Inequality Solver
Solve a linear inequality, a compound inequality (AND or OR), a quadratic inequality, an absolute value inequality, or a rational inequality — each with a distinct visual, full step-by-step reasoning, and a callout explaining what the solution set actually means.
Background
An inequality compares two expressions with <, ≤, >, or ≥ instead of an equals sign. Solving one means finding every value of x that makes the statement true — usually a whole range of values, not just one number. That range can be written as an inequality (like x ≤ 3), as interval notation (like (−∞, 3]), or shaded directly on a number line.
How to use this calculator
- Choose Linear for the standard form ax + b (comparison) c.
- Choose Compound for AND (a bounded "squeeze," like low < ax+b ≤ high) or OR (two separate conditions joined by "or").
- Choose Quadratic for ax² + bx + c compared to 0.
- Choose Absolute Value for |ax + b| compared to a number c.
- Choose Rational for a fraction of two linear expressions compared to 0.
- Click Solve to see the diagram, the full step-by-step math, and a callout explaining what the result means.
How Solving Inequalities Works
Solving an inequality works almost exactly like solving an equation — isolate x using inverse operations — with one crucial exception: multiplying or dividing both sides by a negative number flips the inequality sign.
A compound AND inequality (like 1 < x ≤ 5) requires both conditions to be true at once — the solution is the overlap (intersection) of the two individual solutions.
A compound OR inequality requires only one condition to be true — the solution is everything from either individual solution (the union), even the parts that don't overlap.
A quadratic inequality is solved by finding where the expression equals 0 (its roots), then testing a point in each region those roots create to see where the expression is positive or negative. Whether the parabola opens upward or downward (the sign of a) tells you where its minimum or maximum sits, which helps you sanity-check the answer.
An absolute value inequality splits into a compound inequality: |expr| < c becomes a bounded AND (−c < expr < c), while |expr| > c becomes a disjoint OR (expr > c or expr < −c).
A rational inequality is solved the same way as a quadratic — using critical values to split the number line — except the value that makes the denominator 0 is always excluded, no matter what the comparison symbol is.
Formulas & Equations Used
Linear: ax + b (comparison) c → x (comparison, flipped if a < 0) (c − b)/a
Compound AND: low < expr < high ⟺ (expr > low) AND (expr < high)
Quadratic roots: x = (−b ± √(b² − 4ac)) / 2a
Absolute value (c > 0): |expr| < c ⟺ −c < expr < c, |expr| > c ⟺ expr > c or expr < −c
Rational sign analysis: critical values = zeros of numerator and denominator; denominator zero is always excluded
Example Problems & Step-by-Step Solutions
Example 1 — Linear with a sign flip
Solve: −3x + 4 > 10.
Step: −3x > 6, then divide by −3 and flip: x < −2.
Result: x < −2, or (−∞, −2).
Example 2 — Compound AND
Solve: −3 < 2x + 1 ≤ 7.
Step: Subtract 1 from all three parts: −4 < 2x ≤ 6. Divide by 2: −2 < x ≤ 3.
Result: (−2, 3].
Example 3 — Quadratic
Solve: x² − 4 ≥ 0.
Step: Roots at x = ±2. Since a > 0 the parabola opens upward, so it's above or on the x-axis outside the roots.
Result: (−∞, −2] ∪ [2, ∞).
Example 4 — Rational
Solve: (x + 2)/(x − 3) ≥ 0.
Step: Critical values x = −2 (numerator) and x = 3 (denominator, excluded). Testing each region gives the sign pattern.
Result: (−∞, −2] ∪ (3, ∞).
Frequently Asked Questions
When exactly do I flip the inequality sign?
Only when you multiply or divide both sides by a negative number. Adding or subtracting anything — positive or negative — never flips the sign.
What's the difference between a compound AND and OR?
AND requires both conditions at once, so the solution is the overlap between them. OR requires just one condition, so the solution is everything covered by either one, including parts that don't overlap at all.
Why does |expr| < c become an AND, but |expr| > c become an OR?
Being "close to 0" (within a distance c) is one connected range, which is an AND. Being "far from 0" (farther than c) means you're on one side or the other — two separate, disconnected rays, which is an OR.
Why is the denominator's zero always excluded in a rational inequality?
Dividing by zero is undefined — there's no value there at all, not even a value that happens to fail the inequality. So that x-value is removed from the solution set no matter what the comparison symbol is.
Can a quadratic inequality have no solution, or be true for every real number?
Yes. If the parabola never crosses the x-axis (no real roots), the expression keeps the same sign everywhere — so depending on the comparison, either every real number works, or none do. Whether that sign is positive or negative depends on which way the parabola opens.
What does it mean if the solution is a single point instead of a range?
This happens at a repeated (double) root of a quadratic with a ≤ or ≥ comparison — the expression touches 0 at exactly one point without crossing, so only that one x-value satisfies the inequality. The same thing can happen in a compound AND inequality if the low and high bounds happen to meet exactly.
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