Inequality solver

How do I solve a linear inequality?

Treat it like an equation, with one exception: multiplying or dividing both sides by a negative number reverses the inequality sign. For 13 − 7x ≥ 10x − 4, move the terms to get −17x ≥ −17, then divide by −17 and flip the sign: x ≤ 1. In interval notation that is (−∞, 1] (OpenStax Algebra and Trigonometry 2e, Example 5). Type the inequality above to see the same steps checked by a test point.

How do I solve a quadratic or polynomial inequality?

Move everything to one side so the other side is 0, find where the expression equals 0 (the critical numbers), and test one number in each interval between them. For x² − x − 12 ≥ 0 the zeros are −3 and 4. The parabola opens upward, so the expression is positive outside the zeros and the answer is (−∞, −3] ∪ [4, ∞) (OpenStax Intermediate Algebra 2e, section 9.8). The same sign-chart method works for cubics and higher: x³ − x > 0 gives (−1, 0) ∪ (1, ∞).

How do I solve a rational inequality such as 4x/(x − 6) < 1?

Do not multiply both sides by x − 6, because that expression is negative on one side of 6 and positive on the other, so you cannot know whether to flip the sign. Subtract 1 instead to get 0 on the right: (4x − (x − 6)) ÷ (x − 6) = (3x + 6) ÷ (x − 6) < 0. The critical numbers are −2 (numerator zero) and 6 (denominator zero, never allowed). Testing the three intervals gives (−2, 6) (OpenStax Intermediate Algebra 2e, section 7.6). Write a fraction as (numerator)/(denominator) with brackets when you type it.

How do absolute value inequalities work?

|A| < c means −c < A < c (the “and” case, one interval), while |A| > c means A < −c or A > c (the “or” case, two pieces). For |x − 1| ≤ 3 the answer is [−2, 4]. For |4x − 5| > 6 it is (−∞, −1/4) ∪ (11/4, ∞) (OpenStax Algebra and Trigonometry 2e, Example 11). The calculator splits the problem on the sign of the expression inside the bars, solves each case, and keeps only the values that fit that case, so it also handles harder forms such as −½|4x − 5| + 3 < 0 or |x − 2| < x.

What does it mean when the answer is “no solution” or “all real numbers”?

Some inequalities hold for no x, some for every x. Because x² − 3x + 4 has a negative discriminant (9 − 16 = −7), its graph never touches the x-axis and stays above it: x² − 3x + 4 > 0 is true for all real numbers, (−∞, ∞), while x² − 3x + 4 ≤ 0 has the answer ∅, the empty set (OpenStax Intermediate Algebra 2e, section 9.8).

When is an endpoint included?

Use a square bracket for ≤ and ≥ and a parenthesis for < and >, but a value that makes a denominator zero is never included, whatever the sign. Infinity is always written with a parenthesis. A repeated root can leave a single excluded point: x² − 2x + 1 > 0 is true for every x except 1, which the calculator writes as (−∞, 1) ∪ (1, ∞). A hole behaves the same way: (x² − 1) ÷ (x − 1) ≥ 0 simplifies to x + 1 ≥ 0 away from x = 1, so the answer is [−1, 1) ∪ (1, ∞).

Can I type a compound inequality?

Yes: write 3 ≤ 2x + 2 < 6 with two signs in one line. The calculator solves each half and intersects them, giving [1/2, 2) (OpenStax Algebra and Trigonometry 2e, Example 7). You may use <=, >= or the symbols ≤, ≥, and any single letter for the unknown.

The sign-chart method, step by step

1. Move everything to one side so that the inequality compares an expression with 0, using a common denominator if there are fractions. 2. Find the critical numbers: the real zeros of the numerator (the expression is 0 there) and of the denominator (it is undefined there). 3. They cut the number line into intervals; on each interval the expression has one fixed sign, because it can only change sign at a zero or a break. 4. Test one number in each interval and note whether the expression is positive or negative. 5. Keep the intervals whose sign matches the inequality, then decide each endpoint: a zero of the numerator belongs to the answer only if the sign is ≤ or ≥, and a zero of the denominator never does. The results table lists these steps for your own problem.

Worked examples

Linear: 13 − 7x ≥ 10x − 4 gives (−∞, 1]. Quadratic: x² − x − 12 ≥ 0 gives (−∞, −3] ∪ [4, ∞); the zeros are −3 and 4. Rational: 4x/(x − 6) < 1 gives (−2, 6). Absolute value: |4x − 5| > 6 gives (−∞, −1/4) ∪ (11/4, ∞). A repeated factor: x² − 2x + 1 > 0 gives (−∞, 1) ∪ (1, ∞). Each is a published textbook problem or a variant of one, and the answers are also tested by evaluating the inequality at hundreds of sample points.

Common mistakes

Forgetting to flip the sign when multiplying or dividing by a negative number. Multiplying both sides by an expression that contains x, whose sign changes (use the subtract-and-factor method above). Dropping the roots that make the denominator zero. Writing an “or” answer as a single chain such as x < −2 and x > 3, which is not a valid chain: it is two pieces, written with the union sign ∪. Squaring both sides of an absolute value inequality without checking the sign of both sides.

Limits of this calculator

One unknown, polynomials up to degree 8 (the numerator and denominator after moving everything to one side), and at most one absolute value expression per line. Square roots, exponentials and trigonometric functions are not supported, and neither is a “not equal” sign. Roots that are not rational are shown as exact radicals when they come from a quadratic factor and as rounded decimals otherwise; the test-point check uses exact arithmetic either way. Type a fraction with brackets: (x + 1)/(x − 2), not x + 1/x − 2.

Sources: OpenStax, Algebra and Trigonometry 2e, section 2.7 “Linear Inequalities and Absolute Value Inequalities”; OpenStax, Intermediate Algebra 2e, section 7.6 “Solve Rational Inequalities” and section 9.8 “Solve Quadratic Inequalities” (problems from these sections are automated tests). Related: equation solver, quadratic formula calculator, polynomial calculator, asymptote calculator and linear programming calculator.