Graphing Systems of Linear Inequalities

The definition of a system of linear inequalities is very similar to the definition of a system of linear equations — but it has inequalities instead of equations. To solve a system of linear inequalities, we find the values of the variables that are solutions to both inequalities, and we show that solution set as a shaded region on a graph.

Determine Whether an Ordered Pair Is a Solution of a System of Linear Inequalities

A system of two linear inequalities looks like this:

To determine whether an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system.

Example. Determine whether each ordered pair is a solution to the system $\begin{cases} x + 4y \geq 10 \\ 3x - 2y < 12 \end{cases}$: (a) $(-2, 4)$ (b) $(3, 1)$.

(a) Substituting $x = -2, y = 4$ into both inequalities:

Both inequalities are true, so $(-2, 4)$ is a solution to the system.

(b) Substituting $x = 3, y = 1$:

The ordered pair $(3, 1)$ makes one inequality true but the other false, so it is not a solution to the system.

Solve a System of Linear Inequalities by Graphing

The solution to a single linear inequality is the region on one side of its boundary line containing all the points that make the inequality true. The solution to a system of two linear inequalities is a region containing the solutions to both inequalities. To find it, we graph each inequality separately on the same grid, then locate the region where the two shadings overlap.

Example. Solve the system by graphing: $\begin{cases} y \geq 2x - 1 \\ y < x + 1 \end{cases}$

We graph the boundary line $y = 2x - 1$ as a solid line, since the inequality is $\geq$. Testing $(0, 0)$: is $0 \geq 2(0) - 1$? Since $0 \geq -1$ is true, we shade the side containing $(0, 0)$ — above and to the left of the line.

On the same grid, we graph the boundary line $y = x + 1$ as a dashed line, since the inequality is $<$. Testing $(0, 0)$ again: is $0 < 0 + 1$? Since this is true, we shade the side containing $(0, 0)$ — below and to the right of this line.

The point where the two boundary lines cross, $(2, 3)$, is not itself part of the solution, since it does not satisfy $y < x + 1$. The solution to the system is the darker, doubly-shaded wedge below and to the left of that crossing point.

We check by choosing a test point in that wedge, such as $(-1, -1)$:

Sometimes the boundary lines of a system are parallel. Depending on the direction of the shading, a system like this may have no solution at all.

Example. Solve the system by graphing: $\begin{cases} 4x + 3y \geq 12 \\ y < -\tfrac{4}{3}x + 1 \end{cases}$

We graph the boundary line $4x + 3y = 12$ as a solid line (intercepts $x = 3$, $y = 4$). Testing $(0, 0)$: is $4(0) + 3(0) \geq 12$? Since this is false, we shade the side that does not contain $(0, 0)$.

We graph $y = -\tfrac{4}{3}x + 1$ as a dashed line — it has the same slope, $-\tfrac{4}{3}$, so it is parallel to the first boundary line. Testing $(0, 0)$: is $0 < 1$? Since this is true, we shade the side that does contain $(0, 0)$.

There is no point in both shaded regions, so this system has no solution.

Solve Applications of Systems of Inequalities

To solve an application of a system of inequalities, we translate each condition into an inequality, then graph the system to see the region that contains all the solutions. Many realistic situations restrict both variables to be positive, so the graph shows only Quadrant I.

Example. Christy sells photographs at a booth at a street fair. At the start of the day, she wants to display at least 25 photos. Each small photo she displays costs her $4 and each large photo costs her $10, and she doesn’t want to spend more than $200 on photos to display.

(a) Write a system of inequalities to model this situation. Let $x$ be the number of small photos and $y$ be the number of large photos. She wants at least 25 photos total, so $x + y \geq 25$. The cost — $4 per small photo plus $10 per large photo — must be no more than $200, so $4x + 10y \leq 200$:

(b) Graph the system. We graph $x + y = 25$ as a solid line and shade the side away from the origin (since $(0,0)$ fails $x + y \geq 25$). We graph $4x + 10y = 200$ as a solid line and shade the side containing the origin (since $(0,0)$ satisfies $4x + 10y \leq 200$). Because this is a real-world situation, we only graph Quadrant I, where $x \geq 0$ and $y \geq 0$.

The solution is the darker, doubly-shaded region bounded by the two lines and the axes.

(c) Could she display 10 small and 20 large photos? Testing $(10, 20)$: $10 + 20 = 30 \geq 25$ is true, but $4(10) + 10(20) = 40 + 200 = 240 \leq 200$ is false. Since $(10, 20)$ is not in the solution region, she could not display 10 small and 20 large photos.

(d) Could she display 20 small and 10 large photos? Testing $(20, 10)$: $20 + 10 = 30 \geq 25$ is true, and $4(20) + 10(10) = 80 + 100 = 180 \leq 200$ is also true. Since $(20, 10)$ is in the solution region, she could display 20 small and 10 large photos.

Key terms

system of linear inequalities — two or more linear inequalities grouped together. solutions of a system of linear inequalities — the ordered pairs $(x, y)$ that make every inequality in the system true, shown as a shaded region in the $x$-$y$ coordinate system where the shadings of the individual inequalities overlap.

This section is adapted from Elementary Algebra 2e, Section 5.6: Graphing Systems 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 overlapping shaded-region figures as accessible inline SVGs; condensed the worked examples; omitted the Be Prepared quiz, Media links, Self Check checklist, and Section Exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.