. Divide both sides by $h$ (division property of equality): $b = \frac{2A}{h}$, for $h \neq 0$.
67. Subtract $2w$ from both sides (subtraction property of equality): $P - 2w = 2l$. Divide both sides by 2 (division property of equality): $l = \frac{P - 2w}{2}$. Check with $P = 30$ and $w = 4$: $l = \frac{30 - 8}{2} = 11$, and $2(11) + 2(4) = 30$. Agreed.
68. Subtract $b$ from both sides (subtraction property of equality): $y - b = mx$. Divide both sides by $m$ (division property of equality): $x = \frac{y - b}{m}$, for $m \neq 0$. If $m = 0$ the original equation contains no $x$ at all, so there is nothing to solve for.

Independent practice

69. a) $r = \frac{C}{2\pi}$ b) $h = \frac{V}{lw}$, for $l, w \neq 0$ c) $t = \frac{I}{Pr}$, for $P, r \neq 0$ d) $by = c - ax$, so $y = \frac{c - ax}{b}$, for $b \neq 0$
70. a) Subtract $P$: $A - P = Prt$. Divide by $Pr$: $t = \frac{A - P}{Pr}$, for $P, r \neq 0$. b) Subtract $2\pi r^2$: $S - 2\pi r^2 = 2\pi rh$. Divide by $2\pi r$: $h = \frac{S - 2\pi r^2}{2\pi r}$, for $r \neq 0$.
71. Subtract 32: $F - 32 = \frac{9}{5}C$. Multiply by $\frac{5}{9}$: $C = \frac{5}{9}(F - 32)$. For $F = 77$: $C = \frac{5}{9}(45) = 25$. Check: $\frac{9}{5}(25) + 32 = 45 + 32 = 77$. Agreed, so $77^\circ$F is $25^\circ$C.
72. Subtract $3x$: $4y = 12 - 3x$. Divide by 4: $y = \frac{12 - 3x}{4}$, which in $y = mx + b$ form is $y = -\frac{3}{4}x + 3$. Check with $x = 4$: the original gives $12 + 4y = 12$, so $y = 0$, and $-\frac{3}{4}(4) + 3 = 0$. Agreed.
73. Multiply both sides by 2: $2A = h(b_1 + b_2)$. Divide by $b_1 + b_2$: $h = \frac{2A}{b_1 + b_2}$. With $A = 48$, $b_1 = 6$, and $b_2 = 10$: $h = \frac{96}{16} = 6$ inches. Check: $\frac{1}{2}(6)(6 + 10) = \frac{1}{2}(6)(16) = 48$ square inches. Agreed.
74. $w = \frac{84 - 2(25)}{2} = \frac{84 - 50}{2} = \frac{34}{2} = 17$ cm. Verify in the original: $P = 2(25) + 2(17) = 50 + 34 = 84$ cm. Agreed.
75. Both use the subtraction property of equality to remove the term that does not contain $w$, then the division property of equality to undo the multiplication by 2. Nothing about the mathematics changes. What is harder is psychological: $P - 2l$ does not collapse into a single tidy symbol the way $17 - 12$ collapses into 5, so the answer stays visibly unfinished and it is tempting to "simplify" it illegally. The difficulty is in reading an expression as a single number, not in the algebra.
76. The student divided only the $2w$ term by 2 and left $P - 2l$ untouched. The division property of equality divides the entire side, so every term of $P - 2l$ must be divided: $w = \frac{P - 2l}{2}$. With $P = 84$ and $l = 25$, the correct version gives $w = \frac{34}{2} = 17$, and $2(25) + 2(17) = 84$. The student's version gives $w = 84 - 50 = 34$, and $2(25) + 2(34) = 118 \neq 84$.

Exit ticket 2.5

77. Divide both sides by $h$ (division property of equality): $B = \frac{V}{h}$, for $h \neq 0$.
78. Divide both sides by $x - x_1$ (division property of equality): $m = \frac{y - y_1}{x - x_1}$, for $x \neq x_1$. The restriction matters: if $x = x_1$ the divisor is 0, and the original equation then says $y - y_1 = 0$ without mentioning $m$ at all.
79. Subtract $2x$: $-5y = 20 - 2x$. Divide by $-5$: $y = \frac{2x - 20}{5}$, which is $y = \frac{2}{5}x - 4$. Check with $x = 10$: the original gives $20 - 5y = 20$, so $y = 0$, and $\frac{2}{5}(10) - 4 = 0$. Agreed.
80. Choose convenient numbers for every letter but the one you solved for, evaluate the original formula to find that letter's value, then evaluate your rearranged version with the same numbers and compare. Agreement on one set of numbers is reassuring but not a proof, because a wrong rearrangement can accidentally agree at a particular value — for instance, a version that divides only one term can match when the other term happens to be 0. A proof comes from checking that every step was a property of equality applied to the whole side.

Lesson 2.6 — Verifying Algebraically, Graphically, and with Technology

Workbook fill-ins. Algebraic verification catches arithmetic mistakes; graphical verification catches structural mistakes such as a dropped term or a flipped sign; technology catches what a hand-drawn graph is too coarse to show. The lines $y = 2x + 1$ and $y = x + 4$ meet at $(3, 7)$: the solution is the $x$-coordinate, 3, and 7 is the common value of the two sides. The difference graph $y = (3x - 5) - (x + 1) = 2x - 6$ has $x$-intercept 3. The cost lines meet at $(15, 180)$: below 15 shirts shop B is cheaper, above 15 shop A is cheaper. A disagreement means one of three things is wrong — the solving, the typing, or the reading of the graph.

Guided practice

81. Left side: $5(4) - 3 = 20 - 3 = 17$. Right side: $3(4) + 5 = 12 + 5 = 17$. The sides match, so $x = 4$ is the solution. Graphically, $y = 5x - 3$ and $y = 3x + 5$ meet at $(4, 17)$: the solution is the $x$-coordinate 4, and 17 is what both sides are worth there.
82. Subtract $2x$: $7 = 2x - 1$. Add 1: $8 = 2x$. Divide by 2: $x = 4$. Check: $2(4) + 7 = 15$ and $4(4) - 1 = 15$. True. The graphs of the two sides meet at $(4, 15)$.
83. The equation is $3x - 1 = x + 5$, and its solution is $x = 3$. The number 8 is the $y$-coordinate of the intersection, which is the common value of the two sides when $x = 3$: $3(3) - 1 = 8$ and $3 + 5 = 8$. It is not the solution of the equation.
84. Let $m$ be the number of miles. $2.25m + 3.00 = 1.75m + 5.00$. Subtract $1.75m$: $0.50m + 3.00 = 5.00$. Subtract 3: $0.50m = 2.00$. Divide by 0.5: $m = 4$. Check: $2.25(4) + 3 = 9 + 3 = 12$ and $1.75(4) + 5 = 7 + 5 = 12$. True. At 4 miles both cost $\$12.00$; the taxi is cheaper for shorter trips, since it starts lower, and the rideshare is cheaper for anything longer than 4 miles, since it charges less per mile.

Independent practice

85. a) Left: $4(6) - 5 = 19$. Right: $2(6) + 7 = 19$. The sides match, so yes, $x = 6$ is the solution. b) Left: $5(2) + 1 = 11$. Right: $3(2) + 7 = 13$. The sides do not match, so $x = 2$ is not a solution. Solving: subtract $3x$ to get $2x + 1 = 7$, so $2x = 6$ and $x = 3$. Check: $5(3) + 1 = 16$ and $3(3) + 7 = 16$. True.
86. Subtract $2x$: $4x - 4 = 12$. Add 4: $4x = 16$. Divide by 4: $x = 4$. Check: $6(4) - 4 = 20$ and $2(4) + 12 = 20$. True. The graphs meet at $(4, 20)$.
87. Expand: $3x + 3 = x + 9$. Subtract $x$: $2x + 3 = 9$, so $2x = 6$ and $x = 3$. Check: $3(3 + 1) = 12$ and $3 + 9 = 12$. True. The difference function is $y = (3x + 3) - (x + 9) = 2x - 6$, whose $x$-intercept is $(3, 0)$ — the same 3.
88. The equation is $2x + 5 = 2x - 1$, and it has no solution. The two lines have the same slope 2 and different $y$-intercepts, so they are parallel; there is no $x$ at which the sides are equal. Algebraically, subtracting $2x$ leaves $5 = -1$, which is false and says the same thing.
89. Let $m$ be the number of months. $12m + 75 = 27m$, so $75 = 15m$ and $m = 5$. Check: $12(5) + 75 = 135$ and $27(5) = 135$. True, and the graphs meet at $(5, 135)$. Interpretation: the totals are equal at 5 months, $\$135$ each. A two-year membership is 24 months, where gym A costs $12(24) + 75 = \$363$ and gym B costs $27(24) = \$648$, so gym A is far cheaper for two years — the joining fee has long since paid for itself.
90. Let $t$ be the number of seconds. $120 - 8t = 40 + 2t$. Add $8t$: $120 = 40 + 10t$. Subtract 40: $80 = 10t$. Divide by 10: $t = 8$. Check: $120 - 8(8) = 120 - 64 = 56$ and $40 + 2(8) = 56$. True. After 8 seconds both drones are at 56 meters, and both heights are positive, so the moment really does occur while both are still flying.
91. If substitution into the original equation fails, then $x = 7$ is not a solution, so the algebra and the graph are agreeing with each other and both are wrong about the original problem. The likeliest cause is that both were built from the same mistyped or miscopied equation — a dropped term or a wrong coefficient carried from your first line into both the solving and the graphing. Check in this order: first re-read the original equation against what you wrote down, then re-check the substitution arithmetic, then re-check each solving step. The substitution is the referee, because it is the only one of the three that uses the original equation directly.
92. Substituting $x = 2$: left $4(2 - 1) = 4$, right $2(2) + 6 = 10$. The sides do not match, so $x = 2$ is not a solution and the graph could not have confirmed it. Correctly: $4x - 4 = 2x + 6$ (distributive property), $2x - 4 = 6$ (subtraction property of equality), $2x = 10$ (addition property of equality), $x = 5$ (division property of equality). Check: $4(5 - 1) = 16$ and $2(5) + 6 = 16$. True. The true intersection is $(5, 16)$.

Exit ticket 2.6

93. Subtract $4x$: $3x - 2 = 13$. Add 2: $3x = 15$. Divide by 3: $x = 5$. Left side: $7(5) - 2 = 33$. Right side: $4(5) + 13 = 33$. The sides match, so $x = 5$ is confirmed.
94. The equation is $-x + 8 = 3x$, and its solution is $x = 2$. Check: $-2 + 8 = 6$ and $3(2) = 6$. True; 6 is the common value of the two sides.
95. Let $p$ be the number of prints. $1.50p + 25 = 2.00p$. Subtract $1.50p$: $25 = 0.50p$. Divide by 0.5: $p = 50$. Check: $1.50(50) + 25 = 75 + 25 = 100$ and $2.00(50) = 100$. True. At 50 prints both labs charge $\$100.00$.
96. Algebraic verification substitutes into the original equation, so it catches arithmetic slips made while solving — it is the only check that touches the original problem directly. Graphical verification draws each side, so it catches structural errors: a dropped term, a misread coefficient, or a sign flipped during expansion changes the picture visibly, while it might change a number only slightly. Technology catches what a hand-drawn graph is too coarse to resolve, such as an intersection at $x = 3.02$ that a sketch would report as 3, and it is the check A.EI.1f names by name. Together they fail in different ways, which is exactly why all three are asked for.

Chapter 2 Review

Part A — Writing an equation for a contextual situation (A.EI.1a)

97. Let $p$ be the number of plates. $18.75p + 95 = 1032.50$, so $18.75p = 937.50$ and $p = 50$. Check: $18.75(50) + 95 = 937.50 + 95 = 1032.50$. True.
98. Let $n$ be the number. $5n - 9 = n + 15$, so $4n = 24$ and $n = 6$. Check: $5(6) - 9 = 21$ and $6 + 15 = 21$. True.
99. Let $w$ be the width in meters, so the length is $3w - 4$. $2(3w - 4) + 2w = 72$, which expands to $6w - 8 + 2w = 72$ and simplifies to $8w - 8 = 72$, so $8w = 80$ and $w = 10$. The width is 10 m and the length is 26 m. Check: $2(26) + 2(10) = 52 + 20 = 72$ m. True.
100. Let $m$ be the number of months. $55m + 120 = 75m$, so $120 = 20m$ and $m = 6$. Check: $55(6) + 120 = 450$ and $75(6) = 450$. True. The totals are equal at 6 months, $\$450$ each.

Part B — Solving multistep equations with named properties (A.EI.1b)

101. a) $10x - 12 = 38$ (combining like terms), so $10x = 50$ and $x = 5$; check $7(5) + 3(5) - 12 = 35 + 15 - 12 = 38$. b) $4x - 20 + 6 = 18$ (distributive property), so $4x - 14 = 18$, $4x = 32$, and $x = 8$; check $4(8 - 5) + 6 = 12 + 6 = 18$. c) $-6x - 14 = 10$ (distributive property), so $-6x = 24$ and $x = -4$; check $-2(3(-4) + 7) = -2(-5) = 10$. d) Subtract $5x$: $4x - 4 = 24$, so $4x = 28$ and $x = 7$; check $9(7) - 4 = 59$ and $5(7) + 24 = 59$.
102. a) $\frac{3}{5}x = 15$ (subtraction property of equality), so $x = 25$ (multiplication by $\frac{5}{3}$); check $\frac{3}{5}(25) + 4 = 15 + 4 = 19$. b) Subtract $0.1x$: $0.2x + 1.5 = 3.1$, so $0.2x = 1.6$ and $x = 8$; check $0.3(8) + 1.5 = 2.4 + 1.5 = 3.9$ and $0.1(8) + 3.1 = 0.8 + 3.1 = 3.9$.
103. Expand both products, watching the second sign: $5x + 10 - 3x + 3 = 27$. Combine: $2x + 13 = 27$. Subtract 13: $2x = 14$. Divide by 2: $x = 7$. Check: $5(7 + 2) - 3(7 - 1) = 45 - 18 = 27$. True.
104. Multiply both sides by 3 (multiplication property of equality): $2x - 1 = 3x - 12$. Subtract $2x$: $-1 = x - 12$. Add 12: $x = 11$. Check: $\frac{2(11) - 1}{3} = \frac{21}{3} = 7$ and $11 - 4 = 7$. True.
105. Expand: $6 - 4x + 12 = 2x + 6$, so $18 - 4x = 2x + 6$. Add $4x$: $18 = 6x + 6$. Subtract 6: $12 = 6x$. Divide by 6: $x = 2$. Check: $6 - 4(2 - 3) = 6 + 4 = 10$ and $2(2) + 6 = 10$. True.
106. $3(2x - 5) + 4x = 35$ — given. $6x - 15 + 4x = 35$ — distributive property. $10x - 15 = 35$ — combining like terms, which is the distributive property read backwards, with the commutative and associative properties of addition bringing $6x$ and $4x$ together. $10x = 50$ — addition property of equality, with the additive inverse property clearing the $-15$. $x = 5$ — division property of equality, with the multiplicative identity property leaving $x$ alone. Check: $3(2(5) - 5) + 4(5) = 3(5) + 20 = 35$. True.