Inequality Solver
Exact critical points, a sign chart, interval notation, a number line and shaded regions.
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About the Inequality Solver
Solve an inequality in one variable and see why the answer is right: linear (2x + 1 > 7), quadratic (x² − 4 > 0), polynomial of any degree, rational ((x + 1)/(x − 2) ≥ 0) and absolute-value (|2x − 1| > 5) inequalities, also chained (−3 < 2x + 1 ≤ 7) and joined with and or or. You get the answer in interval notation and set-builder notation, the critical points, a sign chart with the sign of every factor, a number line, and the steps, checked with numbers inside and outside the answer.
A second mode graphs a system of linear inequalities in x and y: it shades the feasible region, finds its corner points exactly and, if you give an expression such as 3x + 2y, its largest and smallest values at the corners. Everything is worked out in exact fractions in your browser.
How to use it
- Choose One variable or System in x and y.
- Type the inequality with <, >, <= (≤), >= (≥) or ≠, ^ for powers and |…| for absolute values, such as
x^2 - 4 > 0,(x + 1)/(x - 2) >= 0or-3 < 2x + 1 <= 7. The buttons insert ≤, ≥, ≠ and the other symbols where the cursor is. - For a system, type one inequality per line, such as
x + y <= 4andx >= 0, and optionally an expression to maximise and minimise. - Read the answer, the sign chart or the corner points, the number line or the shaded region, and the steps.
- With a pass, or after unlocking this result, copy the whole solution or download it as a text file.
Examples
2x + 1 > 7
(3, ∞)
Subtract 1 and divide by 2: x > 3.
−3 < 2x + 1 ≤ 7
(−2, 3]
Both parts at once: −4 < 2x ≤ 6.
x² − 4 > 0
(−∞, −2) ∪ (2, ∞)
Zeros at −2 and 2; positive outside them.
x² − x − 1 < 0
((1 − √5)/2, (1 + √5)/2)
The endpoints are exact surds.
(x + 1)/(x − 2) ≥ 0
(−∞, −1] ∪ (2, ∞)
−1 makes it 0, so it is included; 2 makes it undefined, so it is not.
|2x − 1| > 5
(−∞, −2) ∪ (3, ∞)
2x − 1 < −5 or 2x − 1 > 5.
x³ − x > 0
(−1, 0) ∪ (1, ∞)
x(x − 1)(x + 1) changes sign at −1, 0 and 1.
x + y ≤ 4, 2x + y ≤ 6, x ≥ 0, y ≥ 0; 3x + 2y
corners (0, 0), (3, 0), (2, 2), (0, 4); largest 10 at (2, 2)
The values at the corners are 0, 9, 10 and 8.
Common uses
- Checking inequality homework, from linear ones to rational and absolute-value inequalities, with every step.
- Finding the domain where an expression is positive, negative or defined, before graphing or integrating it.
- Writing an answer in interval and set-builder notation and drawing it on a number line.
- Small linear programming problems: the feasible region, its corners and the best value of an objective.
The critical-point method and the sign chart
Move every term to one side and write it as one fraction N(x)/D(x) compared with 0. Its critical points are the numbers that make it 0 (the zeros of the top) or undefined (the zeros of the bottom): the expression can change sign only there, so they cut the number line into intervals on which the sign stays the same, and one test point per interval decides it (OpenStax Intermediate Algebra 2e, 7.6 Solve Rational Inequalities; 9.8 Solve Quadratic Inequalities). A zero is part of the answer for ≤ and ≥ and not for < and >; a number that makes the bottom 0 is never part of it.
The sign chart lists the sign of every factor on every interval: a factor such as (x − 1)² never changes sign, which is why some critical points do not split the answer. Here the critical points are exact. The polynomials are factored over the fractions and every real root is pinned down by Sturm’s theorem, so roots such as (1 + √5)/2 stay exact and the others are given as decimals, never guessed. Test points are simple fractions, and the answer is checked by putting numbers from inside and outside it back into the original inequality.
Absolute values, chains, “and” and “or”
For B > 0, |A| < B means −B < A < B, and |A| > B means A < −B or A > B; flipping the sign when you multiply or divide both sides by a negative number is the one rule that differs from equations (OpenStax College Algebra 2e, 2.7 Linear Inequalities and Absolute Value Inequalities). When x is on both sides, as in |x − 1| < 2x, the solver splits the line where each absolute value changes from −u to u and uses the right expression on each piece.
A chain such as −3 < 2x + 1 ≤ 7 is two inequalities that must both hold. And keeps the numbers in both solution sets (the intersection, ∩); or keeps the numbers in either (the union, ∪). In interval notation a round bracket leaves the endpoint out and a square bracket includes it; ∞ always takes a round bracket.
Systems in x and y: the feasible region and its corners
Each linear inequality is a half-plane: draw its boundary line solid for ≤ and ≥ and dashed for < and >, test a point that is not on the line, and shade the side where the test holds. The points that satisfy every inequality, where the shadings overlap, form the feasible region (OpenStax College Algebra 2e, 7.3).
Its corners are where two boundary lines meet, kept only if they satisfy all the inequalities; they are solved exactly as fractions. A linear expression such as 3x + 2y takes its largest and smallest values over the region at its corners (Applied Finite Mathematics, 3.1 Maximization Applications), so a bounded region needs only its corners checked. When the region is unbounded, a largest or smallest value may not exist; when the best corner lies on a dashed line, values get as close as you like to it without reaching it. The tool says which. Inequalities that face each other, such as x + y ≤ 4 and x + y ≥ 4, leave no area: the answer is then the segment, ray, line or single point that is left.
Limitations
- One variable: polynomials and fractions of polynomials with rational coefficients (degree up to 60) and up to 4 absolute values that are not nested. Roots such as √x, logarithms, trigonometric functions, exponentials and π are not supported.
- Up to 8 comparisons at once, joined with and or or; each sign of a chain counts as one.
- Endpoints without a simple exact form are given as decimals: 6 significant digits in the answer and 8 in the list of critical points.
- Systems: 2 to 8 linear inequalities in x and y. The graph shows the region around its corners; an unbounded region continues past the edge of the graph.
- Very long calculations stop after 20 seconds; Stop ends them sooner.
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Frequently asked questions
What do I get without a pass?
Without a pass, Inequality Solver shows a watermarked number line or region and the first rows of the sign chart, with the solution, its intervals, the working and every figure hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.
Why does the inequality sign flip when I multiply by a negative number?
Because multiplying by a negative number reverses the order of numbers: 2 < 3, but −2 > −3. So −3x ≤ 12 becomes x ≥ −4 after dividing by −3. Adding or subtracting the same number on both sides never flips the sign.
Why not multiply both sides by the denominator?
Because its sign depends on x: multiplying by x − 2 keeps the sign where x > 2 and flips it where x < 2. Moving everything to one side and reading the signs from a sign chart avoids that. A number that makes the denominator 0 is never part of the answer.
How do I write the answer in interval notation?
Use a round bracket for an endpoint that is left out (< or >), a square bracket for one that is included (≤ or ≥), and always a round bracket at ∞. Join separate pieces with ∪: x < −2 or x ≥ 3 is (−∞, −2) ∪ [3, ∞), and a single number is written as {2}. In set-builder notation the same answer is {x ∈ ℝ | x < −2 or x ≥ 3}.
How do I solve an absolute value inequality?
With a positive number B on the other side: |A| < B becomes −B < A < B, and |A| > B becomes A < −B or A > B. For |2x − 1| > 5 that is 2x − 1 < −5 or 2x − 1 > 5, so x < −2 or x > 3. If B is negative, |A| < B has no solution and |A| > B holds for every x where A is defined.
What happens when the quadratic has no real roots?
Then it never changes sign: x² + 1 is positive for every x, so x² + 1 > 0 holds for all real numbers and x² + 1 < 0 for none. A perfect square such as (x − 2)² ≤ 0 holds only at x = 2.
Where is the maximum of 3x + 2y over a region?
At a corner of the region, if the region is bounded. For x + y ≤ 4, 2x + y ≤ 6, x ≥ 0 and y ≥ 0 the corners are (0, 0), (3, 0), (2, 2) and (0, 4), where 3x + 2y is 0, 9, 10 and 8: the largest value is 10 at (2, 2) and the smallest 0 at (0, 0).