Graphing Inequalities on a Number Line

One scale is the whole point

Drawing each condition on its own little line tells you what you already knew. Drawing them on ONE line, stacked and aligned, tells you something new: where they overlap, and by how much.

x > 2 and x < 7 gives (2, 7). You can see it because a value sits at the same horizontal position on every row, so the overlap is a vertical strip rather than something to compute.

That is why the tool fixes one scale across all rows rather than letting each self-scale to its own endpoints. A row scaled to fit itself looks identical whether it is x > 2 or x > 2000, and comparison becomes impossible.

Filled, hollow, and the arrow

Three conventions carry most of the marks in this topic, and all three are about what the drawing claims at an endpoint:

• A FILLED circle means the endpoint is included — that is ≤ or ≥, and a square bracket in interval notation.
• A HOLLOW circle means the endpoint is excluded — that is < or >, and a round bracket.
• An ARROW rather than a circle means the set runs on without an endpoint. Infinity is never circled, filled or otherwise, because it is not a number the set contains.
• The rows share one scale, so a value is at the same horizontal position on every one of them. That is what makes the overlap readable by eye.

The arrow is the one people get wrong most often, by drawing a circle at the far end of a ray. There is nothing there to circle: the set has no last element, and marking one claims an endpoint that does not exist. This is the same fact that makes [3, ∞] wrong in interval notation.

Reading AND off the picture

A point satisfies every condition exactly when every row is shaded above it. So the AND row is the vertical intersection, and it can be read without any algebra at all.

x >= -3, x <= 4, x != 0 gives [−3, 0) ∪ (0, 4] — a band with a single point removed from the middle, which is much easier to see drawn than to hold in your head from the three conditions.

|x| < 3 and x > 0 shows that the input form does not matter: an absolute value and a plain inequality land on the same scale as ordinary shaded regions, and their overlap is (0, 3).

Conditions that are doing no work

This is the check worth having and the one a combined answer hides. x > 2 and x > 5 gives (5, ∞) — which is just the second condition. The first adds nothing.

All 2 conditions hold together on (5, ∞). 1 of them is doing no work — removing it would give the same answer.

Drawn, it is obvious: one row completely contains the other, so requiring both removes no point that the tighter one had not already removed. All 4 conditions hold together on (3, 8). 2 of them are doing no work — removing them would give the same answer.

Spotting this matters outside homework. A set of constraints with redundant members is a set you can simplify, and knowing WHICH ones are redundant tells you which constraints are actually binding.

When the rows never meet

x < 1 and x > 5 gives ∅. The rows have no common point, so nothing satisfies all 2 conditions at once — the AND row is empty. That is an answer, not a failure to solve.

The picture makes the reason immediate: the two shaded regions are on opposite sides of a gap, with no column where both are shaded. No amount of algebra will produce a solution, because there is nothing to produce.

The tool names the conflicting PAIR rather than only reporting that the whole system is empty. With two conditions that is the same information; with five it is not, and knowing which two are irreconcilable is what tells you which one to relax.

OR is a different row, and usually a bigger one

The union row answers a different question: where does at least one condition hold? For x < 1 and x > 5 that is (−∞, 1) ∪ (5, ∞) — two pieces, even though the AND row is empty.

That contrast is the clearest illustration of the difference between the two operations, and it is why both rows are drawn. The same pair of conditions can have an empty intersection and a union covering most of the line.

For x > 2 and x < 7 the union is (−∞, ∞) — everything, because between them the two rays cover the whole line. The AND row for the same pair is the narrow band (2, 7). Same two conditions, opposite answers, and the picture shows both at once.

The set-theoretic versions of these operations, on finite sets rather than intervals, are on the Union and Intersection Calculator, which verifies the same De Morgan laws that govern them here.

Sources and methodology

The notation and the graphing conventions are standard; the reference below carries the canonical statements.

• 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/graphing-inequalities-number-line.ts, which parses each condition through the same converter the Interval Notation Calculator uses and computes the AND and OR rows with the same exact interval algebra. The drawn rows and the printed notation come from one set of objects, so they cannot drift apart. Redundancy is decided by actual containment rather than by comparing endpoints. That engine is verified on every change against 55 hand-written assertions, including that the AND row is contained in every individual row, that the OR row contains all of them, and that a condition reported as redundant can be removed without changing the intersection. The count and the per-case breakdown are published on the formula verification page.

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Read the guide

Graphing is for comparing; solving comes first. The Inequality to Interval Notation Calculator does the solving with the sign flip marked, and the Interval Notation Calculator converts a finished answer between all four written forms.