# 5.5: Graphs of Linear Inequalities

**Advanced**Created by: CK-12

**Practice**Linear Inequalities in Two Variables

A linear equation is a line when graphed. What about a linear inequality? Can you represent the following linear inequality in the Cartesian plane?

\begin{align*}2x-3y < 6\end{align*}

### Graphing Linear Inequalities

When a linear inequality is graphed on a grid, the solution will appear as a shaded area, as opposed to simply a straight line. The general form of a linear inequality with two variables is \begin{align*}ax+by>c\end{align*}

The inequality symbol \begin{align*}(>)\end{align*}

- All inequalities that have the symbol \begin{align*}(>)\end{align*}
(>) or \begin{align*}(<)\end{align*}(<) are graphed with a dashed line. - All inequalities that have the symbol \begin{align*}(\ge)\end{align*}
(≥) or \begin{align*}(\le)\end{align*}(≤) are graphed with a solid line.

Don't forget that when an inequality is divided or multiplied by a negative number, the direction of the inequality sign is reversed.

#### Let's practice by graphing the following inequalities:

- \begin{align*}3x+4y \le 12\end{align*}
3x+4y≤12

\begin{align*}& 3x+4y \le 12 && \text{Write the inequality in slope intercept form } (y=mx+b).\\
& 3x{\color{red}-3x}+4y \le {\color{red}-3x}+12\\
& 4y \le {\color{red}-3x}+12\\
& \frac{4y}{{\color{red}4}} \le \frac{-3x}{{\color{red}4}}+\frac{12}{{\color{red}4}}\\
& \frac{\cancel{4}y}{\cancel{4}} \le \frac{-3x}{4}+\frac{\overset{{\color{red}3}}{\cancel{12}}}{\cancel{4}}\\
& \boxed{y \le -\frac{3}{4}x+3}\end{align*}

The above graph represents the graph of the inequality before the solution set region is shaded. The line is a solid line because the inequality symbol is \begin{align*}(\le)\end{align*}

\begin{align*}3x+4y &\le 12\\
3({\color{red}1})+4({\color{red}1}) &\le 12 && \text{Substitute} \ (1, 1) \ \text{for} \ x \ \text{and} \ y \ \text{of the original inequality.}\\
{\color{red}3}+{\color{red}4} &\le 12 && \text{Evaluate the inequality.}\\
{\color{red}7} &\le 12 && \text{Is it true?}\end{align*}

Yes, seven is less than 12. The point (1, 1) satisfies the inequality. Therefore, the solution set is all of the area below the line that contains the point (1, 1). The ordered pair that satisfies the inequality will lie within the shaded region.

The solution set for the inequality is the entire shaded region shown in the graph. The solid line means that all of the points on the line will satisfy the inequality. In general, you will shade below the line if the inequality is of the form \begin{align*}y<mx+b\end{align*}

- \begin{align*}2x-4y < -12\end{align*}
2x−4y<−12

\begin{align*}& 2x-4y < -12 && \text{Write the inequality in slope intercept form} \ (y=mx+b).\\
& 2x{\color{red}-2x}-4y < {\color{red}-2x}-12\\
& -4y < {\color{red}-2x}-12\\
& \frac{-4y}{{\color{red}-4}} < \frac{-2x}{{\color{red}-4}} -\frac{12}{{\color{red}-4}}\\
& \frac{\cancel{-4}y}{\cancel{-4}}<\frac{-2x}{-4}-\frac{\overset{{\color{red}-3}}{\cancel{12}}}{\cancel{-4}}\\
& \boxed{y \ {\color{red}>} \ \frac{1}{2}x+3}\end{align*}

Note that the inequality was divided by negative 4 which caused the inequality sign to reverse its direction.

The above graph represents the graph of the inequality before the solution set region is shaded. The line is a dashed line because the inequality symbol is \begin{align*}(<)\end{align*}

\begin{align*}2x-4y &< -12\\
2({\color{red}1})-4({\color{red}1}) &< -12 && \text{Substitute} \ (1, 1) \ \text{for} \ x \ \text{and} \ y \ \text{of the original inequality.}\\
{\color{red}2-4} &< -12 && \text{Evaluate the inequality.}\\
{\color{red}-2} &< -12 && \text{Is it true?}\end{align*}

No, negative two is greater than negative twelve. The point (1, 1) does not satisfy the inequality. Therefore, the solution set is all of the area above the line that does not contain the point (1, 1). The ordered pair does not satisfy the inequality and will not lie within the shaded region.

The solution set for the inequality is the entire shaded region shown in the graph. The dashed or dotted line means that none of the points on the line will satisfy the inequality. In general, you will shade above the line if the inequality is of the form \begin{align*}y>mx+b\end{align*}

#### Now, let's write the appropriate inequality for a given graph:

Begin by determining the slope of the line. The slope of the line is determined by counting 2 units to the right and 3 units upward. The slope of the line for this graph is

\begin{align*}m=\frac{rise}{run}={\color{red}\frac{3}{2}}\end{align*}

The \begin{align*}y\end{align*}

\begin{align*}y=\frac{3}{2}x-2\end{align*}

The solution set is found in the shaded region that is above the line. The line is a solid line. Therefore the inequality symbol that must be inserted is greater than or equal to. The inequality that is modeled by the above graph is:

\begin{align*}\boxed{y \ge \frac{3}{2}x-2}\end{align*}

### Examples

#### Example 1

Earlier, you were asked to graph the following linear inequality:

\begin{align*}2x-3y < 6\end{align*}

The first step is to rearrange the inequality in slope-intercept form. This process is the same as it is for linear equations.

\begin{align*}& 2x-3y < 6\\
& 2x{\color{red}-2x}-3y < {\color{red}-2x}+6\\
& -3y < -2x+6\\
& \frac{-3y}{{\color{red}-3}} < \frac{-2x}{{\color{red}-3}} + \frac{6}{{\color{red}-3}}\\
& \frac{\cancel{-3}y}{\cancel{-3}} < \frac{-2x}{-3} + \frac{\overset{{\color{red}-2}}{\cancel{6}}}{\cancel{-3}}\\
& \boxed{y \ {\color{red}>} \ \frac{2}{3}x-2}\end{align*}

Note that the inequality was divided by negative 3 which caused the inequality sign to reverse its direction.

The graph of the inequality is done the same as it is for a linear equation. In this case the graph will be a dashed or dotted line because the sign is greater than \begin{align*}(>)\end{align*}.

The above graph represents the graph of the inequality before the solution set region is shaded. To determine whether to shade above the line or below the line, choose a point that is not on the line and test its coordinates in the original inequality. If the coordinates of the point satisfy the inequality, then the area above or below the line, containing the point, will be shaded. If the coordinates of the point do not satisfy the inequality, then the area above or below the line, that does not contain the test point, will be shaded.

The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality.

\begin{align*}2x-3y &< 6\\ 2({\color{red}1})-3({\color{red}1}) &< 6 && \text{Substitute} \ (1, 1) \ \text{for} \ x \ \text{and} \ y \ \text{of the original inequality.}\\ {\color{red}2}- {\color{red}3} &< 6 && \text{Evaluate the inequality.}\\ {\color{red}-1} &< 6 && \text{Is it true?}\end{align*}

Yes, negative one is less than six. The point (1, 1) satisfies the inequality. Therefore, the solution set is all of the area above the line that contains the point (1, 1). The ordered pair that satisfies the inequality will lie within the shaded region.

The solution set for the inequality is the entire shaded region shown in the graph. The dashed or dotted line means that none of the points on the line will satisfy the inequality.

#### Example 2

Without graphing, determine if each point is in the shaded region for each inequality.

- (2, –3) and \begin{align*}2y<-3x+1\end{align*}
- (–3, 5) and \begin{align*}-3x > 2y+6\end{align*}

If the point satisfies the inequality, then the point will lie within the shaded region. Substitute the coordinates of the point into the inequality and evaluate the inequality. If the solution is true, then the point is in the shaded area.

\begin{align*}2y &< -3x+1 && \text{Substitute} \ (2, -3) \ \text{for} \ x \ \text{and} \ y \ \text{in the inequality.}\\ 2({\color{red}-3}) &< -3({\color{red}2})+1 && \text{Evaluate the inequality.}\\ {\color{red}-6} &< {\color{red}-6}+1\\ -6 &< {\color{red}-5} && \text{Is it true?}\end{align*}Yes, negative six is less than negative five. The point (2, –3) satisfies the inequality. The ordered pair will lie within the shaded region.

\begin{align*}-3x &> 2y+6 && \text{Substitute} \ (-3, 5) \ \text{for} \ x \ \text{and} \ y \ \text{in the inequality.}\\ -3({\color{red}-3}) &> 2({\color{red}5})+6 && \text{Evaluate the inequality.}\\ {\color{red}9} &> {\color{red}10}+6\\ 9 &> {\color{red}16} && \text{Is it true?}\end{align*}No, nine is not greater than sixteen. The point (–3, 5) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.

#### Example 3

Graph the following inequality:

\begin{align*}5x-3y \ge 15\end{align*}

- \begin{align*}& 5x-3y \ge 15 && \text{Write the inequality in slope intercept form} \ (y=mx+b).\\ & 5x{\color{red}-5x}-3y \ge {\color{red}-5x}+15\\ & -3y \ge -5x+15\\ & \frac{-3y}{{\color{red}-3}} \ge \frac{-5x}{{\color{red}-3}} + \frac{15}{{\color{red}-3}}\\ & \frac{\cancel{-3}y}{\cancel{-3}} \ge \frac{5}{3}x + \frac{\overset{{\color{red}-5}}{\cancel{15}}}{\cancel{-3}}\\ & \boxed{y \ {\color{red}\le} \ \frac{5}{3}x-5}\end{align*}
- Note that the inequality was divided by negative 3 which caused the inequality sign to reverse its direction.
- Is the graph of the inequality shaded correctly?
- The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality.
- \begin{align*}5x-3y &\ge 15\\ 5({\color{red}1})-3({\color{red}1}) &\ge 15 && \text{Substitute} \ (1, 1) \ \text{for} \ x \ \text{and} \ y \ \text{of the original inequality.}\\ {\color{red}5}{\color{red}-3} &\ge 15 && \text{Evaluate the inequality.}\\ {\color{red}2} &\ge 15 && \text{Is it true?}\end{align*}
- No, two is not greater than or equal to fifteen. The point (1, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.

#### Example 4

Determine the inequality that models the following graph:

Begin by determining the slope of the line. The slope of the line is determined by counting 2 units to the left and 4 units downward. The slope of the line for this graph is

- \begin{align*}m=\frac{rise}{run}={\color{red}\frac{-4}{-2}}={\color{red}2}\end{align*}
- The \begin{align*}y\end{align*}-intercept for the line is (0, –3). The equation of the line in slope-intercept form is \begin{align*}y=2x-3\end{align*}
- The solution set is found in the shaded region that is below the line. The line is a solid line. Therefore the inequality symbol that must be inserted is less than or equal to. The inequality that is modeled by the above graph is:
- \begin{align*}\boxed{y \le 2x-3}\end{align*}

### Review

Without graphing, determine if each point is in the shaded region for each inequality.

- (2, 1) and \begin{align*}2x+y>5\end{align*}
- (–1, 3) and \begin{align*}2x-4y \le -10\end{align*}
- (–5, –1) and \begin{align*}y > -2x+8\end{align*}
- (6, 2) and \begin{align*}2x+3y \ge -2\end{align*}
- (5, –6) and \begin{align*}2y< 3x+3\end{align*}

Determine the inequality that is modeled by each of the following graphs.

- .

- .

- .

- .

- .

Graph the following linear inequalities on a Cartesian plane.

- \begin{align*}4x-2y>8\end{align*}
- \begin{align*}4y-3x \le -8\end{align*}
- \begin{align*}x-y < -3\end{align*}
- \begin{align*}3x+y > -1\end{align*}
- \begin{align*}x+3y \ge 9\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 5.5.

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Please Sign In to create your own Highlights / Notes | |||

Show More |

Cartesian Plane

The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.inequality

An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are , , , and .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 how to graph a linear inequality in the Cartesian plane.

## Concept Nodes:

**Save or share your relevant files like activites, homework and worksheet.**

To add resources, you must be the owner of the Modality. Click Customize to make your own copy.