The solution of a system of linear inequalities is represented on a graph as the overlap of the shaded regions that represent the solutions of the separate inequalities.

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**system of inequalities**is a set of inequalities that are used at the same time. A**solution of a system of inequalities**is a set of values of the variables in the system that make every inequality in the system true. The solution set of a system of inequalities is often an infinite set of points. The solution set can be represented by graphing the inequalities in the system on the same coordinate plane and identifying the region where their solutions overlap. In some instances, a system of inequalities may have no solution. On a graph of a system of inequalities with no solution, the solutions of the inequalities in the system do not overlap.Step-By-Step Example

Graphing a System of Linear Inequalities

Solve the system of inequalities.

$\begin{cases}\begin{aligned}y &\leq x+8 \\ 5x-y &< -4 \end{aligned}\end{cases}$

Step 1

Solve the inequalities for $y$ .

The first inequality is already solved for $y$.$y \leq x+8$

$\begin{aligned}5x-y&<-4\\-5x+5x-y&<-5x-4\\-y&<-5x-4\\-1(-y)&>-1(-5x-4)\\y&>5x+4\end{aligned}$

Step 2

Graph the boundary lines.

Use a solid line for:$y \leq x+8$

$y>5x+4$

Solution

The inequality symbol in $y \leq x+8$ is $\leq$, so shade below the boundary line.

The inequality symbol in $y > 5x+4$ is $\gt$, so shade above the boundary line.

The solution is the region where the two shaded areas overlap. Check the solution by selecting a test point from the shaded region that is not on either boundary line. Then substitute the coordinates into both inequalities to verify that each is true.The point $(0, 6)$ is in the overlap of the shaded areas and is not on either line.

First inequality:$\begin{aligned}y&\leq x + 8\\(6)&\stackrel{?}{\leq} (0)+8\\6&\leq 8 & \checkmark\end{aligned}$

$\begin{aligned}5x-y&<-4\\5(0)-(6)&\stackrel{?}{<}-4\\0-6&\stackrel{?}{<}-4\\-6&<-4 & \checkmark\end{aligned}$

Step-By-Step Example

Graphing a System of Linear Inequalities with No Solution

Solve the system of inequalities.

$\begin{cases}\begin{aligned}y &< 4x-4 \\ 4x-y &\leq -6 \end{aligned}\end{cases}$

Step 1

Solve the inequalities for $y$.

The first inequality is already solved for $y$.$y<4x-4$

$\begin{aligned}4x-y&\leq -6\\-4x+4x-y&\leq -4x-6\\-y&\leq -4x-6\\-1(-y)&\geq -1(-4x-6)\\y&\geq 4x+6\end{aligned}$

Step 2

Graph the boundary lines. The lines have the same slope but different $y$-intercepts. This means that they are parallel, so they do not intersect.

Use a dashed line for $y \lt 4x-4$.

Use a solid line for $y \geq 4x+6$.Solution

The inequality symbol in $y < 4x-4$ is $\lt$, so shade below the boundary line.

The inequality symbol in $y \geq 4x+6$ is $\geq$, so shade above the boundary line.

The shaded areas do not overlap, so there is no solution.