Graph a linear inequality in two variables as a half-plane, using a solid or dashed boundary and shading the correct side (LA A1: A-REI.D.12).

What this topic is asking

Standard A1: A-REI.D.12 asks you to graph a linear inequality in two variables: draw the boundary line, decide whether it is solid or dashed, and shade the half-plane of solutions. On LEAP 2025 these are Type I Major Content items, often graphing items or multiple choice about the boundary and shading. The solution is a whole region, not a line or a point.

Step one: the boundary line

Graph the line you get by replacing the inequality with an equals sign. Its style records whether the line itself is part of the solution:

• Solid line for $\le$ or $\ge$: points on the line are solutions (the boundary is included).
• Dashed line for $<$ or $>$: points on the line are not solutions (the boundary is excluded).

This mirrors the open-versus-closed circle from one-variable inequalities, in two dimensions the circle becomes a whole line.

Step two: shade the correct half-plane

The line splits the plane into two halves; one half is the solution set. Use a test point to decide which.

If the test point makes the inequality false, shade the other side.

Why a test point is reliable

Rather than memorizing "above for greater, below for less," substitute a point. A point not on the line is either a solution or not; that single check fixes the entire region, because every point on the same side behaves the same way. The origin $(0, 0)$ is the easiest test point whenever the line does not pass through it.

How LEAP examines this topic

• Graphing item. Draw the boundary (solid or dashed) and shade the half-plane.
• Multiple choice. Identify the correct line style and shading direction.
• Test-point item. Decide whether a given point is a solution by substitution.

A clarifying idea: solving for $y$ first can make shading intuitive, $y >$ shades above the line and $y <$ shades below, but only once $y$ is isolated. Until then, a test point is safer.

Why the solution is a half-plane

A two-variable linear inequality has a half-plane of solutions because the boundary line is the exact set where the two sides are equal, and crossing that line is what flips the comparison from true to false. Every point on one side makes the left side larger than the right; every point on the other side makes it smaller; only on the line are they equal. So the line is a clean dividing wall, and one entire side is the solution set. This is why a single test point determines everything: the inequality cannot change truth value without crossing the boundary, so all points on the same side share the same status. The line's style then records the boundary's own membership, included for $\le$ and $\ge$ (a closed edge), excluded for $<$ and $>$ (an open edge). Understanding the half-plane as "one side of an equality wall" is exactly what you need for systems of inequalities, where two or more half-planes overlap to form a feasible region, the shared solution set of several constraints at once.

Try this

Q1. When graphing $y \ge 3x - 2$, is the boundary solid or dashed, and do you shade above or below? [2 points]

• Cue. Solid (because $\ge$); shade above (test $(0, 0)$: $0 \ge -2$ true, origin is above).

Q2. Is $(2, 0)$ a solution to $x + y < 1$? [1 point]

• Cue. $2 + 0 = 2$, and $2 < 1$ is false, so no.