The absolute value of a number is its distance from zero on the number line. Distance is never negative, so ∣5∣ = 5 and ∣−5∣ = 5. Every absolute value problem comes back to that one idea.
Solving ∣x∣ = a
If ∣x∣ = 7, then x is 7 units from zero. There are two numbers like that: 7 and −7. So an absolute value equation usually turns into two ordinary equations.
The same works when there is an expression inside the bars. ∣x − 3∣ = 7 means x − 3 is 7 units from zero:
- x − 3 = 7, so x = 10
- x − 3 = −7, so x = −4
Both answers check: ∣10 − 3∣ = 7 and ∣−4 − 3∣ = ∣−7∣ = 7.
Isolate the absolute value first
You can only split into two cases once the absolute value is alone on one side. Move everything else away first, the same way you would with any linear equation.
Example Solve 2∣x + 1∣ − 5 = 9
- Add 5: 2∣x + 1∣ = 14.
- Divide by 2: ∣x + 1∣ = 7.
- Split: x + 1 = 7 or x + 1 = −7.
- Solve: x = 6 or x = −8.
Splitting too early
In 2∣x + 1∣ − 5 = 9, writing 2(x + 1) − 5 = −9 as the second case gives the wrong answer. The negative goes on the whole absolute value expression only after it is isolated.
When there is no solution
If the isolated absolute value equals a negative number, stop. A distance cannot be negative.
Example Solve ∣3x − 2∣ + 8 = 4
- Subtract 8: ∣3x − 2∣ = −4.
- No number has an absolute value of −4, so there is no solution.
If it equals zero, there is exactly one solution: ∣x − 5∣ = 0 only when x = 5.
Absolute value on both sides
∣A∣ = ∣B∣ means A and B are the same distance from zero, so either A = B or A = −B.
Example Solve ∣2x − 1∣ = ∣x + 4∣
- Case 1: 2x − 1 = x + 4, so x = 5.
- Case 2: 2x − 1 = −(x + 4) = −x − 4, so 3x = −3 and x = −1.
- Check x = 5: ∣9∣ = ∣9∣. Check x = −1: ∣−3∣ = ∣3∣. Both work.
Checking for extraneous answers
When the right side has a variable, like ∣x + 2∣ = 3x − 6, one of the cases can give an answer that doesn’t work. Always substitute back.
Example Solve ∣x + 2∣ = 3x − 6
- Case 1: x + 2 = 3x − 6, so 8 = 2x and x = 4.
- Case 2: x + 2 = −(3x − 6) = −3x + 6, so 4x = 4 and x = 1.
- Check x = 4: ∣6∣ = 6 and 3(4) − 6 = 6. Works.
- Check x = 1: ∣3∣ = 3, but 3(1) − 6 = −3. Fails.
- The only solution is x = 4.
Inequalities: “less than” means “and”
∣x∣ < 4 means x is less than 4 units from zero. Those numbers sit between −4 and 4:
∣x∣ < 4 → −4 < x < 4
With an expression inside, keep the same shape and solve the middle:
Example Solve ∣2x − 3∣ ≤ 5
- Write it as a compound inequality: −5 ≤ 2x − 3 ≤ 5.
- Add 3 to all three parts: −2 ≤ 2x ≤ 8.
- Divide by 2: −1 ≤ x ≤ 4.
Inequalities: “greater than” means “or”
∣x∣ > 4 means x is more than 4 units from zero. That happens on both ends of the number line:
∣x∣ > 4 → x < −4 or x > 4
Example Solve ∣x + 5∣ > 2
- Split: x + 5 < −2 or x + 5 > 2.
- Solve: x < −7 or x > −3.
| Form | Meaning | Answer shape |
|---|---|---|
| ∣u∣ = a, a > 0 | u is exactly a away from 0 | u = a or u = −a |
| ∣u∣ < a | u is closer than a to 0 | −a < u < a |
| ∣u∣ > a | u is farther than a from 0 | u < −a or u > a |
| ∣u∣ < negative | impossible | no solution |
| ∣u∣ > negative | always true | all real numbers |
Flip the sign when you divide by a negative
If the middle has a negative coefficient, like −3 < 1 − 2x < 3, dividing by −2 flips both inequality signs. The inequalities guide has more on this.
Practice problems
Solve ∣x − 6∣ = 2.
Show answer
x = 8 or x = 4.
Solve 3∣x + 2∣ = 12.
Show answer
x = 2 or x = −6. Divide by 3 first to get ∣x + 2∣ = 4.
Solve ∣4x + 1∣ + 7 = 3.
Show answer
No solution. ∣4x + 1∣ = −4 is impossible.
Solve ∣x − 1∣ < 3.
Show answer
−2 < x < 4.
Solve ∣2x + 1∣ ≥ 7.
Show answer
x ≤ −4 or x ≥ 3.
Solve ∣x + 3∣ = 2x.
Show answer
x = 3. Case 2 gives x = −1, which fails the check because 2(−1) is negative.