Appendix A — Answer Key, Chapter 11: Writing and Solving One-Step Equations

SOL 6.PFA.3 · Covers textbook Chapter 11 and the companion workbook. Item numbers match the textbook; workbook items that repeat textbook problems share the same answers, and workbook-only items are keyed at the end. Where students write their own equations or situations, a sample response is given — accept any answer that matches the required operation and numbers.

Lesson 11.1 — The Language of Algebra

Guided practice

1. An expression. It has no equal sign, so it is a phrase rather than a sentence.
2. $r$
3. $6$
4. Two terms: $3n$ and $8$.
5. Yes. Substituting gives $4 + 6 = 10$, which is true.

Independent practice

6. a) equation b) expression c) expression d) equation
7. Variable $w$; coefficient $10$; constant $3$; two terms.
8. a) $-8$ b) $1$ c) $\tfrac{1}{2}$ d) $\tfrac{1}{6}$
9. Sample: $4t = 20$, with solution $t = 5$, since $4(5) = 20$.
10. You evaluate an expression by substituting a value and computing a single number — evaluating $2x + 1$ at $x = 3$ gives $7$. You solve an equation by finding the value of the variable that makes it true — solving $2x + 1 = 7$ gives $x = 3$.
11. Expression: $9n$. Equation: $9n = 63$. The coefficient is $9$. (The solution is $n = 7$ tickets.)
12. Every variable has a coefficient. When no number is written, the coefficient is $1$, because $x$ means $1 \cdot x$.

Exit ticket 11.1

1. An equation.
2. $\tfrac{1}{3}$
3. Two terms; the constant is $7$.
4. A solution is a number that makes the equation true when you substitute it for the variable. Both sides then name the same value.

Lesson 11.2 — Modeling Equations with Balance Scales and Algebra Tiles

Guided practice

1. It represents the equal sign: the two sides of the equation have the same value.
2. $x + 4 = 9$
3. Remove 4 unit weights from each pan. The left pan holds only $x$; the right holds $9 - 4 = 5$. So $x = 5$.
4. A zero pair is one positive tile and one negative tile of the same size. Together they are worth $0$, so removing them does not change the value on that side.
5. Replace the $x$ tile with three unit tiles. The left mat holds $3 + 6 = 9$ tiles and the right mat holds $9$ tiles, so the model balances and $x = 3$ is confirmed.

Independent practice

6. a) $m + 7 = 10$ b) $3x = 15$
7. a) Remove 7 unit weights from each pan, leaving $m$ on the left and $3$ on the right, so $m = 3$. b) Split both mats into 3 equal groups. Each group pairs one $x$ tile with $15 \div 3 = 5$ unit tiles, so $x = 5$.
8. Left mat: one $x$ tile and four $-1$ tiles. Right mat: two unit tiles. Add four $1$ tiles to each mat. On the left, four zero pairs form and are removed, leaving the $x$ tile. On the right there are $2 + 4 = 6$ tiles. So $x = 6$. Confirm: $6 - 4 = 2$.
9. Left mat: four $x$ tiles. Right mat: twenty unit tiles. Divide each mat into 4 equal groups; each group pairs one $x$ tile with $20 \div 4 = 5$ unit tiles. So $x = 5$. Confirm: $4(5) = 20$.
10. Removing tiles from only one mat breaks the balance, so the two sides are no longer equal and the equation he writes is no longer true. He must remove 2 tiles from both mats, leaving $x = 5$.
11. $b + 3 = 11$. Remove 3 pounds from each pan, leaving the backpack on the left and $11 - 3 = 8$ pounds on the right, so $b = 8$ pounds. Confirm: an 8-pound backpack with a 3-pound weight is 11 pounds, matching the other pan.
12. Solving searches for an answer; confirming tests an answer you already have. A model used to confirm will visibly tip if the value is wrong, so it catches arithmetic slips and sign errors that are easy to make while solving. Doing both means you leave with an answer and evidence that it is right.

Exit ticket 11.2

1. $x + 5 = 12$
2. $x = 7$. Remove 5 unit weights from each pan.
3. A positive tile and a negative tile of the same size, which together are worth $0$.
4. Replace the $x$ tile with seven unit tiles. The left mat then holds $7 + 3 = 10$ tiles and the right mat holds $10$ tiles. The counts match, so the model balances and the solution is confirmed.

Lesson 11.3 — Solving One-Step Equations with Addition and Subtraction

Guided practice

1. Subtracting 9.
2. $x = 8$. Check: $8 + 3 = 11$.
3. $p = 11$. Check: $11 - 5 = 6$.
4. $t = -6$. Check: $-6 + 10 = 4$.
5. The subtraction property of equality.

Independent practice

6. a) $x = 13$; check $13 + 6 = 19$ b) $n = 14$; check $14 - 11 = 3$ c) $w = -7$; check $-7 + 12 = 5$ d) $h = -8$; check $-8 - 7 = -15$
7. $q = -13$. Check: $-13 + 9 = -4$.
8. $r = 0$. Check: $0 - 20 = -20$.
9. $b + 13 = 2$. Subtract 13 from both sides by the subtraction property of equality: $b + 13 - 13 = 2 - 13$. The additive inverse property gives $13 - 13 = 0$, and the additive identity property gives $b + 0 = b$, so $b = -11$. Check: $-11 + 13 = 2$.
10. Sample: $x - 5 = -13$, solved by adding 5 to both sides. Check: $-8 - 5 = -13$.
11. $s - 17 = 28$, so $s = 45$. Check: starting with $45 and spending $17 leaves $45 - 17 = 28$ dollars, which matches the story.
12. He added 9 to both sides instead of subtracting it. Subtracting 9 gives $x = 4 - 9 = -5$. A check would have caught it immediately: substituting $13$ gives $13 + 9 = 22$, not $4$.

Exit ticket 11.3

1. $x = 7$
2. $n = 7$. Check: $7 - 9 = -2$.
3. Yes. Substituting gives $-5 + 12 = 7$, which is true.
4. The equal sign says the two sides have the same value. Changing one side only makes them unequal, like taking weight off one pan of a balance scale. Doing the same thing to both sides keeps the statement true.

Lesson 11.4 — Solving One-Step Equations with Multiplication and Division

Guided practice

1. Dividing by 7.
2. $x = 8$. Check: $4(8) = 32$.
3. $n = 15$. Check: $\tfrac{1}{5}(15) = 3$.
4. $y = -7$. Check: $-3(-7) = 21$.
5. The coefficient is $\tfrac{1}{8}$. Multiply both sides by 8, which gives $m = 16$. Check: $\tfrac{16}{8} = 2$.

Independent practice

6. a) $c = 6$; check $9(6) = 54$ b) $d = 24$; check $\tfrac{1}{6}(24) = 4$ c) $k = -7$; check $-7(-7) = 49$ d) $p = -15$; check $\tfrac{-15}{3} = -5$
7. $m = -5$. Check: $12(-5) = -60$.
8. $v = -90$. Check: $\tfrac{1}{10}(-90) = -9$.
9. $-4x = -28$. Divide both sides by $-4$ using the division property of equality: $\tfrac{-4x}{-4} = \tfrac{-28}{-4}$. The multiplicative identity property gives $1x = x$, so $x = 7$. Check: $-4(7) = -28$.
10. Sample: $\tfrac{1}{6}x = 5$. Multiply both sides by 6 to get $x = 30$. Check: $\tfrac{1}{6}(30) = 5$.
11. $12t = 84$, so $t = 7$. Check: $12(7) = 84$. Seven tables each receiving 12 markers accounts for all 84 markers, so the answer makes sense.
12. Dividing by 4 shrinks the left side further instead of undoing the division already there. The coefficient $\tfrac{1}{4}$ means $x$ was split into 4 equal parts, so you multiply both sides by 4: $x = 32$. The check exposes her answer at once, since $\tfrac{1}{4}(2) = \tfrac{1}{2}$, not $8$, while $\tfrac{1}{4}(32) = 8$ is true.

Exit ticket 11.4

1. $x = 9$
2. $n = 35$
3. $y = -5$. Check: $-6(-5) = 30$.
4. A unit fraction coefficient means the variable has already been divided into equal parts. Dividing again would make the parts smaller still. Multiplying by the denominator rebuilds the whole, which is the inverse operation.

Lesson 11.5 — Writing an Equation from a Situation

Guided practice

1. $n + 6 = 14$; $n = 8$.
2. $p - 8 = 24$; $p = 32$. Check: $32 - 8 = 24$.
3. $\tfrac{1}{4}g = 15$; $g = 60$. (The equivalent $\tfrac{g}{4} = 15$ is also correct.) Check: $\tfrac{1}{4}(60) = 15$.
4. $\tfrac{1}{2}n = 11$; $n = 22$.
5. The amount left, $21, is given, and $9 is given. The only unknown is what Ben had before he spent anything, so that is what the variable stands for.

Independent practice

6. $n - 17 = 5$; $n = 22$.
7. $7n = 63$; $n = 9$.
8. $\tfrac{1}{5}n = 12$; $n = 60$.
9. $d - 30 = -75$; $d = -45$ meters. Check: $-45 - 30 = -75$.
10. $n + 20 = 4$; $n = -16$. Check: $-16 + 20 = 4$.
11. $8p = 96$; $p = 12$. The school ordered 12 pizzas. Check: $8(12) = 96$ slices.
12. Carla's equation adds 5, but Omar gave away 5, which removes marbles from his original amount. The correct equation is $m - 5 = 14$, so $m = 19$. Check: $19 - 5 = 14$.

Exit ticket 11.5

1. $n + 13 = 30$; $n = 17$.
2. $6r = 54$; $r = 9$ chairs per row.
3. $e + 25 = 10$; $e = -15$ meters. Check: $-15 + 25 = 10$.
4. The variable stands for the quantity the problem does not give you — the one the question asks you to find. Every other number in the story is known and belongs in the equation as a number.

Lesson 11.6 — Writing a Situation from an Equation

Accept any situation whose operation, numbers, and result match the equation and whose context makes sense. Samples follow.

Guided practice

1. Leo had some trading cards, bought 3 more, and now has 12. How many did he start with? $x = 9$.
2. Five equal teams have 45 players in all. How many players are on each team? $n = 9$.
3. A jar lost 7 marbles and now holds 2. How many did it hold before? $x = 9$.
4. Half of a bag of flour weighs 9 ounces. How much does the full bag weigh? $m = 18$.
5. The equation adds 6 to the unknown and then reaches 15, so 15 is the amount after the change. If the story started at 15, the equation would have to be $15 + 6 = x$ instead.

Independent practice

6. A theater had some seats, added 11 more, and now has 30. How many were there before? $x = 19$.
7. Nine identical boxes hold 72 crayons altogether. How many crayons are in each box? $k = 8$.
8. A board is cut into 6 equal pieces, and each piece is 5 feet long. How long was the board? $t = 30$.
9. A store sold 12 shirts and has 8 left. How many did it have to begin with? $n = 20$.
10. The morning temperature rose 15 degrees to reach 4°F. What was the temperature before it rose? $w = -11$°F. Check: $-11 + 15 = 4$.
11. A class is divided into 8 equal groups, and each group has 3 students. How many students are in the class? $c = 24$. There are 24 students in the class.
12. Her story adds an unknown amount to 4, which would be $4 + x = 20$. The equation $4x = 20$ describes 4 equal groups of an unknown size. Either fix works: change the equation to $4 + x = 20$, giving $x = 16$; or change the story to "Four friends earned the same amount and together earned $20," giving $x = 5$.

Exit ticket 11.6

1. A playlist had some songs, gained 8 more, and now has 22. How many were there before? $x = 14$.
2. Six equal shelves hold 42 books. How many books are on each shelf? $n = 7$.
3. One third of a pitcher of juice is 7 cups. How many cups does the full pitcher hold? $p = 21$.
4. The variable stands for the unknown quantity your story asks about — the amount before the change, the size of one group, or the whole that was split. Everything else in the equation becomes a known number in the story.

Chapter 11 Review

Part A — Algebraic vocabulary (6.PFA.3a)

1. a) expression b) equation c) expression d) equation
2. Variable $k$; terms $12k$ and $9$; coefficient $12$; constant $9$.
3. a) $-6$ b) $1$ c) $\tfrac{1}{7}$ d) $\tfrac{1}{2}$
4. Sample: expression $3x + 4$; equation $3x + 4 = 19$; solution $x = 5$.

Part B — Modeling with manipulatives (6.PFA.3b)

5. $x + 6 = 14$. Remove 6 unit weights from each pan, leaving $x$ on the left and $8$ on the right, so $x = 8$.
6. Place five $x$ tiles on the left mat and thirty-five unit tiles on the right. Split both mats into 5 equal groups; each group pairs one $x$ tile with $35 \div 5 = 7$ unit tiles, so $x = 7$.
7. Place one $x$ tile and three $-1$ tiles on the left mat and four unit tiles on the right. Add three $1$ tiles to each mat. On the left, three zero pairs form and are removed, leaving the $x$ tile. On the right there are $4 + 3 = 7$ tiles, so $x = 7$.

Part C — Solving with properties (6.PFA.3c)

8. a) $x = 14$; check $14 + 9 = 23$ b) $n = 20$; check $20 - 14 = 6$ c) $y = -7$; check $-7 + 8 = 1$ d) $m = -7$; check $-7 - 5 = -12$
9. a) $a = 9$; check $7(9) = 63$ b) $b = 36$; check $\tfrac{1}{4}(36) = 9$ c) $c = -5$; check $-8(-5) = 40$ d) $d = -30$; check $\tfrac{-30}{5} = -6$
10. $\tfrac{1}{3}x = 12$. Multiply both sides by 3 using the multiplication property of equality: $3 \cdot \tfrac{1}{3}x = 3 \cdot 12$. Since $3 \cdot \tfrac{1}{3} = 1$ and the multiplicative identity property gives $1x = x$, the result is $x = 36$. Check: $\tfrac{1}{3}(36) = 12$.

Part D — Confirming solutions (6.PFA.3d)

11. Yes. Replace the $x$ tile with nine unit tiles: the left mat holds $9 + 6 = 15$ tiles and the right mat holds $15$, so the model balances.
12. Malik is incorrect. Replacing each of the four $x$ tiles with 8 unit tiles gives $4(8) = 32$ tiles on the left, but the right mat holds only $24$. Splitting both mats into 4 equal groups gives $24 \div 4 = 6$, so $x = 6$. Check: $4(6) = 24$.
13. Put the proposed value in place of the variable and load the pans. If the value is wrong, the two pans hold different amounts and the scale tips toward the heavier side, showing directly that the two expressions are not equal.

Part E — Writing an equation from a situation (6.PFA.3e)

14. $n - 9 = 16$; $n = 25$. Check: $25 - 9 = 16$.
15. $12m = 132$; $m = 11$ members. Check: $12(11) = 132$.
16. $\tfrac{1}{4}p = 7$; $p = 28$ pencils. Check: $\tfrac{1}{4}(28) = 7$.
17. $t + 11 = 3$; $t = -8$°F. Check: $-8 + 11 = 3$.

Part F — Writing a situation from an equation (6.PFA.3f)

18. A garden had some tulips, and 7 more were planted, bringing the total to 25. How many were there before? $x = 18$.
19. Eight identical cartons hold 96 eggs altogether. How many eggs are in each carton? $n = 12$.
20. One fifth of a roll of tape measures 4 feet. How long is the whole roll? $p = 20$.

Part G — Mixed application and reasoning

21. $s = 9$. The coefficient 4 represents the four equal sides of the square, each of length $s$, that make up the perimeter.
22. Both equations are true for the same value. Subtracting 5 from each side of $x + 5 = 12$ removes the same amount from both pans of the balance, so the scale still balances and the remaining statement, $x = 7$, describes the same value of $x$. Substituting confirms it: $7 + 5 = 12$.
23. Situation: A savings account fell by $40 and now holds $85. What was the starting balance? Equation: $b - 40 = 85$, so $b = 125$. Check in context: a $125 balance minus $40 leaves $85, which matches the story.
24. Dividing by 2 undoes multiplying by 2, but the coefficient is $\tfrac{1}{2}$, which means $x$ was already halved. Multiply both sides by 2 instead: $x = 20$. Check: $\tfrac{1}{2}(20) = 10$. Yusuf's method would give $x = 5$, and $\tfrac{1}{2}(5) = 2\tfrac{1}{2}$, not 10.

Workbook-only items

Page 2, fill in the blanks. A variable is a letter that stands for a number. An expression is a phrase with no equal sign; you evaluate it. An equation is a sentence with an equal sign; you solve it. A coefficient is the number multiplied by a variable. A constant is a term that is only a number.

Page 2, expression or equation. $8m$ expression; $8m = 24$ equation; $p - 4$ expression; $p - 4 = 10$ equation; $14 = 2h$ equation; $\tfrac{1}{5}p$ expression.

Page 2, coefficients. $9b \to 9$; $y \to 1$; $\tfrac{1}{4}t \to \tfrac{1}{4}$; $-3w \to -3$; $-8k \to -8$; $\tfrac{g}{6} \to \tfrac{1}{6}$.

Page 3, parts of an expression.

Page 3, is it a solution. $x = 4$ in $x + 6 = 10$ gives $4 + 6 = 10$ — yes. $n = 6$ in $n + 5 = 12$ gives $6 + 5 = 11$ — no; the solution is $n = 7$.

Page 3, build your own. Sample: expression $5x + 3$; equation $5x + 3 = 18$.

Page 3, explain. A variable written alone has a coefficient of $1$, since $x$ means $1 \cdot x$.

Page 5, balance scale table.

Page 5, explain. The scale tips, because the two pans no longer hold equal amounts, so the equation is no longer true. He must remove 2 weights from both pans.

Page 6, zero pair blanks. Its value is 0, so removing it changes nothing about the value on that mat.

Page 6, tile model table. 3 $x$ tiles against 15 units: $3x = 15$, $x = 5$. 1 $x$ tile and 4 $-1$ tiles against 2 units: $x - 4 = 2$, $x = 6$. 4 $x$ tiles against 20 units: $4x = 20$, $x = 5$. 1 $x$ tile and 6 units against 9 units: $x + 6 = 9$, $x = 3$.

Page 6, describe your moves for $x - 4 = 2$. Add four $1$ tiles to each mat. Four zero pairs form on the left and are removed, leaving the $x$ tile; the right mat has $2 + 4 = 6$ tiles. So $x = 6$.

Page 7, confirm the solution table.

Page 7, apply it. $b + 3 = 11$; remove 3 pounds from each pan; the backpack weighs 8 pounds. Confirming: 8 pounds plus the 3-pound weight is 11 pounds, matching the other pan.

Page 7, explain. A model used to confirm tips visibly when the value is wrong, so it catches sign errors and arithmetic slips that are easy to miss while solving.

Page 9, solve table. $x + 3 = 11$: subtract 3, $x = 8$, check $8 + 3 = 11$. $p - 5 = 6$: add 5, $p = 11$, check $11 - 5 = 6$. $t + 10 = 4$: subtract 10, $t = -6$, check $-6 + 10 = 4$. $x + 6 = 19$: subtract 6, $x = 13$, check $13 + 6 = 19$. $n - 11 = 3$: add 11, $n = 14$, check $14 - 11 = 3$. $w + 12 = 5$: subtract 12, $w = -7$, check $-7 + 12 = 5$. $h - 7 = -15$: add 7, $h = -8$, check $-8 - 7 = -15$.

Page 10, solve and check. $q = -13$, check $-13 + 9 = -4$. $r = 0$, check $0 - 20 = -20$. $b = -11$, check $-11 + 13 = 2$. $m = 4$, check $4 - 6 = -2$.

Page 10, write your own. Sample: $x - 5 = -13$, solved by adding 5 to both sides, giving $x = -8$.

Page 10, apply it. $s - 17 = 28$; starting amount $45; check $45 - 17 = 28$.

Page 10, find the error. Devin added 9 instead of subtracting it. The solution is $x = -5$. A check catches it, since $13 + 9 = 22$, not $4$.

Page 12, coefficient table.

Page 13, solve and check. $m = -5$, check $12(-5) = -60$. $v = -90$, check $\tfrac{1}{10}(-90) = -9$. $x = 7$, check $-4(7) = -28$. $w = -12$, check $\tfrac{1}{2}(-12) = -6$.

Page 13, write your own. Sample: $\tfrac{1}{6}x = 5$, giving $x = 30$; check $\tfrac{1}{6}(30) = 5$.

Page 13, apply it. $12t = 84$; 7 tables; check $12(7) = 84$.

Page 13, find the error. Nina divided when she should have multiplied. The coefficient $\tfrac{1}{4}$ means $x$ was split into 4 parts, so multiply both sides by 4. The solution is $x = 32$.

Page 15, five steps. Read it all → name the variable → find the action → find the result → write and check the equation.

Page 15, action table. Gained, earned, rose → addition, sample $x + 8 = 20$. Spent, lost, fell → subtraction, sample $x - 8 = 20$. Equal groups, each, times as many → multiplication, sample $8x = 40$. Shared equally, one third of → division or a unit fraction, sample $\tfrac{1}{3}x = 12$.

Page 15, write the equation. $n + 6 = 14$, $n = 8$. $p - 8 = 24$, $p = 32$. $\tfrac{1}{4}g = 15$, $g = 60$. $\tfrac{1}{2}n = 11$, $n = 22$. $n - 17 = 5$, $n = 22$. $7n = 63$, $n = 9$. $\tfrac{1}{5}n = 12$, $n = 60$. $n + 20 = 4$, $n = -16$.

Page 16, real situations. 1. $d - 30 = -75$; $-45$ m. 2. $8p = 96$; 12 pizzas; "The school ordered 12 pizzas." 3. $t - 14 = -6$; 8°F. 4. $\tfrac{1}{6}L = 7$; 42 ft.

Page 16, find the error. Giving marbles away removes them, so the equation must subtract. The correct equation is $m - 5 = 14$, giving $m = 19$.

Page 18, situations. Samples: 1. Leo had some cards, bought 3 more, and now has 12. $x = 9$. 2. Five equal teams have 45 players in all. $n = 9$. 3. A jar lost 7 marbles and now holds 2. $x = 9$. 4. Half a bag of flour weighs 9 ounces. $m = 18$.

Page 19, situations. Samples: 5. A theater added 11 seats and now has 30. $x = 19$. 6. Nine boxes hold 72 crayons in all. $k = 8$. 7. A board cut into 6 equal pieces gives pieces 5 feet long. $t = 30$. 8. A store sold 12 shirts and has 8 left. $n = 20$. 9. The temperature rose 15 degrees to reach 4°F. $w = -11$°F. 10. A class splits into 8 equal groups of 3 students. $c = 24$.

Page 19, find the error. Her story adds an unknown to 4, which is $4 + x = 20$. The equation $4x = 20$ describes 4 equal groups. Either change the equation to $4 + x = 20$ ($x = 16$) or change the story to four friends who each earned the same amount, $20 in total ($x = 5$).

Pages 21 and 22. These repeat textbook Chapter 11 Review items 1–24; use the Chapter 11 Review key above.