Graph linear inequalities in two variables on the coordinate plane and find the solution set of a system of linear inequalities as the overlap of the half-planes (TN A1.A.REI.D.8, A1.A.REI.D.9).

What this topic is asking

Standard A1.A.REI.D.8 asks you to graph a linear inequality in two variables, whose solution is a half-plane (one side of a boundary line). A1.A.REI.D.9 extends this to a system of inequalities, whose solution is the overlap of the half-planes. The graded skills are choosing a solid or dashed boundary, shading the correct side, and identifying points in the solution region.

Graphing one inequality

Two decisions follow from the inequality.

• Boundary type. $\le$ or $\ge$ gives a solid line (boundary points are solutions); $<$ or $>$ gives a dashed line.
• Which side to shade. Solve for $y$ if helpful: $y >$ shades above the line, $y <$ shades below. Or test a point off the line, the origin is easiest, and shade the side where the inequality is true.

Systems of inequalities

For a system, graph each inequality on the same axes, then the solution region is where all the shaded areas overlap. The boundaries can be solid or dashed independently. The overlap may be a wedge, a strip, a bounded polygon, or empty if the half-planes do not share any region.

How TNReady examines this topic

• Multiple choice. Choose the correct boundary type and shading, or identify which point is a solution to a system.
• Graphing. Draw the boundary (solid or dashed) and shade the half-plane or overlap.
• Multiple select. Choose all points lying in the solution region.

A clarifying idea is that testing a point is the most reliable method on the calculator subparts: pick a simple coordinate, substitute it into every inequality, and the point is a solution only if all of them are satisfied. This single check resolves most multiple-choice items quickly.

Why the overlap, and why test points work

A system of inequalities means every condition must hold at once, so the solution is the intersection of the individual solution sets, not their union. Geometrically that is the region covered by all the shading at the same time. Test points work because a linear inequality divides the plane into exactly two half-planes, and the boundary is the only place where the expression switches from "true" to "false." Therefore any single point strictly inside a half-plane represents the truth value of the whole region: if the origin makes $2x + y \le 4$ true, every point on the origin's side does too. This is why one well-chosen test point settles the shading for an entire half-plane, and why checking a candidate against all the inequalities settles membership in a system's region.

Graphing a system step by step

A system item asks you to combine two half-planes, and a clear routine prevents shading mistakes.

A handy partial-credit habit on graphing items is to label one point you have verified inside the region, since it demonstrates you found the correct overlap even if the shading is hard to read.

Try this

Q1. Is the boundary of $y > 3x - 2$ solid or dashed, and do you shade above or below? [2 points]

• Cue. Dashed (strict $>$); shade above ($y$ greater than the line).

Q2. Is $(0, 0)$ a solution of $\begin{cases} y \le x + 1 \\ y \ge -2 \end{cases}$? [1 point]

• Cue. $0 \le 1$ true and $0 \ge -2$ true, so yes.