Graph Inequalities Calculator

Solving an inequality gives a set of answers, and a graph is the honest way to display a set. Which picture you draw depends on how many variables are present.

• One variable ($-1 < x \le 3$) lives on a number line: a segment with an open circle at an excluded endpoint and a filled circle at an included one.
• Two variables ($y \le 2x - 1$) lives on the coordinate plane: a boundary line with one entire side shaded. Every point in the shaded region — thousands of them — satisfies the inequality.

The boundary itself carries information. A solid line means the boundary points are solutions, which happens with $\le$ and $\ge$. A dashed line means they are not, which happens with the strict $<$ and $>$. Drawing the wrong style is the same error as writing a bracket where a parenthesis belongs.

Three steps, always in this order.

1. Draw the boundary

Replace the inequality sign with $=$ and graph that line. Use the intercepts, or slope-intercept form if the equation is already solved for $y$.

2. Choose solid or dashed

3. Shade using a test point

Pick any point not on the boundary — $(0,0)$ whenever the line misses the origin, because the arithmetic is trivial. Substitute it into the original inequality.

• True → shade the side containing that point.
• False → shade the other side.

This test is why you never need to remember rules like "$\le$ means shade below". Those rules only hold once the inequality has been solved for $y$, and they fail the moment the coefficient of $y$ is negative — the very case where a division flipped the sign.

• Forgetting the flip. Dividing $-2y > 4$ by $-2$ gives $y < -2$. Miss this and you shade the wrong half of the plane.
• Testing a point on the boundary. $(0,0)$ is useless for $3x + 2y > 0$; it makes the inequality an equality. Choose $(1,1)$ instead.
• Mixing up circle types. Open circle for $<$ and $>$, filled circle for $\le$ and $\ge$.
• Shading both sides of a system. With two inequalities the answer is only the overlap; shade lightly and outline the common region.
• Assuming the shading is always below for $\le$. True only after isolating $y$ with a positive coefficient.