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7th Grade Math Quiz

7th Grade Math Quiz: Subtract Using Additive Inverse

Practice Subtract Using Additive Inverse in 7th Grade Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

A football team gains 121212 yards, then loses 181818 yards, then gains 777 yards. Using additive inverse for each loss, what expression shows their total yardage change?

Select an answer to continue

What this quiz covers

This quiz focuses on Subtract Using Additive Inverse, giving you a quick way to practice the rules, question types, and explanations that matter most for 7th Grade Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A football team gains 121212 yards, then loses 181818 yards, then gains 777 yards. Using additive inverse for each loss, what expression shows their total yardage change?

  1. 12+18+7=3712 + 18 + 7 = 3712+18+7=37 yards gained total
  2. 12+(−18)+7=112 + (-18) + 7 = 112+(−18)+7=1 yard gained total (correct answer)
  3. 12−18−7=−1312 - 18 - 7 = -1312−18−7=−13 yards lost total
  4. (−12)+(−18)+(−7)=−37(-12) + (-18) + (-7) = -37(−12)+(−18)+(−7)=−37 yards lost total

Explanation: Using additive inverse, gains are positive and losses are negative. The sequence is: +12 yards (gain), -18 yards (loss), +7 yards (gain). So the expression is 12 + (-18) + 7 = 1 yard net gain. Choice B correctly applies additive inverse to the loss. Choice A treats the loss as a gain. Choice C uses subtraction notation instead of additive inverse. Choice D makes all movements negative.

Question 2

On a number line, Maria moves from position 34\frac{3}{4}43​ to position −58-\frac{5}{8}−85​. Using subtraction as adding the additive inverse, what is her displacement?

  1. 34+58=68+58=118\frac{3}{4} + \frac{5}{8} = \frac{6}{8} + \frac{5}{8} = \frac{11}{8}43​+85​=86​+85​=811​ units
  2. −58+34=−58+68=18-\frac{5}{8} + \frac{3}{4} = -\frac{5}{8} + \frac{6}{8} = \frac{1}{8}−85​+43​=−85​+86​=81​ units
  3. 34−58=68−58=18\frac{3}{4} - \frac{5}{8} = \frac{6}{8} - \frac{5}{8} = \frac{1}{8}43​−85​=86​−85​=81​ units
  4. −58−34=−58+(−68)=−118-\frac{5}{8} - \frac{3}{4} = -\frac{5}{8} + (-\frac{6}{8}) = -\frac{11}{8}−85​−43​=−85​+(−86​)=−811​ units (correct answer)

Explanation: When you encounter displacement problems on a number line, you need to calculate the change in position using the formula: final position minus initial position. This tells you both the distance and direction of movement. Maria starts at 34\frac{3}{4}43​ and ends at −58-\frac{5}{8}−85​. Her displacement is: −58−34-\frac{5}{8} - \frac{3}{4}−85​−43​. To subtract fractions, you need a common denominator. Converting 34\frac{3}{4}43​ to eighths: 34=68\frac{3}{4} = \frac{6}{8}43​=86​. Now calculate: −58−68=−58+(−68)=−118-\frac{5}{8} - \frac{6}{8} = -\frac{5}{8} + (-\frac{6}{8}) = -\frac{11}{8}−85​−86​=−85​+(−86​)=−811​. The negative sign indicates Maria moved left on the number line, which makes sense since she went from a positive position to a negative one. Choice A incorrectly adds the absolute values of both positions, ignoring direction entirely. Choice B calculates −58+34-\frac{5}{8} + \frac{3}{4}−85​+43​, which reverses the displacement formula—this would be the calculation if Maria moved from −58-\frac{5}{8}−85​ to 34\frac{3}{4}43​ instead. Choice C uses 34−(−58)\frac{3}{4} - (-\frac{5}{8})43​−(−85​), which calculates the distance between the points but gets the direction wrong by subtracting in the wrong order. Remember: displacement equals final position minus initial position. The order matters because it determines whether your answer is positive (rightward movement) or negative (leftward movement). Always double-check that your answer's sign matches the actual direction of movement on the number line.

Question 3

The temperature difference between two cities is 23.8°F23.8°F23.8°F. If City AAA has a temperature of −12.3°F-12.3°F−12.3°F, and City BBB is warmer, what is the temperature in City BBB?

  1. City B is 11.5°F11.5°F11.5°F, since ∣11.5−(−12.3)∣=∣23.8∣=23.8°F|11.5 - (-12.3)| = |23.8| = 23.8°F∣11.5−(−12.3)∣=∣23.8∣=23.8°F
  2. City B is 36.1°F36.1°F36.1°F, since ∣−12.3+23.8∣=11.5°F|-12.3 + 23.8| = 11.5°F∣−12.3+23.8∣=11.5°F
  3. City B is 11.5°F11.5°F11.5°F, since −12.3+23.8=11.5°F-12.3 + 23.8 = 11.5°F−12.3+23.8=11.5°F (correct answer)
  4. City B is −36.1°F-36.1°F−36.1°F, since −12.3+(−23.8)=−36.1°F-12.3 + (-23.8) = -36.1°F−12.3+(−23.8)=−36.1°F

Explanation: If City B is warmer and the difference is 23.8°F, then City B = City A + 23.8 = -12.3 + 23.8 = 11.5°F. We can verify: |11.5 - (-12.3)| = |11.5 + 12.3| = |23.8| = 23.8°F. Choice C correctly calculates City B's temperature. Choice A gives the right answer but had an incorrect verification calculation. Choice B has calculation errors in the verification. Choice D makes City B colder instead of warmer.

Question 4

A submarine starts at a depth of −125-125−125 feet below sea level. It then rises 787878 feet. Using the concept that subtraction equals adding the additive inverse, which expression correctly models the submarine's final depth?

  1. −125+78=−47-125 + 78 = -47−125+78=−47 feet, representing 474747 feet below sea level (correct answer)
  2. −125−78=−203-125 - 78 = -203−125−78=−203 feet, representing 203203203 feet below sea level
  3. −125+(−78)=−203-125 + (-78) = -203−125+(−78)=−203 feet, representing 203203203 feet below sea level
  4. 125+(−78)=47125 + (-78) = 47125+(−78)=47 feet, representing 474747 feet above sea level

Explanation: The submarine starts at -125 feet and rises 78 feet. Rising means adding a positive value: -125 + 78 = -47 feet. The negative result indicates it's still below sea level. Choice A is correct. Choice B subtracts when it should add (rising means going up, adding positive). Choice C adds the additive inverse of 78 when it should just add 78. Choice D incorrectly starts with positive 125 instead of -125.

Question 5

Which equation correctly shows that subtracting a negative is the same as adding a positive?

  1. 5−(−2)=(−5)+2=−35-(-2)=(-5)+2=-35−(−2)=(−5)+2=−3
  2. 5−(−2)=5−2=35-(-2)=5-2=35−(−2)=5−2=3
  3. 5−(−2)=5+2=75-(-2)=5+2=75−(−2)=5+2=7 (correct answer)
  4. 5−(−2)=5+(−2)=35-(-2)=5+(-2)=35−(−2)=5+(−2)=3

Explanation: This question tests understanding that subtraction p - q equals adding the additive inverse p + (-q), and the distance between numbers as |p - q| (absolute value of the difference). Subtraction as addition: p - q = p + (-q) by definition (subtracting q means adding the opposite -q: 5 - 8 = 5 + (-8) = -3, or 7 - (-3) = 7 + 3 = 10 since subtracting a negative adds the positive). On the number line, p - q starts at p and moves |q| units left if q > 0 (subtracting a positive) or right if q < 0 (subtracting a negative equals adding a positive), while the distance between p and q is |p - q| = |q - p| (absolute value of the difference, always positive: |3 - 10| = |-7| = 7 units apart, or |10 - 3| = 7). For example, 15 - 20 can be rewritten as 15 + (-20) = -5 (like a temperature drop from 15°C to -5°C), or the distance from -4 to 3 is |3 - (-4)| = |3 + 4| = 7 units (or |-4 - 3| = |-7| = 7), and for money, 10 - 25 = 10 + (-25) = -15 (a debt of $15). The correct equation is 5 - (-2) = 5 + 2 = 7, as it shows subtracting a negative becomes adding a positive. A common error is subtracting negative wrong, like in choice A where 5 - (-2) = 5 + (-2) = 3 treats it as adding negative, or in choice B where it stays as 5 - 2 = 3 without converting. Using the additive inverse, rewrite every subtraction as addition (p - q → p + (-q), making all operations additions), and apply addition rules (p + (-q) follows number line interpretation: start at p, move |q| left).

Question 6

A student claims: “p−qp-qp−q is not the same as p+(−q)p+(-q)p+(−q).” Which choice correctly evaluates the claim using p=−2p=-2p=−2 and q=5q=5q=5?

  1. The claim is true because −2−5=−3-2-5=-3−2−5=−3 but −2+(−5)=−7-2+(-5)=-7−2+(−5)=−7.
  2. The claim is false because −2−5=−7-2-5=-7−2−5=−7 and −2+(−5)=−7-2+(-5)=-7−2+(−5)=−7. (correct answer)
  3. The claim is false because −2−5=3-2-5=3−2−5=3 and −2+(−5)=3-2+(-5)=3−2+(−5)=3.
  4. The claim is true because −2−5=−7-2-5=-7−2−5=−7 but −2+(−5)=7-2+(-5)=7−2+(−5)=7.

Explanation: This question tests understanding that subtraction p-q equals adding the additive inverse p+(-q), and the distance between numbers as |p-q| (absolute value of the difference). Subtraction as addition: p-q = p+(-q) by definition (subtracting q means adding the opposite -q: 5-8=5+(-8)=-3, or 7-(-3)=7+3=10 since subtracting a negative adds the positive). On the number line, p-q starts at p and moves distance |q| left if q>0 (subtracting positive), or right if q<0 (subtracting negative = adding positive); distance between p and q is |p-q|=|q-p| (absolute value of the difference, always positive: |3-10|=|-7|=7 units apart, or |10-3|=7). For example, 15-20 can be rewritten as 15+(-20)=-5 (temperature drops from 15°C to -5°C), or distance from -4 to 3: |3-(-4)|=|3+4|=7 units (or |-4-3|=|-7|=7), or 10-25 for money: 10+(-25)=-15 (debt of $15). The claim is false because -2-5=-7 and -2+(-5)=-7 (same), as in choice C, confirming equivalence. A common error is claiming p-q≠p+(-q) with wrong calculations, like choice A with -2+(-5)=7 (sign error), or choice B with -2-5=-3 (arithmetic wrong), or choice D with both as 3 (multiple errors). Using additive inverse: rewrite p-q→p+(-q), apply addition (start p, move left); mistakes like denying equivalence (p-q≠p+(-q) claimed), or subtracting negative wrong (5-(-2)=3 not 7).

Question 7

Two students argue about distance: Student 1 says the distance from 333 to 101010 is ∣10−3∣=7|10-3|=7∣10−3∣=7. Student 2 says the distance from 101010 to 333 is ∣3−10∣=−7|3-10|=-7∣3−10∣=−7. Which statement is correct?​

  1. Both are incorrect because distance depends on the order of subtraction.
  2. Student 2 is correct because 3−10=−73-10=-73−10=−7.
  3. Student 1 is correct, and Student 2 should have ∣3−10∣=∣−7∣=7|3-10|=|-7|=7∣3−10∣=∣−7∣=7. (correct answer)
  4. Both are correct because distance can be negative.

Explanation: This question tests understanding that subtraction p - q equals adding the additive inverse p + (-q), and the distance between numbers as |p - q| (absolute value of the difference). Subtraction as addition: p - q = p + (-q) by definition (subtracting q means adding the opposite -q: 5 - 8 = 5 + (-8) = -3, or 7 - (-3) = 7 + 3 = 10 since subtracting a negative adds the positive). On the number line, p - q starts at p and moves |q| units left if q > 0 (subtracting a positive) or right if q < 0 (subtracting a negative equals adding a positive), while the distance between p and q is |p - q| = |q - p| (absolute value of the difference, always positive: |3 - 10| = |-7| = 7 units apart, or |10 - 3| = 7). For example, 15 - 20 can be rewritten as 15 + (-20) = -5 (like a temperature drop from 15°C to -5°C), or the distance from -4 to 3 is |3 - (-4)| = |3 + 4| = 7 units (or |-4 - 3| = |-7| = 7), and for money, 10 - 25 = 10 + (-25) = -15 (a debt of $15). The correct statement is that Student 1 is right, and Student 2 should have |3 - 10| = |-7| = 7, as distance uses absolute value and is always positive regardless of order. A common error is claiming distance can be negative, like in choice A where Student 2 is correct with -7, or in choice D where both are correct with negative distance. Distance: between any two numbers p and q, calculate p - q and take the absolute value |p - q| (removes the sign, gives positive distance: 3 - 10 = -7, |-7| = 7 units), or reverse |q - p| (order doesn't matter for distance, both give the same).

Question 8

A science lab starts at 15∘C15^\circ\text{C}15∘C and then the temperature drops by 20∘C20^\circ\text{C}20∘C. Which choice correctly models the situation, rewrites subtraction as addition, and gives the final temperature?

  1. 15−20=15−(−20)=−5∘C15-20=15-(-20)=-5^\circ\text{C}15−20=15−(−20)=−5∘C
  2. 15−20=15+(−20)=−5∘C15-20=15+(-20)=-5^\circ\text{C}15−20=15+(−20)=−5∘C (correct answer)
  3. 15−20=15+(−20)=5∘C15-20=15+(-20)=5^\circ\text{C}15−20=15+(−20)=5∘C
  4. 15−20=15+20=35∘C15-20=15+20=35^\circ\text{C}15−20=15+20=35∘C

Explanation: This question tests understanding that subtraction p-q equals adding the additive inverse p+(-q), and the distance between numbers as |p-q| (absolute value of the difference). Subtraction as addition: p-q = p+(-q) by definition (subtracting q means adding the opposite -q: 5-8=5+(-8)=-3, or 7-(-3)=7+3=10 since subtracting a negative adds the positive). On the number line, p-q starts at p and moves distance |q| left if q>0 (subtracting positive), or right if q<0 (subtracting negative = adding positive); distance between p and q is |p-q|=|q-p| (absolute value of the difference, always positive: |3-10|=|-7|=7 units apart, or |10-3|=7). For example, 15-20 can be rewritten as 15+(-20)=-5 (temperature drops from 15°C to -5°C), or distance from -4 to 3: |3-(-4)|=|3+4|=7 units (or |-4-3|=|-7|=7), or 10-25 for money: 10+(-25)=-15 (debt of 15).Thecorrectmodelingfor15−20is15+(−20)=−5°C,asinchoiceB,rewritingsubtractionasadditionforthetemperaturedrop.Acommonerrorisarithmeticwrong,likechoiceAas15+20=35°C(addinginsteadofsubtracting),orchoiceCwithsumas5°C(absolutevalueerror),orchoiceDsubtracting−20whichadds20incorrectly.Contexts:temperature15°drops20°(15−20=−5°,belowzero),money15). The correct modeling for 15-20 is 15+(-20)=-5°C, as in choice B, rewriting subtraction as addition for the temperature drop. A common error is arithmetic wrong, like choice A as 15+20=35°C (adding instead of subtracting), or choice C with sum as 5°C (absolute value error), or choice D subtracting -20 which adds 20 incorrectly. Contexts: temperature 15° drops 20° (15-20=-5°, below zero), money 15).Thecorrectmodelingfor15−20is15+(−20)=−5°C,asinchoiceB,rewritingsubtractionasadditionforthetemperaturedrop.Acommonerrorisarithmeticwrong,likechoiceAas15+20=35°C(addinginsteadofsubtracting),orchoiceCwithsumas5°C(absolutevalueerror),orchoiceDsubtracting−20whichadds20incorrectly.Contexts:temperature15°drops20°(15−20=−5°,belowzero),money10 spend $25 (10-25=-15, debt), elevation 50 m descend 80 m (50-80=-30 m, below sea level 30 m); using additive inverse rewrites to addition for easier calculation.

Question 9

An underwater robot is at an elevation of −32-\tfrac{3}{2}−23​ meters (below sea level). It then moves down another 52\tfrac{5}{2}25​ meters. Which expression gives its new elevation?

  1. −32−52=−32+52=22=1-\tfrac{3}{2}-\tfrac{5}{2}=-\tfrac{3}{2}+\tfrac{5}{2}=\tfrac{2}{2}=1−23​−25​=−23​+25​=22​=1
  2. −32−52=82=4-\tfrac{3}{2}-\tfrac{5}{2}=\tfrac{8}{2}=4−23​−25​=28​=4
  3. −32−52=−82=−4-\tfrac{3}{2}-\tfrac{5}{2}=-\tfrac{8}{2}=-4−23​−25​=−28​=−4 (correct answer)
  4. −32−52=−32+52=4-\tfrac{3}{2}-\tfrac{5}{2}=-\tfrac{3}{2}+\tfrac{5}{2}=4−23​−25​=−23​+25​=4

Explanation: This question tests understanding that subtraction p - q equals adding the additive inverse p + (-q), and the distance between numbers as |p - q| (absolute value of the difference). Subtraction as addition: p - q = p + (-q) by definition (subtracting q means adding the opposite -q: 5 - 8 = 5 + (-8) = -3, or 7 - (-3) = 7 + 3 = 10 since subtracting a negative adds the positive). On the number line, p - q starts at p and moves |q| units left if q > 0 (subtracting a positive) or right if q < 0 (subtracting a negative equals adding a positive), while the distance between p and q is |p - q| = |q - p| (absolute value of the difference, always positive: |3 - 10| = |-7| = 7 units apart, or |10 - 3| = 7). For example, 15 - 20 can be rewritten as 15 + (-20) = -5 (like a temperature drop from 15°C to -5°C), or the distance from -4 to 3 is |3 - (-4)| = |3 + 4| = 7 units (or |-4 - 3| = |-7| = 7), and for money, 10 - 25 = 10 + (-25) = -15 (a debt of $15). The correct expression is -3/2 - 5/2 = -8/2 = -4, as it computes the new elevation after descending. A common error is rewriting incorrectly, like in choice A where -3/2 - 5/2 = -3/2 + 5/2 = 1 adds instead of subtracting, or in choice C where it becomes positive 4 without proper signs. Mistakes: not rewriting as addition (missing connection p - q = p + (-q)), subtracting negative as subtraction (5 - (-2) staying as subtract when should become 5 + 2 = 7), distance without absolute value (negative distance).

Question 10

Calculate: -9-(-4). Use the idea that subtracting is adding the inverse.

  1. -13
  2. -5 (correct answer)
  3. 13
  4. 5

Explanation: Subtracting -4 is the same as adding its additive inverse, +4: -9 - (-4) = -9 + 4 = -5, matching choice B. Choice A (-13) comes from adding instead of subtracting the inverse, effectively computing -9-4. Choices C (13) and D (5) both ignore the negative sign on -9, treating it as if it were positive.

Question 11

The distance between two numbers on a number line is the absolute value of their difference. What is the distance between −4-4−4 and 333?​

  1. ∣−4−3∣=−7|-4-3|=-7∣−4−3∣=−7, so the distance is −7-7−7
  2. ∣3−(−4)∣=∣7∣=7|3-(-4)|=|7|=7∣3−(−4)∣=∣7∣=7 (correct answer)
  3. 3−(−4)=−13-(-4)= -13−(−4)=−1, so the distance is −1-1−1
  4. ∣−4−3∣=∣−7∣=−7|-4-3|=|-7|=-7∣−4−3∣=∣−7∣=−7

Explanation: This question tests understanding that subtraction p - q equals adding the additive inverse p + (-q), and the distance between numbers as |p - q| (absolute value of the difference). Subtraction as addition: p - q = p + (-q) by definition (subtracting q means adding the opposite -q: 5 - 8 = 5 + (-8) = -3, or 7 - (-3) = 7 + 3 = 10 since subtracting a negative adds the positive). On the number line, p - q starts at p and moves |q| units left if q > 0 (subtracting a positive) or right if q < 0 (subtracting a negative equals adding a positive), while the distance between p and q is |p - q| = |q - p| (absolute value of the difference, always positive: |3 - 10| = |-7| = 7 units apart, or |10 - 3| = 7). For example, 15 - 20 can be rewritten as 15 + (-20) = -5 (like a temperature drop from 15°C to -5°C), or the distance from -4 to 3 is |3 - (-4)| = |3 + 4| = 7 units (or |-4 - 3| = |-7| = 7), and for money, 10 - 25 = 10 + (-25) = -15 (a debt of $15). The correct distance is |3 - (-4)| = |7| = 7, as it applies the absolute value to the difference properly. A common error is treating distance as negative without absolute value, like in choice A where |-4 - 3| = |-7| = -7 claims a negative distance, or in choice C where 3 - (-4) = -1 miscalculates the subtraction. Distance: between any two numbers p and q, calculate p - q and take the absolute value |p - q| (removes the sign, gives positive distance: 3 - 10 = -7, |-7| = 7 units), or reverse |q - p| (order doesn't matter for distance, both give the same).

Question 12

Rewrite the subtraction expression as an addition expression using the additive inverse: 5−85-85−8.

  1. (−5)+8(-5)+8(−5)+8
  2. −5−8-5-8−5−8
  3. 5+85+85+8
  4. 5+(−8)5+(-8)5+(−8) (correct answer)

Explanation: Rewriting a subtraction expression as addition means adding the additive inverse of the number being subtracted, so 5 minus 8 becomes 5 plus negative 8, matching choice D. Choice A, negative 5 plus 8, incorrectly flips the sign of the first number instead of the second. Choice B, negative 5 minus 8, is still written as a subtraction and does not represent the additive inverse rule being tested. Choice C, 5 plus 8, adds positive 8 instead of its opposite, negative 8, which changes the value of the expression entirely. The additive inverse rule always keeps the first number the same and changes only the sign of the number being subtracted.

Question 13

Two students argue about distance: Student 1 says the distance from 333 to 101010 is ∣10−3∣=7|10-3|=7∣10−3∣=7. Student 2 says the distance from 101010 to 333 is ∣3−10∣=−7|3-10|=-7∣3−10∣=−7. Which statement is correct?

  1. Both are incorrect because distance depends on the order of subtraction.
  2. Both are correct because distance can be negative.
  3. Student 1 is correct, and Student 2 should have ∣3−10∣=∣−7∣=7|3-10|=|-7|=7∣3−10∣=∣−7∣=7. (correct answer)
  4. Student 2 is correct because 3−10=−73-10=-73−10=−7.

Explanation: This question tests understanding that subtraction p - q equals adding the additive inverse p + (-q), and the distance between numbers as |p - q| (absolute value of the difference). Subtraction as addition: p - q = p + (-q) by definition (subtracting q means adding the opposite -q: 5 - 8 = 5 + (-8) = -3, or 7 - (-3) = 7 + 3 = 10 since subtracting a negative adds the positive). On the number line, p - q starts at p and moves |q| units left if q > 0 (subtracting a positive) or right if q < 0 (subtracting a negative equals adding a positive), while the distance between p and q is |p - q| = |q - p| (absolute value of the difference, always positive: |3 - 10| = |-7| = 7 units apart, or |10 - 3| = 7). For example, 15 - 20 can be rewritten as 15 + (-20) = -5 (like a temperature drop from 15°C to -5°C), or the distance from -4 to 3 is |3 - (-4)| = |3 + 4| = 7 units (or |-4 - 3| = |-7| = 7), and for money, 10 - 25 = 10 + (-25) = -15 (a debt of $15). The correct statement is that Student 1 is right, and Student 2 should have |3 - 10| = |-7| = 7, as distance uses absolute value and is always positive regardless of order. A common error is claiming distance can be negative, like in choice A where Student 2 is correct with -7, or in choice D where both are correct with negative distance. Distance: between any two numbers p and q, calculate p - q and take the absolute value |p - q| (removes the sign, gives positive distance: 3 - 10 = -7, |-7| = 7 units), or reverse |q - p| (order doesn't matter for distance, both give the same).

Question 14

You have \10andspendand spendandspend$25.Thiscanbemodeledby. This can be modeled by .Thiscanbemodeledby10-25$. Which statement correctly rewrites the subtraction and interprets the result?

  1. 10−25=10+25=3510-25=10+25=3510−25=10+25=35, so you have \35$ left.
  2. 10−25=1510-25=1510−25=15, so you have \15$ left.
  3. 10−25=10+(−25)=−1510-25=10+(-25)=-1510−25=10+(−25)=−15, so you are \15$ in debt. (correct answer)
  4. 10−25=10+(−25)=1510-25=10+(-25)=1510−25=10+(−25)=15, so you are \15$ in debt.

Explanation: This question tests understanding that subtraction p - q equals adding the additive inverse p + (-q), and the distance between numbers as |p - q| (absolute value of the difference). Subtraction as addition: p - q = p + (-q) by definition (subtracting q means adding the opposite -q: 5 - 8 = 5 + (-8) = -3, or 7 - (-3) = 7 + 3 = 10 since subtracting a negative adds the positive). On the number line, p - q starts at p and moves |q| units left if q > 0 (subtracting a positive) or right if q < 0 (subtracting a negative equals adding a positive), while the distance between p and q is |p - q| = |q - p| (absolute value of the difference, always positive: |3 - 10| = |-7| = 7 units apart, or |10 - 3| = 7). For example, 15 - 20 can be rewritten as 15 + (-20) = -5 (like a temperature drop from 15°C to -5°C), or the distance from -4 to 3 is |3 - (-4)| = |3 + 4| = 7 units (or |-4 - 3| = |-7| = 7), and for money, 10 - 25 = 10 + (-25) = -15 (a debt of 15).Thecorrectstatementis10−25=10+(−25)=−15,soyouare15). The correct statement is 10 - 25 = 10 + (-25) = -15, so you are 15).Thecorrectstatementis10−25=10+(−25)=−15,soyouare15 in debt, as it rewrites properly and interprets the financial context. A common error is context misapplied, like in choice B where 10 - 25 = 10 + 25 = 35 shows positive balance instead of debt, or arithmetic wrong as in choice D where 10 + (-25) = 15 ignores the signs. Mistakes: not rewriting as addition (missing connection p - q = p + (-q)), subtracting negative as subtraction (5 - (-2) staying as subtract when should become 5 + 2 = 7), distance without absolute value (negative distance).

Question 15

Find the distance between the rational numbers 12\frac{1}{2}21​ and −32-\frac{3}{2}−23​. (Distance is the absolute value of their difference.)

  1. ∣12−(−32)∣=∣42∣=2\left|\frac{1}{2}-\left(-\frac{3}{2}\right)\right|=\left|\frac{4}{2}\right|=2​21​−(−23​)​=​24​​=2 (correct answer)
  2. ∣12−(−32)∣=∣−42∣=−2\left|\frac{1}{2}-\left(-\frac{3}{2}\right)\right|=\left|\frac{-4}{2}\right|=-2​21​−(−23​)​=​2−4​​=−2
  3. ∣12−(−32)∣=∣22∣=1\left|\frac{1}{2}-\left(-\frac{3}{2}\right)\right|=\left|\frac{2}{2}\right|=1​21​−(−23​)​=​22​​=1
  4. ∣12−(−32)∣=∣−22∣=−1\left|\frac{1}{2}-\left(-\frac{3}{2}\right)\right|=\left|\frac{-2}{2}\right|=-1​21​−(−23​)​=​2−2​​=−1

Explanation: This question tests understanding that subtraction p-q equals adding the additive inverse p+(-q), and the distance between numbers as |p-q| (absolute value of the difference). Subtraction as addition: p-q = p+(-q) by definition (subtracting q means adding the opposite -q: 5-8=5+(-8)=-3, or 7-(-3)=7+3=10 since subtracting a negative adds the positive). On the number line, p-q starts at p and moves distance |q| left if q>0 (subtracting positive), or right if q<0 (subtracting negative = adding positive); distance between p and q is |p-q|=|q-p| (absolute value of the difference, always positive: |3-10|=|-7|=7 units apart, or |10-3|=7). For example, 15-20 can be rewritten as 15+(-20)=-5 (temperature drops from 15°C to -5°C), or distance from -4 to 3: |3-(-4)|=|3+4|=7 units (or |-4-3|=|-7|=7), or 10-25 for money: 10+(-25)=-15 (debt of $15). The correct distance is |1/2 - (-3/2)| = |4/2| = 2, as in choice A, using absolute value for rational numbers. A common error is calculation mistake, like choice B with | -2/2 | = -1 (negative and wrong difference), or choice C as |2/2|=1 (incorrect sum), or choice D with | -4/2 | = -2 (wrong signs). Distance: between any two numbers p and q, calculate p-q, take absolute value |p-q| (removes sign, gives positive distance: 3-10=-7, |-7|=7 units), or reverse: |q-p| (order doesn't matter); mistakes include distance negative (without absolute value).

Question 16

Compare distances on the number line: Which distance is greater? Distance 1: between −2-2−2 and 666. Distance 2: between 333 and −5-5−5.

  1. Distance 1 is greater
  2. They are equal (correct answer)
  3. Distance 2 is greater
  4. Not enough information

Explanation: This question tests understanding that subtraction p-q equals adding the additive inverse p+(-q), and the distance between numbers as |p-q| (absolute value of the difference). Subtraction as addition: p-q = p+(-q) by definition (subtracting q means adding the opposite -q: 5-8=5+(-8)=-3, or 7-(-3)=7+3=10 subtracting negative adds positive). Number line: p-q starts at p, moves distance |q| left if q>0 (subtracting positive), or right if q<0 (subtracting negative = adding positive). Distance between p and q: |p-q|=|q-p| (absolute value of difference, always positive: |3-10|=|-7|=7 units apart, or |10-3|=7). Distance 1: |-2-6|=|-8|=8; Distance 2: |3-(-5)|=|8|=8, so they are equal, matching choice C. A common error is thinking order matters without absolute value, leading to A or B. Distance: between any two numbers p and q, calculate p-q, take absolute value |p-q| (removes sign, gives positive distance: 3-10=-7, |-7|=7 units), or reverse: |q-p| (order doesn't matter for distance, both give same). Mistakes: not rewriting as addition (missing connection p-q=p+(-q)), subtracting negative as subtraction (5-(-2) staying as subtract when should become 5+2=7), distance without absolute value (negative distance).

Question 17

Find the distance between 12\tfrac{1}{2}21​ and −52-\tfrac{5}{2}−25​ on a number line.​

  1. ∣12−(−52)∣=∣62∣=3\left|\tfrac{1}{2}-\left(-\tfrac{5}{2}\right)\right|=\left|\tfrac{6}{2}\right|=3​21​−(−25​)​=​26​​=3 (correct answer)
  2. ∣12−(−52)∣=∣−42∣=2\left|\tfrac{1}{2}-\left(-\tfrac{5}{2}\right)\right|=\left|\tfrac{-4}{2}\right|=2​21​−(−25​)​=​2−4​​=2
  3. ∣12−(−52)∣=∣62∣=−3\left|\tfrac{1}{2}-\left(-\tfrac{5}{2}\right)\right|=\left|\tfrac{6}{2}\right|=-3​21​−(−25​)​=​26​​=−3
  4. 12−(−52)=−3\tfrac{1}{2}-\left(-\tfrac{5}{2}\right)=-321​−(−25​)=−3, so the distance is −3-3−3

Explanation: This question tests understanding that subtraction p - q equals adding the additive inverse p + (-q), and the distance between numbers as |p - q| (absolute value of the difference). Subtraction as addition: p - q = p + (-q) by definition (subtracting q means adding the opposite -q: 5 - 8 = 5 + (-8) = -3, or 7 - (-3) = 7 + 3 = 10 since subtracting a negative adds the positive). On the number line, p - q starts at p and moves |q| units left if q > 0 (subtracting a positive) or right if q < 0 (subtracting a negative equals adding a positive), while the distance between p and q is |p - q| = |q - p| (absolute value of the difference, always positive: |3 - 10| = |-7| = 7 units apart, or |10 - 3| = 7). For example, 15 - 20 can be rewritten as 15 + (-20) = -5 (like a temperature drop from 15°C to -5°C), or the distance from -4 to 3 is |3 - (-4)| = |3 + 4| = 7 units (or |-4 - 3| = |-7| = 7), and for money, 10 - 25 = 10 + (-25) = -15 (a debt of $15). The correct distance is |1/2 - (-5/2)| = |6/2| = 3, as it applies absolute value to the difference correctly. A common error is distance negative, like in choice B where 1/2 - (-5/2) = -3 without absolute value, or arithmetic wrong as in choice C where |1/2 - (-5/2)| = |-4/2| = 2 miscalculates the fraction. Distance: between any two numbers p and q, calculate p - q and take the absolute value |p - q| (removes the sign, gives positive distance: 3 - 10 = -7, |-7| = 7 units), or reverse |q - p| (order doesn't matter for distance, both give the same).

Question 18

A student rewrites subtraction as adding the inverse. Which expression correctly rewrites 5−85-85−8 as addition and gives the correct value?

  1. 5−8=5+8=135-8=5+8=135−8=5+8=13
  2. 5−8=5−(−8)=−35-8=5-(-8)=-35−8=5−(−8)=−3
  3. 5−8=5+(−8)=−35-8=5+(-8)=-35−8=5+(−8)=−3 (correct answer)
  4. 5−8=5+(−8)=35-8=5+(-8)=35−8=5+(−8)=3

Explanation: This question tests understanding that subtraction p - q equals adding the additive inverse p + (-q), and the distance between numbers as |p - q| (absolute value of the difference). Subtraction as addition: p - q = p + (-q) by definition (subtracting q means adding the opposite -q: 5 - 8 = 5 + (-8) = -3, or 7 - (-3) = 7 + 3 = 10 since subtracting a negative adds the positive). On the number line, p - q starts at p and moves |q| units left if q > 0 (subtracting a positive) or right if q < 0 (subtracting a negative equals adding a positive), while the distance between p and q is |p - q| = |q - p| (absolute value of the difference, always positive: |3 - 10| = |-7| = 7 units apart, or |10 - 3| = 7). For example, 15 - 20 can be rewritten as 15 + (-20) = -5 (like a temperature drop from 15°C to -5°C), or the distance from -4 to 3 is |3 - (-4)| = |3 + 4| = 7 units (or |-4 - 3| = |-7| = 7), and for money, 10 - 25 = 10 + (-25) = -15 (a debt of $15). The correct rewriting is 5 - 8 = 5 + (-8) = -3, as it properly adds the additive inverse and computes the value accurately. A common error is claiming p - q = p + q instead of p + (-q), like in choice A where 5 - 8 = 5 + 8 = 13 denies the equivalence, or miscalculating the addition as in choice C where 5 + (-8) = 3, or confusing the signs as in choice D. Using the additive inverse, rewrite every subtraction as addition (p - q → p + (-q), making all operations additions), and apply addition rules (p + (-q) follows number line interpretation: start at p, move |q| left).

Question 19

Calculate and interpret: At 6 a.m., the temperature was 15∘C15^\circ\text{C}15∘C. By noon it had dropped 20∘C20^\circ\text{C}20∘C. What is the noon temperature? (Use subtraction as adding the inverse.)

  1. 35∘C35^\circ\text{C}35∘C
  2. −5∘C-5^\circ\text{C}−5∘C (correct answer)
  3. −35∘C-35^\circ\text{C}−35∘C
  4. 5∘C5^\circ\text{C}5∘C

Explanation: This question tests understanding that subtraction p-q equals adding the additive inverse p+(-q), and the distance between numbers as |p-q| (absolute value of the difference). Subtraction as addition: p-q = p+(-q) by definition (subtracting q means adding the opposite -q: 5-8=5+(-8)=-3, or 7-(-3)=7+3=10 subtracting negative adds positive). Number line: p-q starts at p, moves distance |q| left if q>0 (subtracting positive), or right if q<0 (subtracting negative = adding positive). Distance between p and q: |p-q|=|q-p| (absolute value of difference, always positive: |3-10|=|-7|=7 units apart, or |10-3|=7). For 15-20, rewrite as 15+(-20)=-5°C, matching choice A. A common error is choosing B (35°C), which adds instead of subtracting, or D (-35°C), which might double the drop incorrectly. Using the additive inverse: rewrite every subtraction as addition (p-q→p+(-q), makes all operations additions), apply addition rules (p+(-q) follows number line interpretation: start p, move |q| left). Contexts: temperature 15° drops 20° (15-20=-5°, below zero), money 10spend10 spend 10spend25 (10-25=-15, debt), elevation 50 m descend 80 m (50-80=-30 m, below sea level 30 m).

Question 20

A science lab records a change in pH as 34−(54)\frac{3}{4}-\left(\frac{5}{4}\right)43​−(45​). Which option correctly rewrites and evaluates it using p−q=p+(−q)p-q=p+(-q)p−q=p+(−q)?

  1. 34−54=24=12\frac{3}{4}-\frac{5}{4}=\frac{2}{4}=\frac{1}{2}43​−45​=42​=21​
  2. 34+54=84=2\frac{3}{4}+\frac{5}{4}=\frac{8}{4}=243​+45​=48​=2
  3. 34+(−54)=−24=−12\frac{3}{4}+\left(-\frac{5}{4}\right)=\frac{-2}{4}=-\frac{1}{2}43​+(−45​)=4−2​=−21​ (correct answer)
  4. (−34)+(−54)=−84=−2\left(-\frac{3}{4}\right)+\left(-\frac{5}{4}\right)=-\frac{8}{4}=-2(−43​)+(−45​)=−48​=−2

Explanation: This question tests understanding that subtraction p-q equals adding the additive inverse p+(-q), and the distance between numbers as |p-q| (absolute value of the difference). Subtraction as addition: p-q = p+(-q) by definition (subtracting q means adding the opposite -q: 5-8=5+(-8)=-3, or 7-(-3)=7+3=10 subtracting negative adds positive). Number line: p-q starts at p, moves distance |q| left if q>0 (subtracting positive), or right if q<0 (subtracting negative = adding positive). Distance between p and q: |p-q|=|q-p| (absolute value of difference, always positive: |3-10|=|-7|=7 units apart, or |10-3|=7). For (3/4) - (5/4), rewrite as (3/4) + (-5/4) = -2/4 = -1/2, matching choice B. A common error is choosing A, which adds positive 5/4 instead, or C, which doesn't use the inverse. Using the additive inverse: rewrite every subtraction as addition (p-q→p+(-q), makes all operations additions), apply addition rules (p+(-q) follows number line interpretation: start p, move |q| left). Contexts: temperature 15° drops 20° (15-20=-5°, below zero), money 10spend10 spend 10spend25 (10-25=-15, debt), elevation 50 m descend 80 m (50-80=-30 m, below sea level 30 m).