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Virginia SOL Mathematics Textbook

Appendix A — Answer Key, Chapter 3: Multiplying and Dividing Integers; Absolute Value Expressions

SOL 6.CE.2 a, b, c, d · Covers textbook Chapter 3 and the companion workbook. Item numbers match the textbook; workbook items that repeat textbook problems share those answers, and every workbook-only item is keyed in the final section. Reasoning answers show an acceptable response, not the only wording.

Lesson 3.1 — Modeling Integer Multiplication and Division

Guided practice

  1. Three groups of 5 red chips: 3×(−5)=−153 \times (-5) = -15
  2. Remove 4 groups of 2 yellow chips from a pile worth zero: −4×2=−8-4 \times 2 = -8
  3. Remove 2 groups of 6 red chips from a pile worth zero: −2×(−6)=12-2 \times (-6) = 12
  4. Share 18 red chips into 6 equal groups, 3 red chips each: −18÷6=−3-18 \div 6 = -3
  5. Count how many groups of 5 red chips fit in 20 red chips: −20÷(−5)=4-20 \div (-5) = 4

Independent practice

  1. a) −20-20 b) −18-18 c) 2121 d) 00
  2. a) −3-3 b) 55 c) −4-4
  3. 4×(−3)=−124 \times (-3) = -12
  4. −6×4=−24-6 \times 4 = -24; 4×(−6)=−244 \times (-6) = -24; −24÷4=−6-24 \div 4 = -6; −24÷(−6)=4-24 \div (-6) = 4
  5. (−2)+(−2)+(−2)+(−2)+(−2)=−10(-2) + (-2) + (-2) + (-2) + (-2) = -10
  6. 6×(−3)=−186 \times (-3) = -18. The water level is 18 feet lower than when the drought began.
  7. Reading down the pattern, each product increases by 4: −8-8, −4-4, 00. Continuing the same steady increase gives −4×(−1)=4-4 \times (-1) = 4, −4×(−2)=8-4 \times (-2) = 8, and so on, so −4×(−5)=20-4 \times (-5) = 20, a positive number. A pattern that increased for every other line would have no reason to reverse itself here.

Exit ticket 3.1

  1. −12-12
  2. 1515
  3. −3-3
  4. Red chips carry negative value, so taking them away leaves the pile worth more than before. Because the pile can start as zero pairs, you can always remove as many red chips as a problem asks for, and each one removed raises the value by 1.

Lesson 3.2 — Multiplying and Dividing Integers

Guided practice

  1. −36-36
  2. 5656
  3. −8-8
  4. 55
  5. −3×(−2)=6-3 \times (-2) = 6; then 6×(−5)=−306 \times (-5) = -30.

Independent practice

  1. a) −66-66 b) 5252 c) −120-120 d) 2525
  2. a) −12-12 b) 99 c) −4-4
  3. (−4)×(−3)=12(-4) \times (-3) = 12; then 12×(−2)=−2412 \times (-2) = -24.
  4. a) −9-9 b) 3030
  5. Positive, because there are four negative factors and an even count of negatives gives a positive product. The product is 120120.
  6. 7×(−8)=−567 \times (-8) = -56 feet, a descent of 56 feet. To reach −96-96 feet: −96÷(−8)=12-96 \div (-8) = 12 minutes.
  7. Because 9×(−4)=−369 \times (-4) = -36. Division asks what number multiplied by the divisor gives the dividend, and only a positive 99 works here, since a negative times a negative would give a positive product instead of −36-36.

Exit ticket 3.2

  1. −72-72
  2. 7272
  3. 1515
  4. If −24÷(−6)=q-24 \div (-6) = q, then q×(−6)=−24q \times (-6) = -24. A negative value of qq would make that product positive, so qq must be positive; here q=4q = 4.

Lesson 3.3 — Simplifying Expressions with Absolute Value Bars

Guided practice

  1. 44
  2. −4-4
  3. 88
  4. −10-10
  5. 1818

Independent practice

  1. a) 55 b) 55 c) −5-5 d) −5-5
  2. a) 55 b) −5-5 c) 1313 d) −13-13
  3. a) 2020 b) −6-6
  4. 2−11=−92 - 11 = -9, so ∣2−11∣=9|2 - 11| = 9 and −∣2−11∣=−9-|2 - 11| = -9. Graph a solid dot at −9-9, nine units to the left of zero.
  5. ∣−3−4∣=∣−7∣=7|-3 - 4| = |-7| = 7 and −∣3+4∣=−∣7∣=−7-|3 + 4| = -|7| = -7. Since 7>−77 > -7, the expression ∣−3−4∣|-3 - 4| is greater. The bars produce a distance in both problems; the second expression then makes that distance negative.
  6. Size of the change: ∣45−52∣=∣−7∣=7|45 - 52| = |-7| = 7, a change of $7. Signed change: 45−52=−745 - 52 = -7, a drop of $7.
  7. Yes. With a=5a = 5 and b=8b = 8: ∣5−8∣=∣−3∣=3|5 - 8| = |-3| = 3 and ∣8−5∣=∣3∣=3|8 - 5| = |3| = 3. Both expressions measure the distance between the two numbers, and distance does not depend on which end you start from.

Exit ticket 3.3

  1. 55
  2. −5-5
  3. −7-7
  4. Graph a solid dot at −7-7, seven units to the left of zero. Inside the bars, −9+2=−7-9 + 2 = -7; the bars report the distance 77; the negative sign outside is applied to that finished distance, placing the result to the left of zero. The bars only group what is inside them, so a sign outside is never absorbed.

Lesson 3.4 — One- and Two-Step Problems in Context

Guided practice

  1. 4×(−5)=−204 \times (-5) = -20; then −3+(−20)=−23-3 + (-20) = -23°C.
  2. 7×(−12)=−847 \times (-12) = -84, a change of −$84-\$84.
  3. 240÷30=8240 \div 30 = 8 minutes.
  4. −72÷6=−12-72 \div 6 = -12 points per round.
  5. Estimate: −20×30=−600-20 \times 30 = -600. Exact: −19×31=−589-19 \times 31 = -589.

Independent practice

  1. 9×(−6)=−549 \times (-6) = -54; then −15+(−54)=−69-15 + (-54) = -69 feet.
  2. −150÷5=−30-150 \div 5 = -30, so each share is −$30-\$30.
  3. 6×3=186 \times 3 = 18; then −8+18=10-8 + 18 = 10°F.
  4. Estimate: −300÷4=−75-300 \div 4 = -75. Exact: −296÷4=−74-296 \div 4 = -74.
  5. 5×(−4)+30=−20+30=105 \times (-4) + 30 = -20 + 30 = 10 points.
  6. 12+6×(−3)=12+(−18)=−612 + 6 \times (-3) = 12 + (-18) = -6. The elevator ends on level −6-6, six levels below the ground floor.
  7. The student who got −14-14 is correct. Multiplication comes before addition, so −6×4=−24-6 \times 4 = -24 and then −24+10=−14-24 + 10 = -14. The other student added first, computing −6×(4+10)=−6×14=−84-6 \times (4 + 10) = -6 \times 14 = -84.

Exit ticket 3.4

  1. 3×(−5)=−153 \times (-5) = -15; then 2+(−15)=−132 + (-15) = -13°F, or 13 degrees below zero.
  2. −84÷7=−12-84 \div 7 = -12 points per round.
  3. Estimate: −50×20=−1000-50 \times 20 = -1000. Exact: −51×19=−969-51 \times 19 = -969.
  4. Multiply when the same change repeats a known number of times. Divide when a total change is split into equal parts, or when you know a total and a rate and need how many parts there are.

Chapter 3 Review

Part A — Modeling multiplication and division (6.CE.2a)

  1. a) −24-24 b) −15-15 c) 1212
  2. a) −5-5 b) 33
  3. 6×(−2)=−126 \times (-2) = -12
  4. −5×2=−10-5 \times 2 = -10, −5×1=−5-5 \times 1 = -5, −5×0=0-5 \times 0 = 0. Each product increases by 5 as the second factor drops by 1. Keeping that step going gives −5×(−1)=5-5 \times (-1) = 5, a positive product, and every further negative second factor keeps the products positive and growing.

Part B — Multiplying and dividing two integers (6.CE.2b)

  1. a) −70-70 b) 9999 c) −112-112 d) 6060
  2. a) −12-12 b) 88 c) −13-13
  3. (−2)×(−5)=10(-2) \times (-5) = 10; then 10×(−3)=−3010 \times (-3) = -30.
  4. a) 1212 b) −63-63

Part C — Absolute value expressions (6.CE.2c)

  1. a) 88 b) −8-8
  2. a) 1414 b) −14-14
  3. a) 2727 b) −8-8
  4. −∣3−10∣=−∣−7∣=−7-|3 - 10| = -|-7| = -7. Graph a solid dot at −7-7, seven units left of zero.

Part D — Problems in context (6.CE.2d)

  1. 6×(−7)=−426 \times (-7) = -42; then −20+(−42)=−62-20 + (-42) = -62 feet.
  2. −144÷8=−18-144 \div 8 = -18, so each share is −$18-\$18.
  3. Estimate: −40×30=−1200-40 \times 30 = -1200. Exact: −42×29=−1218-42 \times 29 = -1218.
  4. 20÷4=520 \div 4 = 5 hours. Division, because the total rise needed is 0−(−20)=200 - (-20) = 20 degrees and it is being covered in equal 4-degree steps; dividing tells how many steps that takes.

Workbook-only items

Page 2, product table.

Product Model in words Value
3×(−5)3 \times (-5) 3 groups of 5 red chips −15-15
−4×2-4 \times 2 remove 4 groups of 2 yellow chips −8-8
−2×(−6)-2 \times (-6) remove 2 groups of 6 red chips 1212
5×(−4)5 \times (-4) 5 groups of 4 red chips −20-20
−6×3-6 \times 3 remove 6 groups of 3 yellow chips −18-18
−3×(−7)-3 \times (-7) remove 3 groups of 7 red chips 2121

Page 2, repeated addition. (−2)+(−2)+(−2)+(−2)+(−2)=−10(-2) + (-2) + (-2) + (-2) + (-2) = -10

Page 2, pattern. −8-8, −4-4, 00, 44, 88. Each product goes up by 4 as you read down.

Page 3, quotients. a) −3-3 (shared) b) 44 (measured) c) −3-3 (shared) d) 55 (measured) e) −4-4 (shared) f) −3-3 (shared). Either strategy may be named for any item as long as the reasoning fits.

Page 3, fact family. −6×4=−24-6 \times 4 = -24; 4×(−6)=−244 \times (-6) = -24; −24÷4=−6-24 \div 4 = -6; −24÷(−6)=4-24 \div (-6) = 4

Page 3, apply it. 6×(−3)=−186 \times (-3) = -18; the level is 18 feet lower than at the start.

Page 5, fill in the rules. Same signs → positive. Different signs → negative. These rules work for multiplication and for division.

Page 5, find each answer. a) −36-36 b) 5656 c) −8-8 d) 55 e) −66-66 f) 5252 g) −120-120 h) 2525 i) −12-12 j) 99 k) −4-4 l) 6060

Page 6, predict the sign.

Expression Negative factors Sign Product
−3×(−2)×(−5)-3 \times (-2) \times (-5) 33 negative −30-30
(−4)×(−3)×(−2)(-4) \times (-3) \times (-2) 33 negative −24-24
−2×3×(−4)-2 \times 3 \times (-4) 22 positive 2424
(−2)×(−3)×(−4)×(−5)(-2) \times (-3) \times (-4) \times (-5) 44 positive 120120

Page 6, sign only. positive; negative; positive; negative

Page 6, estimate table.

Problem Estimate Exact
−38×21-38 \times 21 −40×20=−800-40 \times 20 = -800 −798-798
−296÷4-296 \div 4 −300÷4=−75-300 \div 4 = -75 −74-74
−19×31-19 \times 31 −20×30=−600-20 \times 30 = -600 −589-589

Page 6, missing number. −9-9; 3030; 1212

Page 8, three-step table.

Expression Inside the bars Absolute value Final value
∣3−7∣\lvert 3 - 7 \rvert −4-4 44 44
−∣3−7∣-\lvert 3 - 7 \rvert −4-4 44 −4-4
∣−2+10∣\lvert -2 + 10 \rvert 88 88 88
−∣−5−5∣-\lvert -5 - 5 \rvert −10-10 1010 −10-10
∣6×(−3)∣\lvert 6 \times (-3) \rvert −18-18 1818 1818
−∣36÷(−6)∣-\lvert 36 \div (-6) \rvert −6-6 66 −6-6

Page 8, compare. 55; 55; −5-5; −5-5. The first two match because 9−49 - 4 and 4−94 - 9 are opposites and opposites have equal absolute value.

Page 9, simplify and plot. 1. −9-9 2. 55 3. −5-5 4. 1313 5. −13-13 6. 2020. Plot solid dots at −13-13, −9-9, −5-5, 55, 1313, and 2020.

Page 9, which is greater. ∣−3−4∣|-3 - 4| is greater. The values are 77 and −7-7.

Page 9, apply it. Size of the change: ∣45−52∣=7|45 - 52| = 7, so $7. Signed change: 45−52=−745 - 52 = -7, a drop of $7.

Page 10, exit ticket 3.3. Same as the textbook exit ticket 3.3 above: 1. 55 2. −5-5 3. −7-7 4. Solid dot at −7-7; the sign outside the bars is applied after the distance is found.

Page 11, choosing the operation table.

Situation Operation(s) Number sentence Answer
Temperature −3-3°C falls 5 degrees per hour for 4 hours multiply, then add −3+4×(−5)-3 + 4 \times (-5) −23-23°C
An account loses $12 each week for 7 weeks multiply 7×(−12)7 \times (-12) −$84-\$84
A balloon 240 ft up descends 30 ft per minute until it lands divide 240÷30240 \div 30 88 minutes
72 points lost, spread equally over 6 rounds divide −72÷6-72 \div 6 −12-12 per round
A diver at −15-15 ft descends 6 ft per minute for 9 minutes multiply, then add −15+9×(−6)-15 + 9 \times (-6) −69-69 ft

Page 12, two-step problems.

  1. −150÷5=−30-150 \div 5 = -30, so each share is −$30-\$30.
  2. 6×3=186 \times 3 = 18; −8+18=10-8 + 18 = 10°F.
  3. 5×(−4)+30=105 \times (-4) + 30 = 10 points.
  4. 12+6×(−3)=−612 + 6 \times (-3) = -6, six levels below the ground floor.
  5. Estimate −300÷4=−75-300 \div 4 = -75; exact −74-74.

Page 12, challenge. The student who got −14-14 is correct. The other multiplied after adding, computing −6×(4+10)=−84-6 \times (4 + 10) = -84 instead of following order of operations.

Page 13, exit ticket 3.4. Same as the textbook exit ticket 3.4 above: 1. −13-13°F 2. −12-12 points per round 3. Estimate −1000-1000, exact −969-969 4. Multiply for a repeated change; divide to split a total into equal parts or to find how many equal steps a total takes.

Pages 14–15, Chapter 3 review. Same items and answers as the textbook Chapter 3 Review above.