Graphs of Linear Inequalities

We have learned how to solve inequalities in one variable. Now we look at inequalities in two variables, which have many applications. If you ran a business, for example, you would want your revenue to be greater than your costs — so that your business would make a profit.

Verify solutions to an inequality in two variables

An inequality in one variable, like $x > 3$, has many solutions — any number greater than $3$ — shown on the number line by shading to the right of $3$ with an open circle at $3$. Similarly, an inequality in two variables has many solutions: any ordered pair $(x, y)$ that makes the inequality true when substituted in is a solution of the inequality.

Example. Determine whether each ordered pair is a solution to the inequality $y > x + 4$: (a) $(0, 0)$ (b) $(1, 6)$ (c) $(2, 6)$ (d) $(-5, -15)$ (e) $(-8, 12)$.

(a) Substituting $x = 0, y = 0$: is $0 > 0 + 4$? Since $0 \not> 4$, $(0, 0)$ is not a solution.

(b) Substituting $x = 1, y = 6$: is $6 > 1 + 4$? Since $6 > 5$ is true, $(1, 6)$ is a solution.

(c) Substituting $x = 2, y = 6$: is $6 > 2 + 4$? Since $6 \not> 6$, $(2, 6)$ is not a solution.

(d) Substituting $x = -5, y = -15$: is $-15 > -5 + 4$? Since $-15 \not> -1$, $(-5, -15)$ is not a solution.

(e) Substituting $x = -8, y = 12$: is $12 > -8 + 4$? Since $12 > -4$ is true, $(-8, 12)$ is a solution.

Recognize the relation between the solutions of an inequality and its graph

Just as the point $x = 3$ separates the number line into the numbers less than $3$ and the numbers greater than $3$, a line $y = x + 4$ separates the plane into two regions. On one side of the line are the points with $y < x + 4$; on the other side are the points with $y > x + 4$. We call the line $y = x + 4$ a boundary line.

For an inequality in one variable, the endpoint is shown with a parenthesis (not included) or a bracket (included). Similarly, for an inequality in two variables, the boundary line is drawn solid or dashed to show whether it is included in the solution.

Points on one side of the boundary line $y = x + 4$ are solutions to $y > x + 4$, and points on the other side are solutions to $y < x + 4$. Any point on the boundary line itself, where $y = x + 4$, is not a solution to $y > x + 4$, so the boundary line is not part of the solution — we draw it dashed. The shaded region shows the solutions to $y > x + 4$.

Example. The boundary line shown is $y = 2x - 1$, drawn as a solid line. Write the inequality shown by the graph.

We test the point $(0, 0)$: is $0 > 2(0) - 1$, or is $0 < 2(0) - 1$? Since $0 > -1$ is true, $(0, 0)$ is on the side of the line where $y > 2x - 1$. Since the boundary line is solid, the inequality includes the equal sign, so the graph shows $y \geq 2x - 1$.

Graph linear inequalities

Now we put this together to graph linear inequalities.

Example. Graph the linear inequality $y \geq \tfrac{3}{4}x - 2$.

We graph the boundary line $y = \tfrac{3}{4}x - 2$. Since the inequality is $\geq$, we draw a solid line. We test $(0, 0)$: is $0 \geq \tfrac{3}{4}(0) - 2$? Since $0 \geq -2$ is true, $(0, 0)$ is a solution, so we shade the side of the boundary line that includes $(0, 0)$.

Example. Graph the linear inequality $x - 2y < 5$.

We graph the boundary line $x - 2y = 5$, drawn dashed since the inequality is $<$. Testing $(0, 0)$: is $0 - 2(0) < 5$? Since $0 < 5$ is true, we shade the side that includes $(0, 0)$.

If the boundary line passes through the origin, $(0, 0)$ cannot be used as a test point — choose any other point not on the line instead.

Example. Graph the linear inequality $y \leq -4x$.

The boundary line $y = -4x$ is in slope-intercept form with $m = -4$ and $b = 0$; since it passes through the origin, we choose a different test point, such as $(1, 0)$. The inequality is $\leq$, so we draw a solid line. Testing $(1, 0)$: is $0 \leq -4(1)$? Since $0 \not\leq -4$, $(1, 0)$ is not a solution, so we shade the side of the boundary line that does not include $(1, 0)$.

Some linear inequalities have only one variable — an $x$ but no $y$, or a $y$ but no $x$. As with equations, the boundary line is then either a vertical line, $x = a$, or a horizontal line, $y = b$.

Example. Graph the linear inequality $y > 3$.

The boundary line $y = 3$ is horizontal, drawn dashed since the inequality is $>$. Testing $(0, 0)$: is $0 > 3$? Since this is false, $(0, 0)$ is not a solution, so we shade the side that does not include $(0, 0)$ — the region above the line.

Key terms

linear inequality — an inequality that can be written as $Ax + By > C$, $Ax + By \geq C$, $Ax + By < C$, or $Ax + By \leq C$, where $A$ and $B$ are not both zero. solution of a linear inequality — an ordered pair $(x, y)$ that makes the inequality true when substituted in. boundary line — the line $Ax + By = C$ that separates the plane into the region where $Ax + By > C$ and the region where $Ax + By < C$; drawn dashed when strict ($<$ or $>$) and solid when the inequality includes equality ($\leq$ or $\geq$).

Practice

Verify solutions to an inequality in two variables

Recognize the relation between the solutions of an inequality and its graph

Graph linear inequalities

This section is adapted from Elementary Algebra 2e, Section 4.7: Graphs of Linear Inequalities by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the boundary-line and shaded-region figures as accessible inline SVGs; condensed the worked examples; omitted the Be Prepared quiz, Media links, Self Check checklist, and unselected Section Exercises; converted the practice problems (“Try Its”) into interactive exercises with instant feedback; and adapted selected end-of-section exercises into the section-final interactive Practice block, using categorical graph descriptions where the interactive graph component cannot represent shaded half-planes.