DWG NO. 7.2 — Lesson 2 of 5

Systems of Inequalities

Unit 7: Systems of Equations and Matrices · ~20 min

Objective Graph systems of linear inequalities and identify the region of points that satisfies every inequality simultaneously.

Lesson 7.1 solved for the single point where two lines cross. Replace one of the equal signs with < or >, and the picture changes: instead of a line, a linear inequality describes an entire half of the plane. A system of inequalities asks where all of those half-planes overlap — the answer is a region, not a point.

Graphing a single linear inequality

Boundary line Treat the inequality as an equation and graph that line. Make it solid for ≤ or ≥ (the boundary is included) and dashed for < or > (the boundary is excluded).
Test point Pick any point not on the line — (0, 0) is usually easiest. Substitute it into the original inequality.
Shade If the test point makes the inequality true, shade the half-plane containing it. If false, shade the other side.

Systems of inequalities

Graph each inequality in the system on the same axes, using a different shading direction for each. The solution set of the system is only the region shaded by every inequality — the overlap. If the boundary lines intersect, the corner points of that overlapping region are found exactly the way Lesson 7.1 found a system's solution: by solving the boundary equations as a system.

The shaded overlap satisfies both inequalities; the dashed edge is excluded, the solid edge is included

Worked Example 1 · Testing a point against a system

ProblemDoes (1, 1) satisfy the system y ≤ −x + 8 and y > 2x − 2?

1Check the first inequality: 1 ≤ −(1) + 8 = 7. True.

2Check the second inequality: 1 > 2(1) − 2 = 0. True.

(1, 1) satisfies both inequalities, so it lies in the solution region.

Worked Example 2 · Finding a vertex of the solution region

ProblemFind the corner point where the boundaries y = −x + 8 and y = 2x − 2 meet.

1Set the expressions equal: −x + 8 = 2x − 2.

2Solve: 10 = 3x → x = \(\frac{10}{3}\) ≈ 3.33.

3y = 2(\(\frac{10}{3}\)) − 2 = \(\frac{20}{3}\) − 2 = \(\frac{14}{3}\) ≈ 4.67.

Vertex ≈ (3.33, 4.67) — note this point is on the dashed boundary, so it is not itself part of the solution set.

Guided practice

1.Is the boundary line of y ≥ 3x − 1 solid or dashed?
Show answer

≥ includes the boundary, so the line is solid.
2.Using the test point (0, 0), which side of x + y < 6 should be shaded?
Show answer

Substitute: 0 + 0 < 6 is true, so shade the side of the line containing the origin.
3.Does (2, 0) satisfy the system y ≤ x + 1 and y ≥ −x + 1?
Show answer

First: 0 ≤ 2 + 1 = 3, true. Second: 0 ≥ −2 + 1 = −1, true. Both hold, so yes, (2, 0) is in the solution region.
4.Does (0, 5) satisfy the system y < x + 2 and y > 0?
Show answer

First: 5 < 0 + 2 = 2 is false. Since one inequality fails, (0, 5) is not in the solution region — it doesn't matter that the second inequality holds.
5.Find the vertex where the boundaries of y = x + 1 and y = −x + 1 meet.
Show answer

Set equal: x + 1 = −x + 1 → 2x = 0 → x = 0, so y = 1. Vertex: (0, 1).

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