Graph a linear inequality in two variables as a half-plane with the correct boundary line and shading (NC.M1.A-REI, A-CED.3).

What this topic is asking

This topic asks you to graph a linear inequality in two variables (like $y \le 2x + 1$) as a half-plane: a boundary line plus a shaded region of all $(x, y)$ that satisfy the inequality. It supports the systems and modeling standards (A-CED.3 constraints), since a constraint is often an inequality whose solution is a region.

The boundary line: solid or dashed

The first decision is the line style, set by the symbol.

So $y < x + 2$ has a dashed line $y = x + 2$, while $y \ge x + 2$ has a solid one.

Choosing the side with a test point

After drawing the boundary, decide which half-plane to shade.

When the line passes through the origin, choose a different test point, such as $(1, 0)$.

Reading the half-plane

The shaded region is the complete solution set: every point in it satisfies the inequality, and no point outside it does. A solid boundary adds the line itself to the solution; a dashed boundary excludes it. This is the two-variable analog of the shaded ray you get for a one-variable inequality on a number line.

How the NC Math 1 EOC examines this topic

• Technology-enhanced. Draw the boundary (solid or dashed) and shade the correct side with a graphing tool.
• Multiple choice. Choose the graph that matches an inequality, or identify whether a point is a solution.
• Calculator-active. Often paired with a context where the region represents allowed options.

Graphing one inequality is the building block for modeling with systems and the boundary uses the same skills as graphing linear equations.

Why a region, not a line, is the solution

An equation in two variables, like $y = 2x + 1$, is satisfied only by points exactly on the line. An inequality loosens that to "on one side," so its solution swells from a line to a whole half-plane. The boundary line marks the dividing edge: on one side the inequality is true, on the other it is false, and the test point tells you which. This is why a single test point settles the entire region, every point on the same side behaves identically. Seeing the solution as a region also explains constraints in modeling: "spend at most \50$" or "use no more than$8$ hours" each carve out a half-plane of feasible choices, and overlapping several of them gives the feasible system.

Try this

Q1. Is the boundary of $y > 3x - 1$ solid or dashed? [1 point]

• Cue. Dashed, because $>$ is strict (boundary not included).

Q2. For $x + y \ge 5$, test whether $(0, 0)$ is a solution. [1 point]

• Cue. $0 + 0 = 0 \ge 5$ is false, so $(0, 0)$ is not a solution; shade the other side.