Graphing Linear Inequalities in Two Variables

From a line to a region

A linear equation in two variables draws a line, and that line is the picture of every solution.

$$y = x + 1$$

Change the equals sign to an inequality and far more points qualify:

$$y \ge x + 1$$

Now $(0, 5)$ works, since $5 \ge 1$. So does $(0, 2)$ and $(-3, 4)$. None of them sits on the line.

The solutions fill a half-plane — everything on one side of the line, plus the line itself.

• y = x + 1

Why one side and not the other

Pick any $x$. The line gives one particular $y$, namely $x + 1$. Every $y$ above that is greater, and every $y$ below is smaller.

So $y \ge x + 1$ holds for the point on the line and everything vertically above it. Do that for every $x$ and the solutions sweep out the whole region above the line.

Below the line, $y$ is smaller than $x + 1$, so the inequality fails at every point. There is no mixing: the line separates the plane into a side where the inequality always holds and a side where it never does.

That is why shading a region is a complete answer rather than a rough one.

Solid or dashed

The boundary itself is a separate question from the shading.

Sign	Boundary	Reason
$\le$ or $\ge$	solid	points on the line satisfy the inequality
$<$ or $>$	dashed	points on the line give equality, which is excluded

For $y > x + 1$, the point $(0, 1)$ gives $1 > 1$, which is false. The line is not part of the solution, and the dashes say so.

• y = x + 1

The test point

Rather than reasoning about which side is “above”, test a point.

1. Draw the boundary line.
2. Pick any point not on the line.
3. Substitute it into the inequality.
4. True means shade that point’s side. False means shade the other side.

$(0, 0)$ is the point to use whenever the line misses the origin, because substituting zeros is a single glance.

Shade $2x + 3y < 6$.

Test $(0, 0)$:

$$2(0) + 3(0) = 0 < 6 \quad\text{true}$$

So shade the side containing the origin, with a dashed boundary since the sign is strict.

The test point works in any form. Reading the sign directly only works once $y$ is alone on the left, and rearranging is where sign errors creep in — dividing by a negative flips the sign.

Systems of inequalities

A system asks for points satisfying every inequality at once. Shade each half-plane, and the solution is where they overlap.

$$y \le -x + 6 \qquad y \ge x - 2$$

Each condition removes half the plane. What survives both is a wedge.

• y = -x + 6
• y = x - 2

Every corner of the overlap is an intersection of two boundary lines, which is why solving systems of equations is the tool for finding them. Here the two lines meet where $-x + 6 = x - 2$, giving $x = 4$ and the corner $(4, 2)$.

An overlap can also be empty. Two parallel lines shaded away from each other share nothing, and the system has no solutions.

Worked examples

Common mistakes

Practice problems