5.6: Graphical Solutions to Systems of Inequalities
Graph the following system of linear inequalities on the same Cartesian grid.
\begin{align*}\begin{Bmatrix} y>\frac{1}{2}x+2 \\ y \le 2x3 \end{Bmatrix}\end{align*}
What is the solution to this system?
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Khan Academy Systems of Linear Inequalities
Guidance
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.
Example A
Solve the following system of linear inequalities by graphing:
\begin{align*}\begin{Bmatrix} 2x6y \le 12\\ x +2y > 4 \end{Bmatrix}\end{align*}
Solution:
Write each inequality in slopeintercept form.
\begin{align*}& 2x6y \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}x2}\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}x2}\end{align*}
Graph: \begin{align*}y \ge \frac{1}{3}x2\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*}2x6y &\le 12\\ 2({\color{red}1})6({\color{red}1}) &\le 12\\ {\color{red}26} &\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}x2\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*}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.
Example B
Solve the following system of linear inequalities by graphing:
\begin{align*}\begin{Bmatrix} x \ge 1\\ y > 2 \end{Bmatrix}\end{align*}
Solution:
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*}
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*}
Now graph \begin{align*}y>2\end{align*}
On the graph, every point above the horizontal line has a \begin{align*}y\end{align*}
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{align*}\begin{Bmatrix} y < x+1\\ y \le 2x+5\\ y > 0 \end{Bmatrix}\end{align*}
Solution:
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*}
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 &\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*}
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.
Concept Problem Revisited
\begin{align*}\begin{Bmatrix} y>\frac{1}{2}x+2 \\ y \le 2x3 \end{Bmatrix}\end{align*}
Both inequalities are in slopeintercept form. Begin by graphing \begin{align*}\boxed{y > \frac{1}{2}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*}& 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 onehalf. 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 2x3\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 \le 2x3\\ & ({\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.
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{align*}\begin{Bmatrix} 4x+5y \le 20\\ 3x + y \le 6\\ \end{Bmatrix}\end{align*}
2. 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*}
3. Determine and prove three points that satisfy the following system of linear inequalities:
\begin{align*}\begin{Bmatrix} y < 2x+7\\ y \ge 3x4\\ \end{Bmatrix}\end{align*}
Answers:
1. \begin{align*}\begin{Bmatrix} 4x+5y \le 20\\ 3x + y \le 6\\ \end{Bmatrix}\end{align*}
 Write each inequality in slopeintercept 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 \le3x+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*}
 The point (1, 1) is not on the graphed line. Test the point (1, 1) to determine the location of the shading.

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

\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 pink.
2. \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 slopeintercept 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*}
 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+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*}
 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+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.
3. \begin{align*}\begin{Bmatrix} y < 2x+7\\ y \ge 3x4 \\ \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 3x4\\
& y < 2x+7 && y \ge 3x4\\
& ({\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 3x4\\
& ({\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 3x4\\
& ({\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 twentytwo. Therefore, the point lies within the feasible region and is a solution for the system of linear inequalities.
Practice
Solve the following systems of linear inequalities by graphing.
\begin{align*}\begin{Bmatrix} 3x+5y>15\\ 2x7y \le 14\\ \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 3x+2y \ge 10\\ xy < 1\\ \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} xy > 4\\ x+y > 6\\ \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} y>3x2\\ y < 2x+5\\ \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 3x6y > 6\\ 5x+9y \ge 18\\ \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 2xy<4\\ x \ge 1\\ y \ge 2 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 2x+y>6\\ x +2y \ge 6\\ x \ge 0\\ y \ge 0 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} x \le 3\\ x \ge 2\\ y \le 4\\ y \ge 1 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} y < x+1\\ y \ge 2x+3\\ y > 0 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} x+y > 1\\ 3x 2y \ge 2\\ x < 3\\ y \ge 0 \end{Bmatrix}\end{align*}
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Description
Learning Objectives
Here you'll learn to solve a system of linear inequalities by graphing.
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Date Created:
Dec 19, 2012Last Modified:
Aug 18, 2015Vocabulary
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