Calculator

Inequality Calculator with Steps

Solve linear, compound, absolute value, quadratic, and rational inequalities, with the sign-flip rule applied explicitly and the answer given in both inequality and interval notation.

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How to use it

Type the inequality with <= and >= for the inclusive versions, and use vertical bars for absolute value: |2x-5|<=7. Compound inequalities can be written as one chain. Ask for interval notation, a number line, or both. If the problem is an equation rather than an inequality, the equation solver is the right tool.

The method behind it

An inequality is solved with the same moves as an equation — add the same thing to both sides, multiply both sides by the same thing — with one exception that changes everything.

a < b ⟹ - a > - b

Multiplying or dividing by a negative number reverses the direction of the inequality. The reason is visible on a number line: multiplying by −1 reflects it, so anything that was to the left is now to the right. Note that $2 < 5$ becomes $- 2 > - 5$ , which is still true only because the sign turned around.

A compound inequality is two statements at once. An and statement, usually written as a chain like $- 4 \leq 3 - 2 x < 7$ , is satisfied only where both parts hold, so the solution is an intersection and normally a single interval. An or statement is a union, and its solution is usually two pieces.

Absolute value inequalities split into those two cases, and which one you get depends only on the direction of the sign. Read the absolute value as distance from zero and both rules become obvious rather than memorised.

∣ X ∣ \leq a ⟺ - a \leq X \leq a, ∣ X ∣ \geq a ⟺ X \leq - a or X \geq a

Quadratic and rational inequalities need a different approach entirely, because you cannot multiply out a denominator whose sign you do not know. Collect everything onto one side against zero, factor, and mark the critical points: the zeros of the numerator and the zeros of the denominator. Those points are the only places the expression can change sign, so test one sample value in each interval between them and the whole sign pattern follows from a handful of arithmetic checks. Being able to factor the expression first is what makes the critical points visible.

Endpoints need care at the end. A zero of the numerator satisfies a $\geq$ inequality and is included; a zero of the denominator never is, because the expression is undefined there. Interval notation records that distinction with the bracket shape, which is why it is worth writing answers that way rather than in words.

Worked examples

Worked example

∣2 x - 5∣ \leq 7

1
The sign is less than or equal to, so this is the single-interval case: the expression inside must lie within 7 of zero.
$- 7 \leq 2 x - 5 \leq 7$
2
Add 5 to all three parts. Whatever you do to the middle you must do to both outer parts.
$- 2 \leq 2 x \leq 12$
3
Divide all three parts by 2. The divisor is positive, so no sign reverses.
$- 1 \leq x \leq 6$
4
Write it as an interval. Both endpoints are included, so both brackets are square.
$[- 1, 6]$
5
Test the endpoints in the original. Both give exactly 7, which satisfies the inclusive inequality.
$∣2 (6) - 5∣ = 7, ∣2 (- 1) - 5∣ = 7$

Answer

- 1 \leq x \leq 6, [- 1, 6]

Worked example

\frac{x - 1}{x + 3} \geq 0

1
Do not multiply by x + 3. Its sign is unknown, so you would not know whether to reverse the inequality. The expression is already compared with zero, which is what a sign chart needs.
$\frac{x - 1}{x + 3} \geq 0$
2
Find the critical points: the numerator is zero at 1 and the denominator is zero at −3. These split the line into three intervals.
$x = - 3 (undefined), x = 1 (zero)$
3
Test one value in each interval. Only the sign of the result matters, not its size.
$x = - 4 : \frac{- 5}{- 1} = 5 > 0; x = 0 : \frac{- 1}{3} < 0; x = 2 : \frac{1}{5} > 0$
4
Keep the intervals where the expression is positive, then decide the endpoints separately.
$x < - 3 or x > 1$
5
At x = 1 the fraction equals zero, which satisfies the inclusive sign, so include it. At x = −3 the fraction is undefined, so it is excluded no matter what the sign says.
$(- \infty, - 3) \cup [1, \infty)$

Answer

x < - 3 or x \geq 1, (- \infty, - 3) \cup [1, \infty)

The sign flip, step by step

This compound inequality contains the move that costs more marks than any other in the topic: dividing by a negative number, twice over, in a chain. Work each step before revealing it.

Solve −4 ≤ 3 − 2x < 7

0 of 6 steps

Try it yourself first — you'll remember it longer.

Common mistakes

Not reversing after dividing by a negative

From $- 2 x > 6$ , dividing by −2 gives $x < - 3$ , not $x > - 3$ . Test it: $x = - 4$ gives $8 > 6$ , which is true, while $x = 0$ gives $0 > 6$ , which is false. One substitution settles the direction every time.

Multiplying by an expression of unknown sign

Clearing the denominator in $\frac{x - 1}{x + 3} \geq 0$ by multiplying by $x + 3$ assumes that factor is positive. For $x < - 3$ it is negative and the inequality should have reversed, which is why that method loses an entire branch of the solution. Use a sign chart instead.

Splitting an absolute value the wrong way

$∣ x ∣ < 3$ means $- 3 < x < 3$ , a single interval. Writing $x < - 3$ or $x > 3$ describes exactly the values that fail. Check the direction of the original sign before splitting, not after.

Practice

Answers are checked here — nothing is sent anywhere.

1
$Solve 5 - 2 x > 11$
2
$Solve ∣ x - 4∣ < 3$
3
$Solve x^{2} - 9 \leq 0$

Frequently asked questions

Can I get the answer in interval notation?

Both notations are shown. Interval notation uses a square bracket for an endpoint that is included and a round one for an endpoint that is excluded, and infinity always takes a round bracket because it is a direction, not a number.

Does it draw the number line?

Yes, with a filled circle for an included endpoint and an open circle for an excluded one. The picture is worth having for compound inequalities, where the union or intersection is easy to get backwards in symbols.

How are quadratic and rational inequalities handled?

By sign chart. Everything is collected on one side against zero, the critical points are found from the zeros of the numerator and denominator, and each resulting interval is tested with a single sample value.

Why do the two absolute value cases behave so differently?

Because absolute value is distance from zero. Less than a means within distance a of zero, which is one interval. Greater than a means further than a from zero in either direction, which is two separate intervals.

What does it cost?

The first solution is free with no account. A free account includes three solutions a day, and Gauth Plus removes the limit for $11.99 a month with a three-day free trial.

Keep going

Algebra solverInequalities alongside equations and systems, with the reasoning at each line.Order of operations beyond PEMDASUndo an expression in the right order so the sign-flip step arrives cleanly.The quadratic formula: when to use itFinding the critical points that a quadratic inequality sign chart needs.Equation solverThe same balance rules without the direction to keep track of.