Solve it first — with the reversal marked at the step that causes it — then write the interval.
Solve it, then write the interval
With the sign flip marked where it actually happens.
Use <= and >= for ≤ and ≥. Linear, chained, or an absolute value. Fractions stay exact.
2x + 3 < 7 — read as a linear inequality
(−∞, 2)
x < 2
Solved
x < 2
the inequality, isolated
Interval notation
(−∞, 2)
1 piece
Set-builder
{x ∈ ℝ | x < 2}
the same set as a condition
Sign flip needed?
no
nothing was multiplied by a negative
The working, step by step
As given — 2x + 3 < 7 The inequality to solve.
Subtract 3 from both sides — 2x < 4 Still no change of direction — that only happens on multiplication and division.
Divide both sides by 2 — x < 2 2 is positive, so the direction is unchanged.
The answer on a number line
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.
Substitute 1 into the original and it should hold; substitute 1 and it should not. Two substitutions catch a reversed inequality immediately, which is why they are worth doing.
Endpoints stay exact: dividing 5 < 2x by 2 gives 5/2, not 2.5, and the fraction goes straight into the bracket.
The solution is (−∞, 2) — x < 2.
What this tool shows
Solve first, then write the interval — with the reversal marked at the step that causes it.
Linear, chained and absolute-value inequalities solved step by step
The reversal marked at the step that causes it — and its absence stated
Exact endpoints: 5/2, not 2.5
The answer in interval, inequality and set-builder form, plus a number line
The sign flip, marked where it happens Exact fractional endpoints All four written forms of the answer Free, no signup
Free, no signup — exact endpoints, and every step shown.
Updated 7 September 2026 · Works in any browser, no installation
Only one operation reverses an inequality: multiplying or dividing by a negative number. -3x >= 12 becomes x ≤ −4 — the ≥ turns into ≤ at the division step — and in interval notation that is (−∞, −4].
At a glance
Formula shown
Adding and subtracting never change the direction. Multiplying or dividing by a negative always does. Nothing else in a linear solve can.
Scenario support
Linear inequalities in one variable, single, chained, or inside absolute-value bars.
Educational estimate
Planning support from the values you enter — not professional advice.
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.
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.
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 55 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:
Interval NotationConvert between inequality, interval notation, set-builder form and a number-line graph in every direction, including absolute values, unions and the empty set.
Graphing Inequalities on a Number LinePlot up to six inequalities on one shared number line, read the overlap and the union straight off it, and see which conditions conflict and which are doing no work at all.
Graphing Quadratic InequalitiesSolve a quadratic inequality by sign chart rather than by four memorised cases, and shade the region for the two-variable form, with the repeated-root case where all four relations differ.
System of EquationsSolve linear systems in up to six unknowns with exact fractions — row operations, the matrix form, Cramer’s rule, and the parametric family when there is no single answer.
Substitution MethodSolve simultaneous equations by substitution, with the unknown to isolate chosen and explained, the bracketed substitution as its own line, and the expansion after it.
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?
Educational use disclaimer
This calculator solves linear inequalities in one variable — single, chained, or inside absolute-value bars — and writes the solution in interval notation. All arithmetic is exact, so an endpoint of 5/2 stays a fraction. A quadratic inequality needs a sign chart rather than isolation and is refused by name, with a pointer to the page that handles it.
Published the inequality-to-interval page: it SOLVES, which the interval notation page says outright that it does not.
The sign flip is marked at the step that causes it rather than warned about in prose, and a solve that never needed one says so — because the flip is a rule about negatives, not about inequalities.
When the variable sits only on the right the engine turns the statement round rather than moving the variable leftwards, so it never reports a reversal the reader would not have made.
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