Inequality to Interval Notation Calculator

The one rule that reverses an inequality

Multiplying or dividing both sides by a NEGATIVE number reverses the inequality. That is the whole rule, and it is the only step in a linear solve that can change the direction.

-3x >= 12: dividing by −3 turns ≥ into ≤, giving x ≤ −4 and the interval (−∞, −4]. 5 - 2x > 11 is the same rule with a constant in the way — move the 5 first, then divide by −2 and reverse, giving x < −3.

Why it is true is worth one line: 2 < 3, but multiply both by −1 and you get −2 > −3. Negating reflects the number line, and a reflection swaps left and right. Nothing about inequalities is special here; it is what negation does to order.

The tool marks the step that does it rather than warning about it in general, because “remember to flip” is advice and “this step flipped it, here” is working.

What does NOT flip, including two things that look like they should

Adding and subtracting never change the direction, whatever their sign. 2x + 3 < 7 solves to x < 2 with no reversal at any step — No step here divided or multiplied by a negative, so the direction never changed. That is worth noticing: the flip is a rule about negatives, not about inequalities in general.

Two cases look like they should flip and do not:

• Subtracting a bigger number. Going from x + 3 < 1 to x < −2 involves a negative result, not a negative multiplier. No flip.
• The variable being on the right. 7 > 2x + 1 solves to x < 3. Turning a < b into b > a is re-reading the same statement, not multiplying by anything — and this tool does it that way round deliberately, so it never reports a reversal the reader did not make.

That second point matters on this page in particular. A solver that moves the variable leftwards would divide by a negative and announce a flip on 7 > 2x + 1, which is arithmetically defensible and pedagogically wrong: it would cry wolf on the one page whose subject is the flip.

A chain is two conditions, joined by AND

-1 < 2x + 3 <= 7 means −1 < 2x + 3 AND 2x + 3 ≤ 7. Both must hold, so the answer is the overlap: (−2, 2].

You can work on a chain by operating on all three parts at once, and that is usually quicker. The tool splits it because the split shows what the chain actually claims — and because it makes the failure mode visible: if a step divides by a negative, BOTH inequalities reverse and the chain reads the other way round.

The other thing a chain settles is what it can never mean. A chain always says AND. There is no way to write an OR condition as a single chain, and 5 < x < 1 does not mean “x < 1 or x > 5” — it describes the empty set, because nothing is both above 5 and below 1.

Two inequalities worked to interval notation, one of them a chain

Both are the calculator’s own output, pinned as tests in its engine suite, so the working here cannot drift from what the tool computes.

3(2 − x) ≤ 4x + 13

Expand the bracket first, multiplying both terms inside it: 3 × 2 = 6 and 3 × (−x) = −3x, so the inequality is −3x + 6 ≤ 4x + 13. Subtract 4x from both sides to get −7x + 6 ≤ 13, then subtract 6 to get −7x ≤ 7. Neither subtraction changes the direction.

The last step divides by −7, and that is the flip: ≤ becomes ≥, giving x ≥ −1. The endpoint is included, so its bracket is square and infinity’s is round: [−1, ∞).

Check with two substitutions. x = 0 is inside the answer: 3 × 2 = 6 and 6 ≤ 13 holds. x = −2 is outside it: 3 × 4 = 12 against −8 + 13 = 5, and 12 ≤ 5 fails. One point holds and one fails, which is what a correct answer produces.

The same check catches the usual slip in the first line. Distribute only to the 6, writing 6 − x instead of 6 − 3x, and the working ends at x ≥ −7/5. Try x = −1.2, which that answer includes: 3 × 3.2 = 9.6 on the left, 8.2 on the right, and 9.6 ≤ 8.2 is false.

−4 ≤ 5 − 3x < 11

A chain is two conditions joined by AND: −4 ≤ 5 − 3x, and 5 − 3x < 11. Solve each. The left half, read the other way round, is −3x + 5 ≥ −4; subtracting 5 gives −3x ≥ −9, and dividing by −3 flips it to x ≤ 3. The right half gives −3x < 6, and the same division flips it to x > −2.

Both must hold, so the answer is the overlap, −2 < x ≤ 3, written (−2, 3]. Notice the chain now reads the other way round: the 3 came from the left side of the original and ends up on the right, because dividing by a negative reversed both halves at once.

The calculator checks it with x = 0.5, which holds (5 − 1.5 = 3.5 sits between −4 and 11), and x = −3, which fails (5 + 9 = 14 is not below 11).

Slips between the working and the interval

• Multiplying only the first term in a bracket. It turned x ≥ −1 into x ≥ −7/5 above. The substitution check exposes it at once.
• Flipping one half of a chain. Dividing by −3 reverses both inequalities. Reverse only one and the result, x ≥ 3 with x > −2 or similar, no longer describes the original.
• Endpoints in the wrong order. (3, −2] is not an interval. The smaller number always comes first, whichever side of the chain it started on.
• Mismatched brackets. ≤ and ≥ give a square bracket, < and > a round one, and infinity is always round because no number reaches it.

Absolute values: inside gives one interval, outside gives two

|expression| < c says the inside is within c of zero. That is one interval, and it is why |2x - 1| < 5 solves to (−2, 3).

Reverse the inequality and you get the outside instead: |x + 3| >= 2 gives (−∞, −5] ∪ [−1, ∞), two rays going opposite ways with a union between them.

So the direction of the inequality decides the SHAPE of the answer, not just its endpoints. The commonest error is keeping one interval when the sign flips, which produces an answer that contains nothing at all — and the two-substitution check below catches it immediately.

A negative bound is the degenerate case and worth recognising: |x| < −1 has no solutions because an absolute value is never negative, and |x| ≥ −1 has every solution for the same reason.

When the variable cancels

Sometimes the x disappears, and what is left is a statement about numbers that is either always true or never true.

3x + 1 > 3x + 5 reduces to a false statement, so the answer is ∅ — nothing satisfies it. 2x < 2x + 1 reduces to a true one, so the answer is (−∞, ∞), every real number.

Both are proper answers rather than signs of a mistake, and both are worth recognising by shape: identical coefficients of x on the two sides mean the variable will cancel, and only the constants decide the outcome.

The exact arithmetic matters here too. 4x <= 2x + 9 solves to x ≤ 9/2 — the endpoint is a fraction, and it stays one. Rounding it to 4.5 loses nothing here but does the moment such an endpoint feeds another step.

Two substitutions, and what they catch

Substitute -5 into the original and it should hold; substitute -5 and it should not. Two substitutions catch a reversed inequality immediately, which is why they are worth doing.

This is the cheapest possible check and it catches the expensive error. A reversed inequality gives an answer that is exactly wrong — every value that should work fails and every value that should fail works — so a single substitution inside the claimed answer exposes it at once.

It also catches an endpoint on the wrong side of a bracket, which a re-read of the working often does not: substitute the endpoint itself, and if the original holds with equality then the bracket should be square.

Once the answer is solved and checked, converting between the written forms is a separate and purely notational job — the Interval Notation Calculator does that in every direction, and the number-line page compares several answers on one scale.

Sources and methodology

The order axioms that make the sign rule true are standard; the reference below covers the notation used for the answer.

• NIST Digital Library of Mathematical Functions — §1.1, Notation and Special Sets
• How we calculate — methodology

Method. Every figure on this page comes from src/lib/inequality-to-interval-notation.ts, which solves over exact rationals and hands the result to the same interval-set representation the notation page uses — so the answers on the two pages are the same objects. When the variable sits only on the right the engine turns the statement round rather than moving the variable leftwards, precisely so it never reports a sign flip the reader would not have made. That engine is verified on every change against 66 hand-written assertions, including that a flip is reported exactly when a step divided by a negative, and that a value inside the returned set always satisfies the original inequality while one outside does not. The count and the per-case breakdown are published on the formula verification page.

Related calculators

Where this goes next:

More in Math, or browse all calculators.

Read the guide

Solving an inequality and writing the answer are two different jobs, and this page does the first. The Interval Notation Calculator converts a finished answer between all four written forms, and for a quadratic — where isolation does not work and a sign chart does — see Which Method Should You Use to Solve a Quadratic?