➕ Algebra 1 · Inequalities
Absolute Value Equations: true or false
Decide fast, then read the reason. Wrong answers are the best teachers, so every statement explains itself.
All statements and answers
The absolute value of a number is never negative.
✓ True. |x| is a distance from 0, and a distance is zero or more. For example, |−7| = 7 and |0| = 0.
The equation |x| = 15 has exactly one solution, x = 15.
✗ False. Both 15 and −15 are 15 units from 0, so x = 15 or x = −15. Check: |−15| = 15.
The equation |2x + 6| = 0 has exactly one solution, x = −3.
✓ True. Only 0 has absolute value 0, so 2x + 6 = 0, which gives 2x = −6 and x = −3. Check: |2(−3) + 6| = |0| = 0.
The equation |x + 3| = −1 has no solution.
✓ True. An absolute value is a distance and cannot be −1. Solving x + 3 = −1 would give x = −4, but |−4 + 3| = 1, not −1.
To solve 2|x + 8| = 14, you can write x + 8 = 14 or x + 8 = −14.
✗ False. The bars must be alone first. Divide by 2: |x + 8| = 7, so x + 8 = 7 or x + 8 = −7, which gives x = −1 or x = −15. The wrong split gives x = 6, and 2|6 + 8| = 28, not 14.
The solutions of |x + 10| = 3 are x = −7 and x = −13.
✓ True. x + 10 = 3 gives x = −7. x + 10 = −3 gives x = −13. These are the two numbers 3 units from −10.
The solutions of |x − 10| = 3 are x = −7 and x = −13.
✗ False. |x − 10| is the distance from 10, so the solutions are 3 units from 10: x = 7 and x = 13. The numbers −7 and −13 solve |x + 10| = 3.
The solutions of |3x − 1| = 8 are x = 3 and x = −7/3.
✓ True. 3x − 1 = 8 gives 3x = 9, so x = 3. 3x − 1 = −8 gives 3x = −7, so x = −7/3. Check: |3(−7/3) − 1| = |−8| = 8.
|x| = 4 means exactly the same thing as x = 4.
✗ False. |x| = 4 is true for x = 4 and for x = −4. The equation x = 4 allows only one of them.
The equation 6 − |x| = 9 has no solution.
✓ True. Subtract 6: −|x| = 3. Divide both sides by −1: |x| = −3. A distance cannot be −3, so there is no solution.
If you solve |2x − 7| = 5 correctly, both of your answers will pass the check in the original equation.
✓ True. The answers are x = 6 and x = 1. Check: |2(6) − 7| = 5 and |2(1) − 7| = |−5| = 5. When the bars are alone and the right side is positive, both cases give true solutions, so a failed check means an arithmetic slip.
The equation −3|x| = 12 has the solutions x = 4 and x = −4.
✗ False. Divide both sides by −3: |x| = −4, which no number satisfies, so there is no solution. Check x = 4: −3|4| = −12, not 12.
On a number line, the solutions of |x − 20| = 6 are the two numbers that are 6 units from 20.
✓ True. They are 20 − 6 = 14 and 20 + 6 = 26.