Graphs of Linear Inequalities and Systems — SAT Math

Introduction

A linear inequality carves the xy-plane into two halves. Rather than asking which x makes an inequality true (that's the job of the sibling note Solving and Modeling Linear Inequalities), this skill asks: given a graph of shaded regions and boundary lines, can you match it to an inequality — and can you tell whether a specific point satisfies a system? These questions appear regularly in the Algebra domain, which makes up about 35% of SAT Math questions.

Core Concept

From equation to inequality graph: three decisions

When you graph a linear inequality in two variables, every point on the plane is either a solution or not. The graph has three features you must control:

The boundary line — graph $y = mx + b$ (or the equivalent) exactly as you would any line.
Solid vs. dashed — if the inequality is $\leq$ or $\geq$ , draw a solid line (points on it are solutions). If it is $<$ or $>$ , draw a dashed line (boundary excluded).
Which half-plane to shade — substitute a test point not on the line. If it satisfies the inequality, shade that side; if not, shade the other.

Key idea

Quick test-point trick: Use $(0, 0)$ whenever the line doesn't pass through the origin. Plug in and check.

Rewriting to "slope-intercept form for inequalities"

When the inequality is not already solved for $y$ , isolate $y$ — but flip the inequality sign when you multiply or divide by a negative number.

3x - y > 6 \;\Longrightarrow\; -y > -3x + 6 \;\Longrightarrow\; y < 3x - 6

Now you can read directly: dashed line, slope 3, y-intercept $-6$ , shading below.

Systems of linear inequalities

The solution to a system of two (or more) linear inequalities is the intersection of their individual solution regions — the set of all points that satisfy every inequality simultaneously. On a graph it is the doubly-shaded overlap region. A point is a solution to the system if and only if it satisfies all inequalities.

To check a point algebraically: substitute its coordinates into each inequality one at a time. If even one fails, the point is not a solution.

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