# 5.6: Graphical Solutions to Systems of Inequalities

**Advanced**Created by: CK-12

**Practice**Graphs of Systems of Linear Inequalities in Two Variables

Graph the following system of linear inequalities on the same Cartesian grid.

\begin{align*}\begin{Bmatrix} y>-\frac{1}{2}x+2 \\ y \le 2x-3 \end{Bmatrix}\end{align*}

What is the **solution** to this system?

### Graphical Solutions to Systems of Inequalities

When a linear inequality is graphed it appears as a shaded region on the Cartesian plane. Therefore, when a system of linear inequalities is graphed, you will see two shaded regions that most likely overlap in some places. The solution to a system of linear inequalities is this ** region** of intersection. The solution to a system of linear inequalities is also known as the

**feasible region**.

#### Solve by graphing

Solve the following system of linear inequalities by graphing:

\begin{align*}\begin{Bmatrix} -2x-6y \le 12\\ -x +2y > -4 \end{Bmatrix}\end{align*}

Write each inequality in slope-intercept form.

\begin{align*}& -2x-6y \le 12\\ & -2x{\color{red}+2x}-6y \le {\color{red}2x}+12\\ & -6y \le {\color{red}2x}+12\\ & \frac{-6y}{{\color{red}-6}} \le \frac{2x}{{\color{red}-6}} + \frac{12}{{\color{red}-6}}\\ & \frac{\cancel{-6}y}{\cancel{-6}} \le -\frac{2}{6}x+\frac{\overset{{\color{red}-2}}{\cancel{12}}}{\cancel{-6}} \quad \text{Simplify the slope to lowest terms.} \ \boxed{-\frac{2}{6}=-\frac{1}{3}}\\ & \boxed{y \ {\color{red}\ge} \ -\frac{1}{3}x-2}\end{align*}

\begin{align*}& -x+2y > -4\\ & -x{\color{red}+x}+2y > {\color{red}x}-4\\ & 2y > {\color{red}x}-4\\ & \frac{2y}{{\color{red}2}} > \frac{x}{{\color{red}2}} -\frac{4}{{\color{red}2}}\\ & \frac{\cancel{2}y}{\cancel{2}} > \frac{1}{2}x - \frac{\overset{{\color{red}2}}{\cancel{4}}}{\cancel{2}}\\ & \boxed{y > \frac{1}{2}x-2}\end{align*}

**Graph: **\begin{align*}y \ge -\frac{1}{3}x-2\end{align*}.

The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.

\begin{align*}-2x-6y &\le 12\\ -2({\color{red}1})-6({\color{red}1}) &\le 12\\ {\color{red}-2-6} &\le 12\\ {\color{red}-8} &\le 12 \quad \text{Is it true?}\end{align*}

Yes, negative eight is less than or equal to twelve. The point (1, 1) satisfies the inequality and will lie within the shaded region.

Now graph \begin{align*}y>\frac{1}{2}x-2\end{align*} on the same Cartesian grid.

The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.

\begin{align*}-x+2y &> -4\\ -({\color{red}1})+2({\color{red}1}) &> -4\\ {\color{red}-1} + {\color{red}2} &> -4\\ {\color{red}1} &> -4 \quad \text{Is it true?}\end{align*}

Yes, one is greater than negative four. The point (1, 1) satisfies the inequality and will lie within the shaded region.

The region that is indicated as the **feasible region** is the area on the graph where the shading from each line overlaps. This region contains all the points that will satisfy both inequalities.

The **feasible region** is the area shaded in blue.

#### Solve by graphing

Solve the following system of linear inequalities by graphing:

\begin{align*}\begin{Bmatrix} x \ge 1\\ y > 2 \end{Bmatrix}\end{align*}

These are the special lines that need to be graphed. The first line is a line that has an undefined slope. The graph is a vertical line parallel to the \begin{align*}y\end{align*}-axis. The second line is a line that has a slope of zero. The graph is a line parallel to the \begin{align*}x\end{align*}-axis.

**Graph:** \begin{align*}x \ge 1\end{align*}.

*On the graph, every point to the right of the vertical line has an \begin{align*}x\end{align*}-value that is greater than one. Therefore, the graph must be shaded to the right of the vertical line.*

Now graph \begin{align*}y>2\end{align*} on the same Cartesian grid.

*On the graph, every point above the horizontal line has a \begin{align*}y\end{align*}-value that is greater than two. Therefore, the graph must be shaded above the horizontal line.*

The region that is indicated as the **feasible region** is the area on the graph where the shading from each line overlaps. This region contains all the points that will satisfy both inequalities.

The **feasible region** is the area shaded in purple.

#### Solve by graphing

More than two inequalities can be shaded on the same Cartesian plane. The solution set is all of the coordinates that lie within the shaded regions that overlap. When more than two inequalities are being shaded on the same grid, the shading must be done accurately and neatly. Solve the following system of linear inequalities by graphing:

\begin{align*}\begin{Bmatrix} y < x+1\\ y \ge -2x+4\\ y > 0 \end{Bmatrix}\end{align*}

Graph \begin{align*}y<x+1\end{align*}:

The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.

\begin{align*}y &< x+1\\ ({\color{red}1}) &< ({\color{red}1}) +1\\ {\color{red}1} &< {\color{red}1}+1\\ 1 &< {\color{red}2} \quad \text{Is it true?}\end{align*}

Yes, one is less than two. The point (1, 1) satisfies the inequality and will lie within the shaded region.

\begin{align*}y \ge -2x+4\end{align*}

\begin{align*}y &\ge -2x+4\\ ({\color{red}1}) &\ge -2({\color{red}1})+4\\ {\color{red}1} &\ge {\color{red}-2}+4\\ 1 &\ge {\color{red}2} \quad \text{Is it true?}\end{align*}

No, one is not greater than or equal to two. The point (1, 1) does not satisfy the inequality and will not lie within the shaded region.

\begin{align*}y>0\end{align*}

The graph of \begin{align*}y>0\end{align*} is a horizontal line along the \begin{align*}x\end{align*}-axis. Every point above the horizontal line has a \begin{align*}y\end{align*}-value greater than zero. Therefore, the shaded area will be above the graphed line.

The region that is indicated as the **feasible region** is the area on the graph where the shading from each line overlaps. This region contains all the points that will satisfy both inequalities.

The **feasible region** is the area on the right shaded in pink.

### Examples

#### Example 1

Earlier, you were asked to graph a set of inequalities.

\begin{align*}\begin{Bmatrix} y>-\frac{1}{2}x+2 \\ y \le 2x-3 \end{Bmatrix}\end{align*}

Both inequalities are in slope-intercept form. Begin by graphing \begin{align*}\boxed{y > -\frac{1}{2}x+2}\end{align*}.

\begin{align*}& y > -\frac{1}{2}x+2\\ & ({\color{red}1}) > -\frac{1}{2}({\color{red}1})+2\\ &{\color{red}1} > {\color{red}-\frac{1}{2}}+2\\ & \boxed{1>1\frac{1}{2}} \quad \text{Is it true?}\end{align*}

No, one is not greater than one and one-half. Therefore, the point (1, 1) does not satisfy the inequality and will not lie in the shaded area. The shaded area is above the dashed line.

Now, the inequality \begin{align*}y \le 2x-3\end{align*} will be graphed on the same Cartesian grid.

\begin{align*}& y \le 2x-3\\ & ({\color{red}1}) \le 2({\color{red}1})-3\\ & {\color{red}1} \le {\color{red}2}-3\\ & {\color{red}1} \le {\color{red}-1}\\ & \boxed{1 \le -1} \quad \text{Is it true?}\end{align*}

No, one is not less than or equal to negative one. Therefore, the point (1, 1) does not satisfy the inequality and will not lie in the shaded area. The shaded area is below the solid line.

The region that is indicated as the **feasible region** is the area on the graph where the shading from each line overlaps. This region contains all the points that will satisfy both inequalities. Another way to indicate the feasible region is to shade the entire region a different color. If you were to do this exercise in your notebook using colored pencils, the feasible region would be very obvious.

The **feasible region** is the area shaded in yellow.

#### Example 2

Solve the following system of linear inequalities by graphing:

\begin{align*}\begin{Bmatrix} 4x+5y \le 20\\ 3x + y \le 6\\ \end{Bmatrix}\end{align*}

\begin{align*}\begin{Bmatrix}
4x+5y \le 20\\
3x + y \le 6\\
\end{Bmatrix}\end{align*}

**Write each inequality in slope-intercept form.**- \begin{align*}& 4x+5y \le 20 && 3x+y \le 6\\ &4x{\color{red}-4x}+5y \le {\color{red}-4x}+20 && 3x{\color{red}-3x}+y \le {\color{red}-3x}+6\\ &5y \le {\color{red}-4x}+20 && y \le {\color{red}-3x}+6\\ & \frac{5y}{{\color{red}5}} \le \frac{-4x}{{\color{red}5}}+\frac{20}{\color{red}5} && \boxed{y \le-3x+6}\\ & \frac{\cancel{5}y}{\cancel{5}} \le -\frac{4}{5}x+\frac{\overset{{\color{red}4}}{\cancel{20}}}{\cancel{5}}\\ & \boxed{y \le -\frac{4}{5}x+4}\end{align*}
**Graph:**\begin{align*}y \le -\frac{4}{5}x+4\end{align*}- \begin{align*}4x+5y &\le 20\\ 4({\color{red}1})+5({\color{red}1}) &\le 20\\ {\color{red}4}+{\color{red}5} &\le 20\\ {\color{red}9} &\le 20 \quad \text{Is it true?}\end{align*}
- Yes, nine is less than or equal to twenty. The point (1, 1) satisfies the inequality and will lie within the shaded region.
**Graph:**\begin{align*}y \le -3x+6\end{align*}- \begin{align*}3x+y &\le 6\\ 3({\color{red}1})+({\color{red}1}) &\le 6\\ {\color{red}3}+{\color{red}1} &\le 6\\ {\color{red}4} &\le 6 \quad \text{Is it true?}\end{align*}
- Yes, four is less than or equal to six. The point (1, 1) satisfies the inequality and will lie within the shaded region.
- The
**feasible region**is the area shaded in purple.

#### Example 3

Solve the system of linear inequalities by graphing:

\begin{align*}\begin{Bmatrix} 2x+y \le 8\\ 2x+3y < 12\\ x \ge 0\\ y \ge 0\\ \end{Bmatrix}\end{align*}

\begin{align*}\begin{Bmatrix}
2x+y \le 8\\
2x + 3y < 12 \\
x \ge 0\\
y \ge 0
\end{Bmatrix}\end{align*}

**Write the first two inequalities in slope-intercept form.**- \begin{align*}& 2x+3y <12 && 2x+y \le 8\\ & 2x{\color{red}-2x}+3y < {\color{red}-2x}+12 && 2x{\color{red}-2x}+y \le {\color{red}-2x}+8\\ & 3y < {\color{red}-2x}+12 && y \le {\color{red}-2x}+8\\ & \frac{3y}{{\color{red}3}} < \frac{-2x}{{\color{red}3}} + \frac{12}{{\color{red}3}} && \boxed{y \le -2x+8}\\ & \frac{\cancel{3}y}{\cancel{3}} < -\frac{2}{3}x + \frac{\overset{{\color{red}4}}{\cancel{12}}}{\cancel{3}}\\ & \boxed{y < -\frac{2}{3}x+4}\end{align*}
**Graph:**\begin{align*}y<-\frac{2}{3}x+4\end{align*}- \begin{align*}2x+3y &< 12\\ 2({\color{red}1})+3({\color{red}1}) &< 12\\ {\color{red}2}+{\color{red}3} &< 12\\ {\color{red}5} &< 12 \quad \text{Is it true?}\end{align*}
- Yes, five is less than twelve. The point (1, 1) satisfies the inequality and will lie within the shaded region.
**Graph:**\begin{align*}y \le -2x+8\end{align*}- \begin{align*}2x+y &\le 8\\ 2({\color{red}1})+({\color{red}1}) &\le 8\\ {\color{red}2}+{\color{red}1} &\le 8\\ {\color{red}3} &\le 8 \quad \text{Is it true?}\end{align*}
- Yes, three is less than or equal to eight. The point (1, 1) satisfies the inequality and will lie within the shaded region.
**Graph:**\begin{align*}x \ge 0\end{align*}**The graph will be a vertical line that will coincide with the \begin{align*}y\end{align*}-axis.**All \begin{align*}x\end{align*}-values to the right of the line are greater than or equal to zero. The shaded area will be to the right of the vertical line.**Graph:**\begin{align*}y \ge 0\end{align*}**The graph will be a horizontal line that will coincide with the \begin{align*}x\end{align*}-axis**. All \begin{align*}y\end{align*}-values above the line are greater than or equal to zero. The shaded area will be above the horizontal line.- The
**feasible region**is the area shaded in green.

#### Example 4

Determine and prove three points that satisfy the following system of linear inequalities:

\begin{align*}\begin{Bmatrix} y < 2x+7\\ y \ge -3x-4\\ \end{Bmatrix}\end{align*}

\begin{align*}\begin{Bmatrix}
y < 2x+7\\
y \ge -3x-4 \\
\end{Bmatrix}\end{align*}

- Graph the system of inequalities to determine the feasible region.
- Three points in the feasible region are (–1, 3); (4, –2); and (6, 5). These points will be tested in each of the linear inequalities. All of these points should satisfy both inequalities.
- Test (–1, 3)
- \begin{align*}& y < 2x+7 \qquad \qquad \qquad and && y \ge -3x-4\\ & y < 2x+7 && y \ge -3x-4\\ & ({\color{red}3})<2({\color{red}-1})+7 && ({\color{red}3}) \ge -3({\color{red}-1})-4\\ & {\color{red}3} < {\color{red}-2}+7 && {\color{red}3} \ge {\color{red}3}-4\\ & 3 < {\color{red}5} && 3 \ge {\color{red}-1}\end{align*}
- The point (–1, 3) satisfies both inequalities. In the first inequality, three is less than five. In the second inequality three is greater than or equal to negative one. Therefore, the point lies within the feasible region and is a solution for the system of linear inequalities.
- Test (4, –2)
- \begin{align*}& y < 2x+7 && y \ge -3x-4\\ & ({\color{red}-2}) < 2({\color{red}4})+7 && ({\color{red}-2}) \ge -3({\color{red}4})-4\\ & {\color{red}-2}<{\color{red}8}+7 && {\color{red}-2} \ge {\color{red}-12}-4\\ & -2 < {\color{red}15} && -2 \ge {\color{red}-16}\end{align*}
- The point (4, –2) satisfies both inequalities. In the first inequality, negative two is less than fifteen. In the second inequality negative two is greater than or equal to negative sixteen. Therefore, the point lies within the feasible region and is a solution for the system of linear inequalities.
- Test (6, 5)
- \begin{align*}& y < 2x+7 && y \ge -3x-4\\ & ({\color{red}5})< 2({\color{red}6})+7 && ({\color{red}5}) \ge -3({\color{red}6}) -4\\ & {\color{red}5} < {\color{red}12}+7 && {\color{red}5} \ge {\color{red}-18}-4\\ & 5 < {\color{red}19} && 5 \ge {\color{red}-22}\end{align*}
- The point (6, 5) satisfies both inequalities. In the first inequality, five is less than nineteen. In the second inequality five is greater than or equal to negative twenty-two. Therefore, the point lies within the feasible region and is a solution for the system of linear inequalities.

### Review

\begin{align*}\begin{Bmatrix} 3x+5y>15\\ 2x-7y \le 14\\ \end{Bmatrix}\end{align*}

- Solve the above system of linear inequalities by graphing.
- Determine three points that satisfy the system of linear inequalities.

\begin{align*}\begin{Bmatrix} 3x+2y \ge 10\\ x-y < -1\\ \end{Bmatrix}\end{align*}

- Solve the above system of linear inequalities by graphing.
- Determine three points that satisfy the system of linear inequalities.

\begin{align*}\begin{Bmatrix} x-y > 4\\ x+y > 6\\ \end{Bmatrix}\end{align*}

- Solve the above system of linear inequalities by graphing.
- Determine three points that satisfy the system of linear inequalities.

\begin{align*}\begin{Bmatrix} y>3x-2\\ y < -2x+5\\ \end{Bmatrix}\end{align*}

- Solve the above system of linear inequalities by graphing.
- Determine three points that satisfy the system of linear inequalities.

\begin{align*}\begin{Bmatrix} 3x-6y > -6\\ 5x+9y \ge -18\\ \end{Bmatrix}\end{align*}

- Solve the above system of linear inequalities by graphing.
- Determine three points that satisfy the system of linear inequalities.

\begin{align*}\begin{Bmatrix} 2x-y<4\\ x \ge -1\\ y \ge -2 \end{Bmatrix}\end{align*}

- Solve the above system of linear inequalities by graphing.
- Determine three points that satisfy the system of linear inequalities.

\begin{align*}\begin{Bmatrix} 2x+y>6\\ x +2y \ge 6\\ x \ge 0\\ y \ge 0 \end{Bmatrix}\end{align*}

- Solve the above system of linear inequalities by graphing.
- Determine three points that satisfy the system of linear inequalities.

\begin{align*}\begin{Bmatrix} x \le 3\\ x \ge -2\\ y \le 4\\ y \ge -1 \end{Bmatrix}\end{align*}

- Solve the above system of linear inequalities by graphing.
- Determine three points that satisfy the system of linear inequalities.

\begin{align*}\begin{Bmatrix} x+y > -1\\ 3x -2y \ge 2\\ x < 3\\ y \ge 0 \end{Bmatrix}\end{align*}

- Solve the above system of linear inequalities by graphing.
- Determine three points that satisfy the system of linear inequalities.

\begin{align*}\begin{Bmatrix} y < x+1\\ y \ge -2x+3\\ y > 0 \end{Bmatrix}\end{align*}

- Solve the above system of linear inequalities by graphing.
- Determine three points that satisfy the system of linear inequalities.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 5.6.

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Feasible region

The feasible region is the inner region within a set of four inequalities on a graph.Linear Inequality

Linear inequalities are inequalities that can be written in one of the following four forms: , or .Slope-Intercept Form

The slope-intercept form of a line is where is the slope and is the intercept.### Image Attributions

Here you'll learn to solve a system of linear inequalities by graphing.