Absolute Value Inequality Calculator — solve and write the interval

This absolute value inequality calculator solves inequalities of the form |ax + b| compared to a constant, and returns the answer as an interval, a union of two intervals, all real numbers, or the empty set. It also prints the two cases it split the problem into, so you can follow the reasoning rather than copying a result. Every field recalculates live, and a test box lets you drop a value of x in and confirm the answer is the right way round.

Arb Digital publishes this because absolute-value inequalities are where a lot of otherwise confident algebra students first go wrong, and the mistake is almost never arithmetic. The difficulty is structural: the same symbol produces two completely different shapes of answer depending on which way the inequality points, and the standard shortcut people remember covers only one of those shapes. Seeing both cases written out every time is a much better habit than memorising a rule that works half the time.

What This Absolute Value Inequality Calculator Does

It takes a linear expression inside absolute value bars and a comparison against a single number, then produces the exact solution set. Under the hood it finds the two critical values where the expression inside the bars equals plus or minus the right-hand constant, sorts them, and then decides whether the solution lies between them or outside them based on the direction of the inequality. It reports whether each endpoint is included, which is the difference between a strict inequality and a non-strict one, and it handles every degenerate case rather than silently returning nonsense.

That is a strictly larger job than our absolute value equation calculator, which handles equalities and returns at most two isolated values of x. An equation asks where the expression hits a level; an inequality asks where it stays below or above that level, and the answer is a region rather than a point. This page is the one to use whenever the relation is anything other than an equals sign. If you already have the region and only need to write it down correctly, our interval notation calculator handles the notation side on its own.

How to Use It

1. Rearrange your inequality first so the absolute value stands alone on the left. Anything added, subtracted or multiplied outside the bars must be moved before you start, and remember that multiplying or dividing by a negative flips the inequality sign.
2. Enter a and b. These are the coefficient and constant inside the bars. For |x + 4|, a is 1 and b is 4.
3. Choose the comparison from the dropdown and enter the constant c on the right-hand side.
4. Read the interval in the headline, then read the working panel underneath to see the two cases the problem was split into.
5. Test a value. Pick a number that should be in the solution and one that should not, and check the fourth grid item agrees with you both times.

The Method and How It Is Calculated

An absolute value measures distance from zero, and every absolute-value inequality is really a statement about distance. |ax + b| < c says the expression sits within c of zero. |ax + b| > c says it sits further than c from zero. Reading it that way makes both answer shapes obvious before any algebra happens: "within" is a band, and "further than" is everything outside a band.

Formally, for a positive c the less-than case unfolds into a double inequality, −c < ax + b < c, which is then solved as a single chain. The greater-than case unfolds into a disjunction, ax + b < −c or ax + b > c, which is solved as two separate inequalities joined by the word "or". Those two patterns are set out with worked examples in Paul Dawkins' absolute value inequalities notes from Lamar University, which is the clearest short treatment available.

Take the default. |2x − 3| < 7 becomes −7 < 2x − 3 < 7. Add 3 to all three parts: −4 < 2x < 10. Divide all three by 2: −2 < x < 5. In interval notation that is (−2, 5), with round brackets because the inequality is strict and the endpoints themselves are excluded. Check it: x = 0 gives |−3| = 3, which is indeed less than 7, and x = 6 gives |9| = 9, which is not. Change the comparison to greater-than and the same two critical values produce (−∞, −2) ∪ (5, ∞) instead — the complement, exactly as the distance reading predicts.

The Cases Nobody Teaches: When c Is Zero or Negative

Most textbook exercises use a positive constant on the right, which lets students get away with a rule that breaks the moment c is zero or negative. An absolute value is never negative, so these cases resolve without any algebra at all, and this calculator handles each of them explicitly.

If c is negative, |ax + b| < c has no solutions whatsoever, because nothing non-negative can be smaller than a negative number. The answer is the empty set. In the same situation |ax + b| > c is satisfied by every real number, because every absolute value is already larger than any negative. If c is exactly zero, the greater-than case excludes precisely one point — the x where the inside becomes zero — while the greater-than-or-equal case admits every real number. The less-than case with c equal to zero is empty, and the less-than-or-equal case collapses to that single point. Five inequalities that look almost identical produce five genuinely different answers, and a memorised shortcut gets most of them wrong.

Why the Two Answer Shapes Are Opposites

There is a symmetry worth internalising: the solution set of a less-than inequality and the solution set of the matching greater-than inequality are complements. Together they cover the whole number line, and they overlap only at the critical values themselves. This is a fast and reliable self-check. Solve the less-than version, take everything you did not include, and you have the greater-than version.

The practical use of the check is catching sign errors. If a is negative, the algebra involves dividing by a negative number and flipping the inequality signs, which is where most errors creep in. Enter a as −2 instead of 2 in the calculator above and watch what happens: the two critical values swap position on the number line, but the solution set for the less-than case is still the interval between them. That is because the tool sorts the critical values before deciding which region to report, and it is why the answer stays correct even when the intermediate algebra looks upside down.

Reading the Result as a Tolerance

Outside the classroom, absolute-value inequalities are almost always tolerance statements. A machined part must be within 0.05 mm of a 40 mm target, which is |x − 40| ≤ 0.05. A batch of drinks must be within 5 ml of 330 ml. A page must load within 200 ms of a target time. In each case the target is the midpoint of the interval and the tolerance is its half-width, so the solution comes out as target plus or minus tolerance without any need to think about cases at all.

That reading also runs backwards, which is the more useful direction. Given an interval such as [37.5, 42.5], you can write it as an absolute-value inequality by taking the midpoint, 40, and the half-width, 2.5: |x − 40| ≤ 2.5. Being able to move between the two forms is what makes this topic worth learning rather than merely passing. Quality-control limits, confidence intervals and error bounds are all written in one form and used in the other, and our margin of error calculator uses precisely this midpoint-and-half-width structure on survey data.

When the Absolute Value Is Not Alone

The tool solves the standard form, so anything else needs a rearrangement first. That rearrangement is where an inequality can quietly flip. If you have 3|x + 1| − 4 ≤ 11, add 4 to get 3|x + 1| ≤ 15 and divide by 3 to get |x + 1| ≤ 5, which is now enterable as a of 1, b of 1, c of 5. Dividing by a positive number leaves the direction alone, so nothing flips here.

Now consider −2|x − 1| > −8. Dividing both sides by −2 flips the sign, giving |x − 1| < 4. Skipping that flip produces the exact complement of the right answer, which is the single most common error on this topic and one that looks perfectly plausible on paper. Two more cases resist the standard form entirely: an inequality with an absolute value on both sides, and one where a variable appears on the right. Both need squaring or a sign-based case analysis rather than this template, and the wider Solving Equations and Inequalities chapter at Lamar University covers those alongside rational and polynomial inequalities. The same limitation applies once the expression inside the bars is quadratic — for that, start with our quadratic equation solver to locate the roots first.

Common Mistakes to Avoid

• Using "and" when the inequality points the other way — less-than gives one interval joined by "and", greater-than gives two joined by "or", and swapping them inverts the whole answer.
• Forgetting to flip when dividing by a negative — this applies to the coefficient inside the bars and to any negative factor you clear from outside them.
• Assuming a negative right-hand side is impossible — it is perfectly legal, and it produces either the empty set or all real numbers depending on the direction.
• Mixing up open and closed brackets — strict inequalities exclude the endpoints and use round brackets, non-strict ones include them and use square brackets.
• Solving before isolating the bars — anything outside the absolute value must be cleared first, or the two cases you generate will not correspond to the original problem.

Related Free Tools From Arb Digital

Solve the equality version with the absolute value equation calculator, write a solution set correctly with the interval notation calculator, handle a squared expression with the quadratic inequality calculator, work with straight lines using the slope calculator, or run the arithmetic with our scientific calculator. The full free online tools hub lists every algebra tool we publish.

Frequently Asked Questions

This page explains an algebraic method for educational purposes. Check any rearrangement you made before entering values, because a sign flipped outside the bars changes the answer entirely.