Modulus Equations and Inequalities Solved Step by Step

The modulus |x|, also called absolute value, measures a number's distance from zero on the number line: |5| = 5 and |−5| = 5. Because distance is never negative, |x| ≥ 0 for every real x.

Modulus equations and inequalities appear wherever only the size of a deviation matters, not its direction — measurement error, manufacturing tolerance, or how far a value strays from a target value.

Key Concepts

• Piecewise definition: |x| = x when x ≥ 0, and |x| = −x when x < 0 (the minus sign makes the result positive).
• Distance reading: |x − a| is the distance between x and a on the number line.
• Modulus equation: |x| = b with b ≥ 0 gives the two solutions x = b and x = −b.
• Less-than form: |x| < a is equivalent to −a < x < a, a single interval around zero.
• Greater-than form: |x| > a is equivalent to x < −a or x > a, two separate rays.
• Shifted form: |x − a| ≤ b means a − b ≤ x ≤ a + b.

Formulas

The whole topic rests on the piecewise definition plus three translations. Note the strict requirement b ≥ 0 in |x| = b: an equation like |x| = −5 has no solution, because no distance is negative.

• |x| = x if x ≥ 0; |x| = −x if x < 0
• |x| < a ⇔ −a < x < a (for a > 0)
• |x| > a ⇔ x < −a or x > a (for a > 0)
• |x − a| ≤ b ⇔ a − b ≤ x ≤ a + b

Worked Example

Solve |2x − 3| = 5 by splitting on the sign of 2x − 3. Case 1: 2x − 3 ≥ 0, so the modulus drops away and 2x − 3 = 5, giving x = 4. Check the case condition: 2(4) − 3 = 5 ≥ 0, valid. Case 2: 2x − 3 < 0, so |2x − 3| = −(2x − 3) and −(2x − 3) = 5 gives 2x − 3 = −5, so x = −1. Check: 2(−1) − 3 = −5 < 0, valid. The solution set is {−1, 4}. Distance check: |2(4) − 3| = |5| = 5 and |2(−1) − 3| = |−5| = 5, both correct.

Common Misconceptions

• "|x| = ±x." The modulus of −5 is the single value 5, not ±5 — the ± belongs to solving |x| = 5, giving x = ±5.
• Solving |x − 2| < 3 as only x < 5. The lower bound x > −1 is equally required; the answer is the interval (−1, 5).
• Believing |x| = −5 has solutions. No number sits at negative distance from zero, so the solution set is empty.

Real-World Applications

• Manufacturing tolerance, e.g., |diameter − target| ≤ 0.1 mm for a machine part.
• Error bounds on scientific measurements expressed as x ± Δx.
• Acceptable temperature or pH ranges around a set point.
• Quality control: flagging items whose deviation from standard exceeds a limit.

Modulus equations and inequalities are distance statements in disguise. Split on the sign inside the bars, or translate |x − a| < b directly into the interval (a − b, a + b), and always remember that a modulus can never be negative.