College Algebra Quiz: Absolute Value Equations
4 questions · exam conditions
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Absolute Value EquationsQuestion 1 of 4

The equation ∣3x−1∣=∣2x+5∣|3x - 1| = |2x + 5| has solution(s):

x=−6x = -6 only
x=45x = \frac{4}{5} only
x=−6x = -6 and x=45x = \frac{4}{5}
x=−45x = -\frac{4}{5} and x=6x = 6
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College Algebra Quiz

College Algebra Quiz: Absolute Value Equations

Practice Absolute Value Equations in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Absolute Value Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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

The equation ∣3x−1∣=∣2x+5∣|3x - 1| = |2x + 5| has solution(s):

  1. x=−6x = -6 only
  2. x=45x = \frac{4}{5} only
  3. x=−6x = -6 and x=45x = \frac{4}{5}
  4. x=−45x = -\frac{4}{5} and x=6x = 6 (correct answer)
Explanation: For ∣3x−1∣=∣2x+5∣|3x - 1| = |2x + 5|, we have two cases: Case 1: 3x−1=2x+53x - 1 = 2x + 5, which gives x=6x = 6. Case 2: 3x−1=−(2x+5)=−2x−53x - 1 = -(2x + 5) = -2x - 5, so 3x−1=−2x−53x - 1 = -2x - 5, giving 5x=−45x = -4, so x=−45x = -\frac{4}{5}. Let's verify: For x=6x = 6: ∣3(6)−1∣=∣17∣=17|3(6) - 1| = |17| = 17 and ∣2(6)+5∣=∣17∣=17|2(6) + 5| = |17| = 17 ✓. For x=−45x = -\frac{4}{5}: ∣3(−45)−1∣=∣−125−1∣=∣−175∣=175|3(-\frac{4}{5}) - 1| = |-\frac{12}{5} - 1| = |-\frac{17}{5}| = \frac{17}{5} and ∣2(−45)+5∣=∣−85+5∣=∣175∣=175|2(-\frac{4}{5}) + 5| = |-\frac{8}{5} + 5| = |\frac{17}{5}| = \frac{17}{5} ✓. Both solutions are valid.

Question 2

If ∣x−2∣2=9|x - 2|^2 = 9, then the product of all solutions is:

  1. −5-5 (correct answer)
  2. 55
  3. −1-1
  4. 11
Explanation: From ∣x−2∣2=9|x - 2|^2 = 9, we get ∣x−2∣=3|x - 2| = 3 (since absolute value is non-negative). This gives us two equations: x−2=3x - 2 = 3 or x−2=−3x - 2 = -3. Solving: x=5x = 5 or x=−1x = -1. The product of all solutions is 5×(−1)=−55 \times (-1) = -5.

Question 3

The solution set of ∣∣x∣−3∣=2||x| - 3| = 2 is:

  1. {−5,−1,1,5}\{-5, -1, 1, 5\} (correct answer)
  2. {−5,1,5}\{-5, 1, 5\}
  3. {−1,1}\{-1, 1\}
  4. {−5,5}\{-5, 5\}
Explanation: For ∣∣x∣−3∣=2||x| - 3| = 2, we first solve ∣x∣−3=2|x| - 3 = 2 or ∣x∣−3=−2|x| - 3 = -2. From ∣x∣−3=2|x| - 3 = 2: ∣x∣=5|x| = 5, so x=5x = 5 or x=−5x = -5. From ∣x∣−3=−2|x| - 3 = -2: ∣x∣=1|x| = 1, so x=1x = 1 or x=−1x = -1. Let's verify: For x=5x = 5: ∣∣5∣−3∣=∣5−3∣=∣2∣=2||5| - 3| = |5 - 3| = |2| = 2 ✓. For x=−5x = -5: ∣∣−5∣−3∣=∣5−3∣=∣2∣=2||-5| - 3| = |5 - 3| = |2| = 2 ✓. For x=1x = 1: ∣∣1∣−3∣=∣1−3∣=∣−2∣=2||1| - 3| = |1 - 3| = |-2| = 2 ✓. For x=−1x = -1: ∣∣−1∣−3∣=∣1−3∣=∣−2∣=2||-1| - 3| = |1 - 3| = |-2| = 2 ✓. All four solutions are valid.

Question 4

The number of solutions to ∣x+3∣=2x−5|x + 3| = 2x - 5 is:

  1. 0
  2. 1 (correct answer)
  3. 2
  4. infinitely many
Explanation: For ∣x+3∣=2x−5|x + 3| = 2x - 5 to have solutions, we need 2x−5≥02x - 5 \geq 0, so x≥52x \geq \frac{5}{2}. Case 1: If x+3≥0x + 3 \geq 0 (i.e., x≥−3x \geq -3), then x+3=2x−5x + 3 = 2x - 5, giving x=8x = 8. Since 8≥528 \geq \frac{5}{2} and 8≥−38 \geq -3, this solution is valid. Case 2: If x+3<0x + 3 < 0 (i.e., x<−3x < -3), then −(x+3)=2x−5-(x + 3) = 2x - 5, so −x−3=2x−5-x - 3 = 2x - 5, giving 3x=23x = 2, so x=23x = \frac{2}{3}. But 23≮−3\frac{2}{3} \not< -3, so this contradicts our case assumption. Therefore, there is exactly 1 solution: x=8x = 8.