Inequality Calculator

Solution

Solve ax + b ⧋ c for x.

Next

Inequalities feel like equations until you divide by a negative and have to flip the symbol. This calculator solves linear (ax + b < c), quadratic (ax² + bx + c ≤ 0) and absolute value |ax + b| > c inequalities step by step, returns the solution in interval notation, set notation and plotted form, and reminds you of every sign-flip along the way.

How to solve an inequality

  1. 1

    Enter the inequality

    Type as you would write it: `2x + 3 > 7`, `x^2 - 4 <= 0`, `|3x + 1| < 5`.

  2. 2

    The tool classifies the form

    Linear, quadratic or absolute-value: different strategies apply.

  3. 3

    Step-by-step solution

    Every algebraic step is shown, with sign-flips highlighted.

  4. 4

    Read the result

    Interval notation, set-builder notation, and a number-line plot.

Solving each inequality type

Linear inequalities

ax + b < c (or , >, )

  1. Subtract b from both sides.
  2. Divide by a.
  3. If a is negative, flip the inequality.

Example: -2x + 5 > 9-2x > 4x < -2 (division by -2 flips > to <).

Quadratic inequalities

ax² + bx + c ≤ 0

  1. Find the roots with the quadratic formula.
  2. Classify by a’s sign and inequality direction:
Case ≤ 0 solution ≥ 0 solution
a > 0 between roots outside roots
a < 0 outside roots between roots
  1. Strict (<, >) excludes roots; non-strict (, ) includes them.

Absolute value inequalities

  • |ax + b| < c (with c > 0) → -c < ax + b < c. Solve as a double inequality.
  • |ax + b| > cax + b > c OR ax + b < -c. Solve each branch separately and union the results.
  • If c is negative, the “< c” form has no solution and the “> c” form is all real numbers.

Interval notation conventions

  • (a, b), open interval, endpoints excluded.
  • [a, b], closed interval, endpoints included.
  • (a, b], half-open.
  • (-∞, 3), unbounded on the left.
  • (-∞, 3) ∪ (5, ∞), union of two disjoint intervals.

Common traps

  • Forgetting to flip when multiplying or dividing by a negative.
  • Using a strict symbol (<) but including the endpoint in the answer.
  • Solving |x - 3| > 5 and combining the two cases with AND instead of OR.

Frequently Asked Questions

Whenever you multiply or divide both sides by a negative number. Also when you take reciprocals of both sides (if both sides have the same sign). Never when adding or subtracting.

A quadratic with a > 0 (parabola opening up) is negative only between its roots, so ax² + bx + c < 0 gives one interval. But > 0 gives the two outer intervals, written as a union.

Yes, enter them separated by commas. The tool intersects (AND) the solution sets by default; use the “union” toggle if you want OR semantics.

Yes. The step-by-step view explains every algebraic operation and justifies each sign-flip, in a form suitable for homework submission or exam practice.

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