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# 5.5: Solving Inequalities on the Cartesian Plane

Difficulty Level: At Grade Created by: CK-12

## Solving a Linear Inequality by Graphing

Introduction

Many graphs that you have created have involved linear equations of the form y=mx+b\begin{align*}y=mx+b\end{align*}. All points on the line satisfy the linear equation and, thus, form the solution set for the equation. In this lesson you will learn to determine the solution set of an inequality by graphing the inequality on a Cartesian plane.

When an inequality is graphed on a grid, the solution will appear as a shaded area, as opposed to simply a straight line. The shaded area will contain the solution set which consists of all the points which satisfy the inequality.

Objectives

The lesson objectives for Solving Inequalities on the Cartesian Plane are:

• Determining the type of line needed to graph the inequality.
• Testing to determine where the shaded area is located on the graph.
• Shading the correct region to produce the solution set for the inequality.

Introduction

An inequality refers to a statement that shows a relationship between two expressions that are not always equal. The general form of an inequality with two variables is ax+by>c\begin{align*}ax+by>c\end{align*} or ax+by<c\begin{align*}ax+by, where ‘a\begin{align*}a\end{align*}’ and ‘b\begin{align*}b\end{align*}’ are the coefficients of the variables and are not both equal to zero. The constant is ‘c\begin{align*}c\end{align*}’. The inequality symbol (>)\begin{align*}(>)\end{align*} is read “greater than” and the inequality symbol (<)\begin{align*}(<)\end{align*} is read “less than”. A hint to use to help distinguish between the two symbols is L-E-S-S points L-E-F-T. However, inequalities can also be written as ax+byc\begin{align*}ax+by \ge c\end{align*} or ax+byc\begin{align*}ax+by \le c\end{align*}. The inequality symbol ()\begin{align*}(\ge)\end{align*} is read “greater than or equal to” and the inequality symbol ()\begin{align*}(\le)\end{align*} is read “less than or equal to”. The solution set to this type of inequality includes all the points on the line as well as those in the shaded region.

Before we begin, there are a few things to consider when graphing inequalities:

• When an inequality is divided or multiplied by a negative number, the direction of the inequality sign is reversed.
• 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.
• The area or region which contains the ordered pairs that satisfy the inequality is indicated by shading.

Watch This

Guidance

To better understand the things that must be considered when graphing inequalities, the inequality \begin{align*}2x-3y < 6\end{align*} will be graphed.

\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*}

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 A

Graph the inequality \begin{align*}3x+4y \le 12\end{align*}.

\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*}. 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*}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.

Example B

Graph the inequality \begin{align*}2x-4y < -12\end{align*}.

\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*}

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*}. 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-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.

Example C

In this example, the graph of the inequality will be given and the task will be to determine the inequality that is modeled by the graph.

For the following, determine the inequality, in slope-intercept form, that is graphed.

This process is the same as determining the equation of the line that is graphed. The next step is then to decide the appropriate inequality symbol to insert.

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*}intercept for the line is (0, -2). The equation of the line in slope-intercept form is

\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*}

Vocabulary

Inequality
An inequality is a statement that shows a relationship between two expressions that are not always equal. An inequality is written using one of the following inequality symbols: greater than \begin{align*}(>)\end{align*}; less than \begin{align*}(<)\end{align*}; greater than or equal to \begin{align*}(\ge)\end{align*}; less than or equal to \begin{align*}(\le)\end{align*}. The solution to an inequality is indicated by a shaded region that contains all the ordered pairs that satisfy the inequality.

Guided Practice

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

i) (2, -3) and \begin{align*}2y<-3x+1\end{align*}

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

2. Graph the following inequality: \begin{align*}5x-3y \ge 15\end{align*}

3. Determine the inequality that models the following graph:

1. 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.

i) \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.

ii) \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.

2. \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*}

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.

3.

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*}

Summary

In this lesson you have learned to graph a linear inequality in two variables on a Cartesian plane. The solution set was all points in the area that was in the shaded region of the graph. To determine where the shaded area should be with respect to the line, a point, which was not on the line, was tested in the original inequality. If the point made the inequality true, then the area containing the point was shaded. If the point did not make the inequality true, then the tested point did not lie within the shaded region.

You also learned that the inequality symbol determined whether the line was dashed or solid. A dashed line was graphed for all inequalities that had a \begin{align*}>\end{align*} symbol or a \begin{align*}<\end{align*} symbol. The dashed line indicated that the points on the line did not satisfy the inequality. A solid line was graphed for all inequalities that had a \begin{align*}\ge\end{align*} symbol or a \begin{align*}\le\end{align*}symbol. The solid line indicated that the points on the line did satisfy the inequality.

In the last example, you learned to determine the inequality that was modeled by a given graph. The inequality was determined by calculating the slope and the \begin{align*}y-\end{align*}intercept of the line in the same way that you would determine the linear equation for the graph of a straight line. The last step was to determine the inequality sign to be inserted.

Problem Set

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

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

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

Without graphing, determine...

1. (2, 1) and \begin{align*}2x+y>5\end{align*}

\begin{align*}2(2)+(1)&>5\\ {\color{red}4}+{\color{red}1}&>5\\ {\color{red}5}&>5\end{align*}

No, five is not greater than five. The point (2, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.

1. (-5, -1) and \begin{align*}y > -2x+8\end{align*}

\begin{align*}(-1)&>-2(-5)+8\\ {\color{red}-1}&>{\color{red}10}+8\\ -1&>{\color{red}18}\end{align*}

No, negative one is not greater than eighteen. The point (-5, -1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.

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

\begin{align*}2(-6)&<3(5)+3\\ {\color{red}-12}&<{\color{red}15}+3\\ -12&<{\color{red}18}\end{align*}

Yes, negative twelve is less than eighteen. The point (5, -6) does satisfy the inequality. The ordered pair does satisfy the inequality and will lie within the shaded region.

Determine the inequality...

1. \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 dashed line. Therefore the inequality symbol that must be inserted is less than. The inequality that is modeled by the above graph is: \begin{align*}\boxed{y<-2x-3}\end{align*}
1. \begin{align*}m=\frac{rise}{run}={\color{red}\frac{-1}{2}}={\color{red}-\frac{1}{2}}\end{align*} The \begin{align*}y-\end{align*}intercept for the line is (0, 2). The equation of the line in slope-intercept form is \begin{align*}y=-\frac{1}{2}x+2\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 -\frac{1}{2}x+2}\end{align*}
1. \begin{align*}m=\frac{rise}{run}={\color{red}\frac{-1}{2}}={\color{red}-\frac{1}{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=-\frac{1}{2}x-3\end{align*} The solution set is found in the shaded region that is above the line. The line is a dashed line. Therefore the inequality symbol that must be inserted is greater than. The inequality that is modeled by the above graph is: \begin{align*}\boxed{y>-\frac{1}{2}x-3}\end{align*}

Graph the following linear...

\begin{align*}4x-2y&>8\\ 4x-2y&>8\\ 4x+4x-2y&>4x+8\\ -2y&>4x+8\\ \frac{-2y}{-2}&>\frac{4x}{-2}+\frac{8}{-2}\\ \frac{\cancel{-2}y}{\cancel{-2}} &> \frac{\overset{{\color{red}-2}}{\cancel{4}x}}{\cancel{-2}}+\frac{\overset{{\color{red}-4}}{\cancel{8}}}{\cancel{-2}}\\ y \ &{\color{red}<-2x-4}\end{align*}

Test the point (1, 1) to determine the shaded region.

\begin{align*}4x-2y &> 8\\ 4(1)-2(1) &> 8\\ {\color{red}4}-{\color{red}2}&>8\\ {\color{red}2}&>8 \end{align*}

No, two is not greater than eight. 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.

\begin{align*}x-y &< -3\\ x-y &< -3\\ x+x-y &< x-3\\ -y &< x-3\\ \frac{-1y}{-1} &< \frac{1x}{-1} - \frac{3}{-1}\\ \frac{\cancel{-1}y}{\cancel{-1}} &< \frac{\overset{{\color{red}-1}}{\cancel{1}}x}{\cancel{-1}} - \frac{\overset{{\color{red}-3}}{\cancel{3}}}{\cancel{-1}}\\ y \ & {\color{red}>-1x+3}\end{align*}

Test the point (1, 1) to determine the shaded region.

\begin{align*}x-y &< -3\\ {\color{red}1}-{\color{red}1} &< -3\\ {\color{red}0}&< -3\end{align*}

No, zero is not less than negative three. 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.

\begin{align*}x+3y &\ge 9\\ x+3y &\ge 9\\ x-x+3y &\ge -x+9\\ 3y &\ge -x+9\\ \frac{3y}{3} &\ge -\frac{1x}{3}+\frac{9}{3}\\ \frac{\cancel{3}y}{\cancel{3}} &\ge - \frac{1}{3}x + \frac{\overset{{\color{red}3}}{\cancel{9}}}{\cancel{3}}\\ y \ &{\color{red}\ge -\frac{1}{3}x+3}\end{align*}

Test the point (1, 1) to determine the shaded region.

\begin{align*}x+3y &\ge 9\\ (1)+3(1) &\ge 9\\ {\color{red}1}+{\color{red}3} &\ge 9\\ {\color{red}4} &\ge 9\end{align*}

No, four is not greater than or equal to nine. 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.

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