Absolute value inequalities appear in two structural forms. When the absolute value is less than c, the solution is a bounded interval — x lies between two values. When the absolute value is greater than c, the solution is a union of two rays — x lies outside an interval. This calculator handles both forms and all four inequality types.
How to Use This Calculator
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Enter coefficient a (the multiplier of x inside the absolute value bars). Use 1 for a plain x.
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Enter constant b (added inside the absolute value). For |x − 3|, enter b = −3.
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Select the inequality type — <, ≤, >, or ≥.
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Enter the right-hand side c.
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Read the solution in interval notation and inequality form below.
How to Solve Absolute Value Inequalities
For less-than inequalities (|ax + b| < c or ≤ c): rewrite as a compound inequality without the absolute value bars. The expression inside must lie between −c and +c. This gives a bounded interval as the solution.
For greater-than inequalities (|ax + b| > c or ≥ c): split into two separate inequalities. The expression inside must be either less than −c or greater than +c. This gives a union of two unbounded rays as the solution.
The sign of c matters. If c is negative, any less-than inequality has no solution (absolute value is always ≥ 0). Any greater-than inequality with c < 0 is always true — the solution is all real numbers.
Solution Rules
For a ≠ 0 and c > 0, let L = min((−c−b)/a, (c−b)/a) and U = max((−c−b)/a, (c−b)/a).
|ax + b| < c → x ∈ (L, U)
|ax + b| ≤ c → x ∈ [L, U]
|ax + b| > c → x ∈ (−∞, L) ∪ (U, +∞)
|ax + b| ≥ c → x ∈ (−∞, L] ∪ [U, +∞)
Worked Example: |x − 3| < 5
a = 1, b = −3, c = 5. Since c > 0 and the inequality is <, split into a compound inequality.
Step 1: −5 < x − 3 < 5
Step 2: add 3 to all parts: −5 + 3 < x < 5 + 3
Step 3: −2 < x < 8
Solution: x ∈ (−2, 8). All values strictly between −2 and 8 satisfy the inequality.
Worked Example: |3x − 6| ≥ 9
a = 3, b = −6, c = 9. Since the inequality is ≥, split into two cases.
Case 1: 3x − 6 ≤ −9 → 3x ≤ −3 → x ≤ −1
Case 2: 3x − 6 ≥ 9 → 3x ≥ 15 → x ≥ 5
Solution: x ∈ (−∞, −1] ∪ [5, +∞). Any x at or below −1, or at or above 5, satisfies the inequality.
Solution Types Summary
| Condition | Type | Solution set |
|---|---|---|
| c < 0, inequality is < or ≤ | No solution | ∅ (empty set) |
| c < 0, inequality is > or ≥ | All real numbers | (−∞, +∞) |
| c = 0, inequality is < | No solution | ∅ |
| c = 0, inequality is ≤ | One point | {−b/a} |
| c = 0, inequality is > | All except one point | x ≠ −b/a |
| c = 0, inequality is ≥ | All real numbers | (−∞, +∞) |
| c > 0, < or ≤ | Bounded interval | (L, U) or [L, U] |
| c > 0, > or ≥ | Union of two rays | (−∞, L) ∪ (U, +∞) or with brackets |
Interval Notation Quick Reference
Parenthesis ( or ) means the endpoint is not included (strict inequality < or >).
Bracket [ or ] means the endpoint is included (non-strict inequality ≤ or ≥).
Infinity is always written with a parenthesis: (−∞, 5] means all x up to and including 5.
Union ∪ joins two disjoint solution sets: (−∞, −4) ∪ (4, +∞) means x < −4 or x > 4.
This calculator solves absolute value inequalities of the form |ax + b| < c, ≤ c, > c, or ≥ c. Select the inequality type, enter the coefficients, and get the complete solution set in interval notation and inequality form.