An equation in two variables draws a line. Replace the equals sign with an inequality sign and the answer swells from a line into half of the plane — and a whole system of inequalities is answered not by a point but by a region: the patch where every shaded half agrees. The one trick that makes it easy is that a single test point decides an entire half of the plane. Move the sliders, tap anywhere on the grid to test a point, and watch the overlap form.
Two half-planes, one overlap
The edge: in or out?
The routine never changes, no matter how many inequalities there are. For each one: (1) swap the inequality sign for an equals sign and graph that boundary line; (2) make it dashed for < or >, solid for ≤ or ≥; (3) pick any test point that is not on the line — the origin is easiest whenever the line misses it — and substitute; (4) if the statement comes out true, shade the side your test point is on; if false, shade the other side. Then the solution of the whole system is the region where every shading overlaps.
Why one point can speak for a whole half-plane. Fix an x. As you slide straight up, your y grows while mx + b sits still, so the difference y − (mx + b) grows steadily and passes through zero exactly once — right at the line. Above the line that difference is always positive; below it, always negative. It can never sneak back across without crossing the line. So the statement cannot change from true to false anywhere inside one side, and one test point settles the entire half.
Four ways this section is misread. 1. Shading toward the inequality sign. “It says ≥, so shade up” only works once the inequality is solved for y with a positive coefficient on y. Given −2y ≥ 4x + 6, dividing by −2 reverses the sign to y ≤ −2x − 3, and the shading goes down. Test points never lie about this; sign-watching does. 2. Testing a point that sits on the boundary. For y ≥ 2x the origin gives 0 ≥ 0, which is true — but the origin is on the line, so it belongs to both sides and decides nothing. Move to (1, 0) or (0, 1) and try again. 3. Turning in the overlap of the boundaries instead of the overlap of the regions. The corner point where two boundary lines cross is only one point of the answer. The answer is the whole shaded patch — usually infinitely many points, often unbounded. 4. Forgetting that a system can have no solution at all.y ≥ x + 2 together with y ≤ x − 3 asks for a y that is at least x + 2 and at most x − 3 — so x + 2 ≤ x − 3, which simplifies to 2 ≤ −3. That is false for every x, so the shadings never meet and the solution set is empty.
move a slider to begin
Inequality 1
slope m11
intercept b11
Inequality 2
slope m21
intercept b21
Tap anywhere on the grid to move the test point. It snaps to whole-number coordinates so every check stays exact.