ACT Math study guide · Chapter 10
Absolute Value
An absolute value equation splits into two separate cases, since the expression inside the bars could be positive or negative before the bars were applied. |A| = B, where B is positive, becomes A = B or A = −B. Both cases must be solved, and both solutions are typically valid, unless the right side of the original equation contains a variable, in which case each candidate solution must be checked.
An absolute value inequality splits differently depending on the inequality symbol. |A| < B becomes a single compound inequality, −B < A < B. |A| > B becomes two separate inequalities joined by "or": A > B or A < −B. Confusing which form goes with which symbol is one of the most common errors on this topic.
When the right side of an absolute value equation contains a variable, that side must be nonnegative for a solution to be valid, since an absolute value can never equal a negative number. Every candidate solution from both cases must be substituted back into the original equation to confirm it actually works.
Where this appears as a model: a manufacturing tolerance. A machined part required to be 10 centimeters long, within 0.05 centimeters, is described by |x−10|≤0.05, giving an acceptable range from 9.95 to 10.05 centimeters.
▸ Absolute value equations with a variable on both sides
Worked Example Solve |x + 2| = 3x − 6.
Split into two cases, then check each candidate solution.
Case 1: x + 2 = 3x − 6 → x = 4
Case 2: x + 2 = −(3x − 6) → x = 1
Check x = 4: |6| = 6, and 3(4)−6 = 6. Valid.
Check x = 1: |3| = 3, and 3(1)−6 = −3. 3 ≠ −3. Rejected.
Only x = 4 is valid.
▸ Absolute value inequalities
Worked Example Solve |3x − 1| ≥ 11.
A ≥ form splits into two separate inequalities joined by "or."
Case 1: 3x − 1 ≥ 11 → x ≥ 4
Case 2: 3x − 1 ≤ −11 → x ≤ −10/3
▸ Practice
Question 1
- AA. x = 13, −3
- BB. x = 13, 3
- CC. x = −13, 3
- DD. x = −13, −3
Question 2
- FF. x = −2, 2.5
- GG. x = −2, −2.5
- HH. x = 2, −2.5
- JJ. x = 2, 2.5
Question 3
- AA. −3 < x < 9
- BB. −9 < x < 3
- CC. −6 < x < 3
- DD. x < −9 or x > 3
Question 4
- FF. −1 ≤ x ≤ 6
- GG. x ≥ 6 or x ≤ −1
- HH. x ≥ 1 or x ≤ −6
- JJ. x ≥ 7 or x ≤ −2
Question 5
- AA. x = −5
- BB. x = 1 or x = −5
- CC. x = 1
- DD. no solution
Question 6
- FF. x = −6
- GG. x = −8/3
- HH. no solution
- JJ. x = −8/3 or x = −6
Question 7
- AA. −4, 3.2
- BB. 4, −3.2
- CC. 4, 3.2
- DD. −4, −3.2
Question 8
- FF. −14 < x < 6
- GG. −6 < x < 14
- HH. −10 < x < 10
- JJ. 4 < x < 24
Question 9
- AA. x ≤ 4 or x ≥ −14/3
- BB. x ≥ 4 or x ≤ −14/3
- CC. −14/3 ≤ x ≤ 4
- DD. x ≥ 14/3 or x ≤ −4
Question 10
- FF. −1 only
- GG. −1 or 3
- HH. no solution
- JJ. 3 only
Question 11
- AA. 8, −1
- BB. −8, 1
- CC. 8, 1
- DD. −8, −1
Question 12
- FF. 4 ≤ x ≤ 8
- GG. −8 ≤ x ≤ 4
- HH. −4 ≤ x ≤ 8
- JJ. −4 ≤ x ≤ −8
Question 13
- AA. x < 2 or x > −6.5
- BB. −6.5 ≤ x ≤ 2
- CC. x > 6.5 or x < −2
- DD. x > 2 or x < −6.5
Question 14
- FF. −5 or −1
- GG. −5 only
- HH. −1 only
- JJ. no solution
Question 15
- AA. 3.5, −5
- BB. −3.5, 5
- CC. 3.5, 5
- DD. −3.5, −5
Question 16
- FF. −3 < x < 15
- GG. −15 < x < 3
- HH. 3 < x < 15
- JJ. −15 < x < −3
Question 17
- AA. x ≤ 3 or x ≥ −12
- BB. −12 ≤ x ≤ 3
- CC. x ≥ 12 or x ≤ −3
- DD. x ≥ 3 or x ≤ −12
Question 18
- FF. −4 only
- GG. 4 or −4
- HH. no solution
- JJ. 4 only
Question 19
- AA. 6 ≤ x ≤ 14
- BB. −14 ≤ x ≤ 6
- CC. −6 ≤ x ≤ 14
- DD. −14 ≤ x ≤ 14
Question 20
- FF. 5 or −1.5
- GG. 5 only
- HH. −1.5 only
- JJ. no solution
▸ Answers and Explanations
Show the answers and explanations
1: A. Case 1: x−5=8→x=13. Case 2: x−5=−8→x=−3. Choice B drops the negative sign on the second case, computing x−5=−8 as if it gave x=3.
2: H. Case 1: 4x+1=9→x=2. Case 2: 4x+1=−9→x=−2.5. Choice F reverses the sign of both solutions.
3: B. Rewrite as −6<x+3<6, then subtract 3 from every part: −9<x<3. Choice A adds 3 to both boundaries instead of subtracting it.
4: G. Case 1: 2x−5≥7→x≥6. Case 2: 2x−5≤−7→x≤−1. Choice F treats this as a single compound inequality instead of two separate cases joined by "or."
5: C. Case 2 gives x=1, and 2(1)+1=3≥0 checks; Case 1 gives x=−5, but 2(−5)+1=−9<0, which is impossible for an absolute value, rejected. Choice B reports both candidate solutions without checking that x=−5 is extraneous.
6: G. Case 2 gives x=−8/3, and 2(−8/3)+7=5/3≥0 checks, matching the left side. Case 1 gives x=−6, but 2(−6)+7=−5<0. Rejected. Choice F reports the extraneous solution instead of checking which case actually satisfies the original equation.
7: B. Case 1: 5x−2=18→x=4. Case 2: 5x−2=−18→x=−3.2. Choice A reverses the sign of both solutions.
8: G. Rewrite as −10<x−4<10, then add 4 to every part: −6<x<14. Choice F subtracts 4 from the boundaries instead of adding it.
9: B. Case 1: 3x+1≥13→x≥4. Case 2: 3x+1≤−13→x≤−14/3. Choice A reverses both inequality directions.
10: J. Case 2 gives x=3, and 3(3)−1=8≥0 checks, matching |3+5|=8. Case 1 gives x=−1, but 3(−1)−1=−4<0, impossible for an absolute value; rejected. Only x=3 is valid. Choice G reports both candidate solutions without checking that x=−1 is extraneous.
11: A. Case 1: 2x−7=9→x=8. Case 2: 2x−7=−9→x=−1. Choice C drops the negative sign on the second case.
12: H. Rewrite as −6≤x−2≤6, then add 2 to every part: −4≤x≤8. Choice F drops the negative sign on the left boundary after adding 2.
13: D. Case 1: 4x+9>17→x>2. Case 2: 4x+9<−17→x<−6.5. Choice A reverses both inequality directions.
14: F. Case 1: x−1=2x+4→x=−5; check: |−6|=6 and |−6|=6, matches. Case 2: x−1=−(2x+4)→x=−1; check: |−2|=2 and |2|=2, also matches. Both cases are valid here. Choice H reports only one of the two solutions, missing that both cases check out in the original equation.
15: A. Case 1: 4x+3=17→x=3.5. Case 2: 4x+3=−17→x=−5. Choice C drops the negative sign on the second case.
16: F. Rewrite as −9<x−6<9, then add 6 to every part: −3<x<15. Choice H drops the negative sign on the left boundary after adding 6.
17: D. Case 1: 2x+9≥15→x≥3. Case 2: 2x+9≤−15→x≤−12. Choice A reverses both inequality directions.
18: J. Case 2 gives x=4, and 2(4)−1=7≥0 checks, matching |4+3|=7. Case 1 gives x=−4, but 2(−4)−1=−9<0, impossible for an absolute value; rejected. Only x=4 is valid. Choice G reports both candidate solutions without checking that x=−4 is extraneous.
19: C. Rewrite as −10≤x−4≤10, then add 4 to every part: −6≤x≤14. Choice A drops the negative sign on the left boundary after adding 4.
20: F. Case 1: 3x−2=x+8→x=5; check: |13|=13 and |13|=13, matches. Case 2: 3x−2=−(x+8)→x=−1.5; check: |−6.5|=6.5 and |6.5|=6.5, also matches. Both cases are valid here. Choice H reports only one of the two solutions, missing that both cases check out in the original equation.