<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Honors Go to the latest version.

# 5.6: Solving System of Inequalities on the Cartesian Plane

Difficulty Level: At Grade Created by: CK-12

## Solving a System of Linear Inequalities

Introduction

In a previous lesson you learned how to solve a system of linear equations graphically. The solution was determined by identifying the intersection point of the two linear equations. When the system of equations was solved by graphing, the system could have one point of intersection, no point of intersection (the lines were parallel), or an infinite number of intersection points (the equations were multiples of each other).

In this lesson you will learn to solve a system of linear inequalities by identifying the regions of intersection. The region of intersection or the common solution of a system of linear equation is where the shading of the inequalities overlaps. This area of overlapping is known as the feasible region and this region contains the common solution of the system of inequalities.

Objectives

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

• Plotting two or more inequalities on the same Cartesian grid.
• Showing the overlap of the shading of the inequalities.
• Identifying the feasible region on the graph.

Introduction

When a linear inequality is graphed on a Cartesian plane, the region that contains the points that will satisfy the inequality is identified by shading. If two inequalities are graphed on the same grid and if they share common points that will satisfy both linear equations, an area will be created that contains shading from both inequalities. The area that contains this shading is the region that contains the common solution for the system of inequalities.

Watch This

Guidance

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

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

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

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

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

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 $y \le 2x-3$ will be graphed 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.

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

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 A

Solve the following system of linear inequalities by graphing:

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

Write each inequality in slope-intercept form.

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

$& -x+2y > -4\\& -x{\color{red}+x}+2y > {\color{red}x}-4\\& 2y > {\color{red}x}-4\\& \frac{2y}{{\color{red}2}} > \frac{x}{{\color{red}2}} -\frac{4}{{\color{red}2}}\\& \frac{\cancel{2}y}{\cancel{2}} > \frac{1}{2}x - \frac{\overset{{\color{red}2}}{\cancel{4}}}{\cancel{2}}\\& \boxed{y > \frac{1}{2}x-2}$

Graph: $y \ge -\frac{1}{3}x-2$.

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

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

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 $y>\frac{1}{2}x-2$ 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.

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

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.

Example B

Solve the following system of linear inequalities by graphing:

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

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 $y-$axis. The second line is a line that has a slope of zero. The graph is a line parallel to the $x-$axis.

Graph: $x \ge 1$.

On the graph, every point to the right of the vertical line has an $x-$value that is greater than one. Therefore, the graph must be shaded to the right of the vertical line.

Now graph $y>2$ on the same Cartesian grid.

On the graph, every point above the horizontal line has a $y-$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.

Example C

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:

$\begin{Bmatrix} y < x+1\\ y \le -2x+5\\ y > 0 \end{Bmatrix}$

Graph: $y:

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

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

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

$y \ge -2x+4$

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

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

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.

$y>0$

The graph of $y>0$ is a horizontal line along the $x-$axis. Every point above the horizontal line has a $y-$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 shaded in pink.

Vocabulary

Feasible region
The feasible region is the part on the graph where the shaded areas of the inequalities overlap. This area contains the all the solution sets for the inequalities.

Guided Practice

1. Solve the following system of linear inequalities by graphing:

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

2. Solve the system of linear inequalities by graphing:

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

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

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

1. $\begin{Bmatrix} 4x+5y \le 20\\ 3x + y \le 6\\\end{Bmatrix}$

Write each inequality in slope-intercept form.

$& 4x+5y \le 20 && 3x+y \le 6\\&4x{\color{red}-4x}+5y \le {\color{red}-4x}+20 && 3x{\color{red}-3x}+y \le {\color{red}-3x}+6\\&5y \le {\color{red}-4x}+20 && y \le {\color{red}-3x}+6\\& \frac{5y}{{\color{red}5}} \le \frac{-4x}{{\color{red}5}}+\frac{20}{\color{red}5} && \boxed{y \le-3x+6}\\& \frac{\cancel{5}y}{\cancel{5}} \le -\frac{4}{5}x+\frac{\overset{{\color{red}4}}{\cancel{20}}}{\cancel{5}}\\& \boxed{y \le -\frac{4}{5}x+4}$

Graph: $y \le -\frac{4}{5}x+4$

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

$4x+5y &\le 20\\4({\color{red}1})+5({\color{red}1}) &\le 20\\{\color{red}4}+{\color{red}5} &\le 20\\{\color{red}9} &\le 20 \quad \text{Is it true?}$

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

Graph $y \le -3x+6$

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

$3x+y &\le 6\\3({\color{red}1})+({\color{red}1}) &\le 6\\{\color{red}3}+{\color{red}1} &\le 6\\{\color{red}4} &\le 6 \quad \text{Is it true?}$

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 pink.

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

Write the first two inequalities in slope-intercept form.

$& 2x+3y <12 && 2x+y \le 8\\& 2x{\color{red}-2x}+3y < {\color{red}-2x}+12 && 2x{\color{red}-2x}+y \le {\color{red}-2x}+8\\& 3y < {\color{red}-2x}+12 && y \le {\color{red}-2x}+8\\& \frac{3y}{{\color{red}3}} < \frac{-2x}{{\color{red}3}} + \frac{12}{{\color{red}3}} && \boxed{y \le -2x+8}\\& \frac{\cancel{3}y}{\cancel{3}} < -\frac{2}{3}x + \frac{\overset{{\color{red}4}}{\cancel{12}}}{\cancel{3}}\\& \boxed{y < -\frac{2}{3}x+4}$

Graph: $y<-\frac{2}{3}x+4$

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

$2x+3y &< 12\\2({\color{red}1})+3({\color{red}1}) &< 12\\{\color{red}2}+{\color{red}3} &< 12\\{\color{red}5} &< 12 \quad \text{Is it true?}$

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

Graph: $y \le -2x+8$

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

$2x+y &\le 8\\2({\color{red}1})+({\color{red}1}) &\le 8\\{\color{red}2}+{\color{red}1} &\le 8\\{\color{red}3} &\le 8 \quad \text{Is it true?}$

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

Graph: $x \ge 0$

The graph will be a vertical line that will coincide with the $y-$axis. All $x-$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: $y \ge 0$

The graph will be a horizontal line that will coincide with the $x-$axis. All $y-$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.

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

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)

$& y < 2x+7 \qquad \qquad \qquad and && y \ge -3x-4\\ & y < 2x+7 && y \ge -3x-4\\& ({\color{red}3})<2({\color{red}-1})+7 && ({\color{red}3}) \ge -3({\color{red}-1})-4\\& {\color{red}3} < {\color{red}-2}+7 && {\color{red}3} \ge {\color{red}3}-4\\& 3 < {\color{red}5} && 3 \ge {\color{red}-1}$

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)

$& y < 2x+7 && y \ge -3x-4\\& ({\color{red}-2}) < 2({\color{red}4})+7 && ({\color{red}-2}) \ge -3({\color{red}4})-4\\& {\color{red}-2}<{\color{red}8}+7 && {\color{red}-2} \ge {\color{red}-12}-4\\& -2 < {\color{red}15} && -2 \ge {\color{red}-16}$

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)

$& y < 2x+7 && y \ge -3x-4\\& ({\color{red}5})< 2({\color{red}6})+7 && ({\color{red}5}) \ge -3({\color{red}6}) -4\\& {\color{red}5} < {\color{red}12}+7 && {\color{red}5} \ge {\color{red}-18}-4\\& 5 < {\color{red}19} && 5 \ge {\color{red}-22}$

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.

Summary

In this lesson you have learned to solve a system of linear inequalities on a Cartesian plane. The solution set was all points in the area that was in the overlapping, shaded region of the graphs. To determine where the shaded area should be with respect to each 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.

Problem Set

Solve the following systems of linear inequalities by graphing.

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

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

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

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

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

Solve the following systems of linear inequalities by graphing.

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

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

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

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

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

Solve the following systems...

$& \begin{Bmatrix} 3x+5y>15\\ 2x-7y \le 14\\\end{Bmatrix}\\& 3x+5y>15 && 2x-7y \le 14\\& 3x-3x+5y > -3x+15 && 2x-2x-7y \le -2x+14\\& 5y > -3x+15 && -7y \le -2x+14\\& \frac{5y}{5}>\frac{-3x}{5}+\frac{15}{5} && \frac{-7y}{-7} \le \frac{-2x}{-7}+\frac{14}{-7}\\& \frac{\cancel{5}y}{\cancel{5}} > -\frac{3}{5}x + \frac{\overset{{\color{red}3}}{\cancel{15}}}{\cancel{5}} && \frac{\cancel{-7}y}{\cancel{-7}} \le \frac{2}{7}x + \frac{\overset{{\color{red}-2}}{\cancel{14}}}{\cancel{-7}}\\& \boxed{y>-\frac{3}{5}x+3} && \boxed{y \ge \frac{2}{7}x-2}$

Test (1, 1)

$3x+5y &>15\\3(1)+5(1) &> 15\\3+5 &> 15\\8 &> 15$

It is not true. The point (1, 1) will not lie within the shaded region.

Test (1, 1)

$2x-7y &\le 14\\2(1)+7(1) &\le 14\\2-7 &\le 14\\-5 &\le 14$

It is true. The point (1, 1) will lie within the shaded region.

The feasible region is the area shaded in purple.

$& \begin{Bmatrix} x-y > 4\\ x+y > 6\\\end{Bmatrix}\\& x-y > 4 && x+y>6\\& x-x-y > -x+4 && x-x+y>-x+6\\& -y > -x+4 && \boxed{y>-x+6}\\& \frac{-y}{-1}> \frac{-x}{-1}+\frac{4}{-1}\\& \frac{\cancel{-1}y}{\cancel{-1}}> \frac{\cancel{-1}x}{\cancel{-1}}+\frac{\overset{{\color{red}-4}}{\cancel{4}}}{\cancel{-1}}\\& \boxed{y < x-4}$

Test (1, 1)

$x+y &> 4\\(1)+(1) &> 4\\1+1 &> 4\\2 &> 4$

It is not true. The point (1, 1) will not lie within the shaded region.

Test (1, 1)

$x+y &> 6\\(1)+(1) &> 6\\1+1 &> 6\\2 &> 6$

It is not true. The point (1, 1) will not lie within the shaded region.

The feasible region is the area shaded in gold.

$& \begin{Bmatrix} 3x-6y > -6\\ 5x+9y \ge -18\\\end{Bmatrix}\\& 3x-6y > -6 && 5x+9y \ge -18\\& 3x-3x-6y > -3x-6 && 5x-5x+9y \ge -5x-18\\& -6y > -3x-6 && 9y \ge -5x-18\\& \frac{-6y}{-6} > \frac{-3x}{-6} -\frac{6}{-6} && \frac{9y}{9} \ge \frac{-5x}{9}-\frac{18}{9}\\& \frac{\cancel{-6}y}{\cancel{-6}} > \frac{3}{6}x - \frac{\overset{{\color{red}-1}}{\cancel{6}}}{\cancel{-6}} && \frac{\cancel{9}y}{\cancel{9}} \ge -\frac{5}{9}x - \frac{\overset{{\color{red}2}}{\cancel{18}}}{\cancel{9}}\\& \boxed{y < \frac{1}{2}x+1} && \boxed{y \ge -\frac{5}{9}x-2}$

Test (1, 1)

$3x-6y &> -6\\3(1)-6(1) &> -6\\3-6 &> -6\\-3 &> -6$

It is true. The point (1, 1) will lie within the shaded region.

Test (1, 1)

$5x+9y &\ge -18\\5(1)+9(1) &\ge -18\\5+9 &\ge -18\\14 &\ge -18$

It is true. The point (1, 1) will lie within the shaded region.

The feasible region is the area shaded in blue.

Solve the following systems...

$& \begin{Bmatrix} 2x-y<4\\ x \ge -1\\ y \ge -2\end{Bmatrix}\\& 2x-y < 4\\& 2x-2x-y < -2x+4\\& -y < -2x+4\\& \frac{-1y}{-1} < \frac{-2x}{-1}+\frac{4}{-1}\\& \frac{\cancel{-1}y}{\cancel{-1}} < \frac{\overset{{\color{red}2}}{\cancel{-2}x}}{\cancel{-1}} + < \frac{\overset{{\color{red}-4}}{\cancel{4}}}{\cancel{-1}}\\& \boxed{y > 2x-4}$

The feasible region is the area shaded in red.

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

The feasible region is the area shaded in yellow.

$& \begin{Bmatrix} x+y > -1\\ 3x -2y \ge 2\\ x < 3\\ y \ge 0\end{Bmatrix}\\& x+y > -1 && 3x-2y \ge 2\\& x-x+y > -x-1 && 3x-3x-2y \ge -3x+2\\& \boxed{y>-x-1} && -2y \ge -3x+2\\& && \frac{-2y}{-2} \ge \frac{-3x}{-2}+\frac{2}{-2}\\& && \frac{\cancel{-2}y}{\cancel{-2}} \ge \frac{3}{2}x+\frac{\overset{{\color{red}-1}}{\cancel{2}}}{\cancel{-2}}\\& && \boxed{y \le \frac{3}{2}x-1}$

The feasible region is the area shaded in gold.

## Date Created:

Jan 16, 2013

Jan 14, 2015
Files can only be attached to the latest version of section