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All courses Math I · A-REI.12 8 of 59
Graph linear inequalities and systems of linear inequalities as half-plane solution regions.

Graph one linear inequality in slope-intercept form

Problem
For y≤2x+1, enter boundary equation, style, and shading direction.
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Hint

Graph the boundary line y = 2x + 1 first, then decide whether it should be solid or dashed.

Because the inequality uses <=, the boundary line is included, so it must be solid. Values less than or equal to y = 2x + 1 are below the line.

Solution walkthrough

01

Draw the boundary line

\[y~=~2x~+~1\]

Use the line y = 2x + 1 as the boundary for the inequality.

02

Choose solid or dashed

\[\le\]

Because the inequality includes equals, points on the line are part of the solution set, so the boundary is solid.

03

Choose the shading

\[y~\le~2x~+~1\]

The inequality says y is less than or equal to the line's value, so shade below the line.

04

State the graph description

\[\text{boundary}~=~y~=~2x~+~1;~\text{style}~=~\text{solid};~\text{shading}~=~\text{below}~\text{or}~\text{on}\]

The graph has a solid boundary line y = 2x + 1 and shading below the line.

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Another way

  1. You can test a point like (0, 0): since 0 <= 1 is true, the region containing (0, 0) should be shaded.

!

Common mistake

A common mistake is to use a dashed line even though <= includes the boundary.

Coordinate graph showing the solid boundary line for y less than or equal to 2x plus 1, labeled to show the solution region is below the line.
Solid line y = 2x + 1; shade below the line.