How do I know which side of the line to shade?

Test a point by substituting it into the original inequality. If the point satisfies the inequality and it is below the line, shade below; if it satisfies the inequality and it is above the line, shade above. If the test point does not satisfy the inequality, shade the opposite side from where it is located.

Does the boundary line itself get shaded?

Only if the inequality includes equality: \leq or \geq mean the boundary line is part of the solution, shown with a solid line. Strict inequalities exclude the line itself, shown with a dashed line to indicate 'points on this line are NOT included.'

What if the line passes through the origin, so I cannot test that point?

Pick a different test point like (1, 0), (0, 1), or (2, 1)—any point not on the line works. This way you can still determine which region satisfies the inequality without worrying about points on the boundary.

What if the y-intercept is negative, like y > -2x - 3?

The boundary line still uses the same method: it goes through (0, -3) on the y-axis (below the origin), and you use the slope m = -2 to find another point. Test any point like the origin to determine which side to shade—the process is identical to positive y-intercepts.

Graphing Linear Inequalities

Graphing linear inequalities shows which coordinate points satisfy a condition like y > mx + b by shading solution regions with appropriate line styles.

y > mx + b \implies \text{shade above a dashed line}

Solve a problem with Graphing Linear Inequalities

Type the problem. The solver will use Graphing Linear Inequalities where Graphing Linear Inequalities is the right tool, and tell you when it is not.

What each symbol means

What Graphing Linear Inequalities takes

Symbol	Meaning
$m$	The slope of the boundary line; positive values slope upward, negative slope downward. Confusing $m$ with $b$ (the y-intercept) produces the wrong line entirely.
$b$	The y-intercept; the line crosses the y-axis at point $(0, b)$ . Mixing up $b$ with the slope $m$ shifts your line to a completely different position.
$x$	The horizontal variable on the x-axis representing independent input. Swapping $x$ and $y$ reverses the coordinate system and flips the solution region.
$y$	The vertical variable on the y-axis compared to $mx + b$ in the inequality. Treating $y$ as independent instead of dependent inverts the inequality's meaning.

When to use it

Use this when you need to visualize all solutions to an inequality and see which region of the coordinate plane contains valid points.

Level

Usually taught in: Algebra I · Appears on: SAT

Worked examples

1. Graph a linear inequality with positive slope

Problem

Graph

y > 2x + 1

$y = 2x + 1$
The boundary line is found by replacing the inequality with an equals sign.
$(0, 1) \text{ and } (1, 3)$
The y-intercept is $(0, 1)$ ; with slope $m = 2$ , the next point is $(1, 3)$ by moving up 2 and right 1.
$0 \stackrel{?}{>} 2(0) + 1 = 0 \not> 1$
Test the point $(0, 0)$ : since $0 > 1$ is false, the origin is not in the solution region.
$\text{Dashed boundary, shade above}$
Use a dashed line because $>$ is strict (not $\geq$ ); shade above since regions above the line satisfy the inequality.

Answer: $\text{Region above the dashed line } y = 2x + 1$

The boundary line is dashed because the original inequality uses $>$ , not $\geq$ . We shade above the line because the test point $(0, 0)$ , which is below the line, does not satisfy the inequality.

2. Graph a linear inequality with negative slope and non-strict inequality

Problem

Graph

y \leq -\frac{1}{2}x + 2

$y = -\frac{1}{2}x + 2$
The boundary line is found by replacing the inequality with an equals sign.
$(0, 2) \text{ and } (2, 1)$
With slope $m = -\frac{1}{2}$ , starting from the y-intercept $(0, 2)$ , move right 2 and down 1 to reach $(2, 1)$ .
$0 \stackrel{?}{\leq} -\frac{1}{2}(0) + 2 = 0 \leq 2 \text{ (true)}$
Test the point $(0, 0)$ : since $0 \leq 2$ is true, the origin is in the solution region.
$\text{Solid boundary, shade below}$
Use a solid line because $\leq$ includes the boundary ( $=$ ). Shade below the line since the test point $(0, 0)$ , which is below it, satisfies the inequality.

Answer: $\text{Region below and on the solid line } y = -\frac{1}{2}x + 2$

The boundary line $y = -\frac{1}{2}x + 2$ is solid because $\leq$ includes points on the line itself. We test $(0, 0)$ and find it satisfies the inequality, so we shade the region below the line containing this point.

3. Graph a linear inequality from a real-world scenario

Problem

On a road trip, you drive at 60 mph for the first leg and 45 mph for the second leg, with a maximum of 7 hours total. If

x

is the time (hours) for the first leg and

y

is the time for the second leg, graph

x + y \leq 7

$x + y = 7$
The boundary line comes from replacing the inequality with an equals sign.
$y = -x + 7$
Solve for $y$ to get slope-intercept form: the slope is $m = -1$ and y-intercept is $b = 7$ .
$(0, 7) \text{ and } (7, 0)$
Plot the y-intercept at $(0, 7)$ and the x-intercept at $(7, 0)$ to draw the boundary line.
$0 + 0 \stackrel{?}{\leq} 7 \implies 0 \leq 7 \text{ (true)}$
Test the origin $(0, 0)$ : since the inequality is true, the origin is in the solution region.
$\text{Solid line, shade the region containing } (0, 0)$
Use a solid line because $\leq$ includes points on the boundary. Shade the region below and left of the line to show all time combinations within the 7-hour limit.

Answer: $x + y \leq 7 \text{ (with } x \geq 0 \text{ and } y \geq 0\text{)}$

The solution region represents all valid combinations of driving times that total at most 7 hours. The solid boundary line $x + y = 7$ represents the trip taking exactly 7 hours, and shading the region containing the origin shows all scenarios completing in less time.

Common mistakes

Where Graphing Linear Inequalities usually goes wrong

The mistake	Why it is wrong	The fix
Using a solid line for $y > 2x + 1$ instead of a dashed line.	Solid lines represent $\geq$ or $\leq$ (non-strict inequalities that include the boundary), while strict inequalities like $>$ or $<$ require dashed lines to show the boundary is excluded.	Check the inequality symbol: if it is $<$ or $>$ , draw a dashed line; if it is $\leq$ or $\geq$ , draw a solid line.
Shading below the line when the answer requires shading above, or vice versa.	Without testing a point, students guess the wrong region. It is tempting to assume $>$ always means shade above, but with negative slopes or other configurations, this leads to the wrong answer.	Always test a point not on the boundary line (such as the origin if it is not on the line) by substituting into the original inequality; shade the side where the test point is located if it satisfies the inequality, or shade the opposite side if it does not.
Attempting to identify the solution region without first graphing the boundary line.	The boundary line is the reference that separates the plane into regions; without it, the shading has no clear structure and you cannot determine which region is the solution.	Always start by writing the boundary line equation by replacing the inequality symbol with an equals sign, then graph that line first before deciding which side to shade.

Tips and when to use something else

Always test a point to determine shading direction; the origin $(0, 0)$ works unless it lies on the boundary line, in which case pick any other point like $(1, 0)$ or $(0, 1)$ .
Remember the line type: strict inequalities ( $<, >$ ) use dashed boundaries; non-strict ( $\leq, \geq$ ) use solid lines. This is the most common mistake students make.
For systems of inequalities, graph each inequality separately on the same plane, then shade only the region where all conditions overlap; this extends the technique to multiple inequalities at once.
If the inequality is given in standard form like $Ax + By > C$ , convert to slope-intercept form $y = ...$ first, since this makes finding the slope and y-intercept much easier.

Related formulas

Slope-Intercept FormSlope-intercept form directly reveals a line's slope and y-intercept, making it ideal for graphing and comparing linear relationships.Systems of InequalitiesFind the region satisfying both inequalities by graphing on the same axes; use when a problem has multiple constraints that must all be satisfied.Slope FormulaThe Slope Formula measures a line's steepness by dividing the vertical change between two points by their horizontal change.Parallel and Perpendicular SlopesFind whether two lines are parallel or perpendicular by comparing their slopes: identical for parallel, negative reciprocals for perpendicular.

Frequently asked questions

How do I know which side of the line to shade?: Test a point by substituting it into the original inequality. If the point satisfies the inequality and it is below the line, shade below; if it satisfies the inequality and it is above the line, shade above. If the test point does not satisfy the inequality, shade the opposite side from where it is located.
Does the boundary line itself get shaded?: Only if the inequality includes equality: $\leq$ or $\geq$ mean the boundary line is part of the solution, shown with a solid line. Strict inequalities $<$ or $>$ exclude the line itself, shown with a dashed line to indicate 'points on this line are NOT included.'
What if the line passes through the origin, so I cannot test that point?: Pick a different test point like $(1, 0)$ , $(0, 1)$ , or $(2, 1)$ —any point not on the line works. This way you can still determine which region satisfies the inequality without worrying about points on the boundary.
What if the y-intercept is negative, like $y > -2x - 3$ ?: The boundary line still uses the same method: it goes through $(0, -3)$ on the y-axis (below the origin), and you use the slope $m = -2$ to find another point. Test any point like the origin to determine which side to shade—the process is identical to positive y-intercepts.

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Reviewed 2026-09-18