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Absolute Value Equation & Inequality Solver

Equations · Inequalities · Double Absolute Values · Step-by-Step

Solve |ax+b| = c, less-than and greater-than inequalities, and double absolute value equations with full algebraic steps and a number line visualization.

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Quick Examples

|ax + b| = c

What Is Absolute Value?

The absolute value of a real number x, denoted |x|, measures its distance from zero on the number line — always a non-negative quantity. The formal definition is piecewise:

|x| = x if x ≥ 0     |x| = −x if x < 0

Equivalently, |x| = max(x, −x) = √(x²). The geometric interpretation is direct: |x| is the number of units between x and 0, so |5| = 5 and |−5| = 5. For expressions, |ax + b| measures the distance between ax + b and 0, making it central to tolerance, error analysis, and constraint problems.

Solving |ax + b| = c (Equation)

The key principle is that |expression| = c has solutions only when c ≥ 0. Split into two cases using the definition:

  • Case 1 (positive branch): ax + b = c  →  x = (c − b) / a
  • Case 2 (negative branch): ax + b = −c  →  x = (−c − b) / a

When a = 0 the equation becomes |b| = c, which is true for all x if |b| = c, and false for all x otherwise. Always verify each solution by substituting back.

Solving |ax + b| < c (Less-Than Inequality)

A less-than absolute value inequality means "the expression is within c units of zero." It is equivalent to the compound inequality:

−c < ax + b < c

Solve both parts simultaneously by isolating x. The solution is always a bounded open interval centred at x = −b/a with radius c/|a|. If c ≤ 0, there is no solution. The number line shows a shaded segment between two open endpoints.

Solving |ax + b| > c (Greater-Than Inequality)

A greater-than inequality means "the expression is more than c units from zero." Split into two separate inequalities:

ax + b > c   OR   ax + b < −c

The solution is a union of two rays: (−∞, x₁) ∪ (x₂, +∞). If c < 0, every real number is a solution since absolute value is always ≥ 0. The number line shows shading extending outward in both directions from two open points.

Solving |ax + b| = |cx + d| (Double Absolute Value)

When two absolute values are set equal, use the fact that |A| = |B| means A = B or A = −B:

  • Case 1: ax + b = cx + d  →  (a − c)x = d − b
  • Case 2: ax + b = −(cx + d)  →  (a + c)x = −d − b

Each case yields at most one solution. Special handling is needed when a = c (Case 1 becomes trivial) or a = −c (Case 2 becomes trivial).

Interval Notation Quick Reference

InequalityMeaningInterval NotationNumber Line
|x| = 3x is 3 from 0{-3, 3}Two filled dots
|x| < 3within 3 of 0(-3, 3)Open-interval shade
|x| ≤ 3at most 3 from 0[-3, 3]Closed-interval shade
|x| > 3more than 3 from 0(-∞,-3) ∪ (3,+∞)Two outward rays
|x| ≥ 3at least 3 from 0(-∞,-3] ∪ [3,+∞)Two outward closed rays

Real-World Applications

  • Engineering tolerances: A manufactured part is acceptable if |dimension − target| ≤ ε, directly an absolute value inequality.
  • Error bounds: Measurement uncertainty is expressed as |measured − true| ≤ error margin.
  • Temperature control: A thermostat keeps a room within ±2°C of the setpoint: |T − T₀| ≤ 2.
  • Signal processing: Keeping a signal's deviation from a reference within a threshold.
  • Quality control: Rejecting items whose weight differs from spec by more than a given amount: |w − w₀| > tol.
  • Finance: Stop-loss rules: exit if |price − entry| > risk tolerance.

Frequently Asked Questions

What is absolute value and how is it defined?
The absolute value of x, written |x|, is its distance from zero on the number line. It is always non-negative: |x| = x when x ≥ 0, and |x| = −x when x < 0. Equivalently, |x| = max(x, −x) = √(x²). So |5| = 5 and |−5| = 5 — both are 5 units from zero.
How do you solve |ax + b| = c?
First check if c < 0 — if so, there is no solution. If c ≥ 0, split into two cases: Case 1: ax + b = c, giving x = (c − b)/a; Case 2: ax + b = −c, giving x = (−c − b)/a. When a ≠ 0 you typically get two distinct solutions. Verify each by substituting back into the original equation.
How do you solve |ax + b| < c?
If c ≤ 0, there is no solution. If c > 0, rewrite as −c < ax + b < c and solve both inequalities simultaneously. Subtract b from all parts then divide by a (reversing inequalities if a < 0). The answer is a bounded open interval: (−c−b)/a < x < (c−b)/a (for a > 0).
How do you solve |ax + b| > c?
If c < 0, every real number is a solution. If c ≥ 0, split into two separate inequalities: ax + b > c OR ax + b < −c. Solve each independently for x. The combined solution is a union of two rays: (−∞, x₁) ∪ (x₂, +∞), shown as outward-pointing arrows on the number line.
How do you solve |ax + b| = |cx + d|?
Two absolute values are equal when either the expressions inside are equal or they are opposites. Split: Case 1: ax + b = cx + d, solving (a−c)x = d−b; Case 2: ax + b = −(cx + d), solving (a+c)x = −d−b. Each case gives at most one x-value. Verify each in the original equation to confirm.
What is interval notation for absolute value solutions?
Parentheses () denote strict exclusion (<, >) and brackets [] denote inclusion (≤, ≥). Examples: |x| < 3 → (−3, 3); |x| ≤ 3 → [−3, 3]; |x| > 3 → (−∞, −3) ∪ (3, +∞); |x| ≥ 3 → (−∞, −3] ∪ [3, +∞). Set notation for equations: {x₁, x₂}.
What are real-world applications of absolute value equations?
Absolute value inequalities model any "within tolerance" constraint: engineering specs (|dimension − target| ≤ ε), measurement error bounds, temperature control (± setpoint), quality control (reject if deviation exceeds limit), financial stop-losses, and signal processing threshold detection. Whenever a quantity must stay within a certain distance of a target value, an absolute value inequality captures the constraint.