Ms. Walochek assigns the following system of linear inequalities for homework. She asks her students to find which quadrant(s) the solution lies in.
\begin{align*}x \ge 0\\ y \le 1\\ \frac{1}{2}x  3y < 2\end{align*}
Caren says the solution falls only in the first quadrant. Bahir says it falls in all four quadrants. Who is correct?
Watch This
James Sousa: Ex 1: Graph a System of Linear Inequalities
Guidance
In this section we will be graphing two and three linear inequalities on the same grid and identifying where the shaded regions overlap. This overlapping region is the solution to the system. Note: If the shaded regions do not overlap, there is no solution as shown in Example B.
Example A
Graph and identify the solution to the system:
\begin{align*}y &> 2x3\\ y & \le 4x+1\end{align*}
Solution:
Since both of these inequalities are given in slope intercept form, we can use the \begin{align*}y\end{align*}
Example B
Graph and identify the solution to the system:
\begin{align*}y & \ge \frac{2}{3}x+2\\ y & \le \frac{2}{3}x5\end{align*}
Solution:
Since inequality 1 has “\begin{align*}y \ge\end{align*}”, we will make a solid line to indicate that the line is included in the solution and shade above the line where \begin{align*}y\end{align*} is “greater” (where the \begin{align*}y\end{align*}axis is above) the line. Since inequality 2 has “\begin{align*}y \le\end{align*}” we will make a dashed line to indicate that the line is not included in the solution set and shade below the line where \begin{align*}y\end{align*} is “less than” (where the \begin{align*}y\end{align*}axis is below) the line. Inequality 1 is graphed in \begin{align*}{\color{blue}\mathbf{blue}}\end{align*} and inequality 2 is graphed in \begin{align*}{\color{red}\mathbf{red}}\end{align*}. In this case the regions do not overlap. This indicates that there is no solution to the system.
Example C
Graph and identify the solution to the system:
\begin{align*}3xy & <6\\ 8x+5y & \le 40\end{align*}
Solution: This time, let’s use a different graphing technique. We can identify the intercepts for each equation and graph the lines using these points:
For \begin{align*}3xy<6\end{align*}, the intercepts are (2, 0) and (0, 6).
For \begin{align*}8x+5y \le 40\end{align*}, the intercepts are (5, 0) and (0, 8).
For the first inequality, the symbol is < so the line is dashed. Now, use a test point to determine which way to shade. (0, 0) is an easy point to test. \begin{align*}\Rightarrow 3(0)(0)<6 \end{align*} Since this is a true statement, (0, 0) is a solution to the inequality and we can shade on the side of the line with (0, 0).
For the second inequality, the symbol is \begin{align*}\le\end{align*} so the line is solid. Using the same test point, \begin{align*}(0, 0), \Rightarrow 8(0)+5(0) \le 40 \end{align*} This is a true statement so (0, 0) is a solution to the inequality and we can shade on the side of the line with (0, 0).
Again, Inequality 1 is graphed in \begin{align*}{\color{blue}\mathbf{blue}}\end{align*} and inequality 2 is graphed in \begin{align*}{\color{red}\mathbf{red}}\end{align*}. The overlap of the shaded regions (\begin{align*}{\color{magenta}\mathbf{purple}}\end{align*} shading) represents the solution.
Example D
Graph the system of linear inequalities:
\begin{align*}y & < \frac{2}{3}x+3\\ y & \ge 1\\ x & \ge 4\end{align*}
Solution:
As in the previous concept, we will graph the lines and determine whether each line should be dashed or solid and which way to shade.
\begin{align*}y<\frac{2}{3}x+3 \Rightarrow\end{align*} This inequality has a \begin{align*}y\end{align*}intercept of 3 and slope of \begin{align*}\frac{2}{3}\end{align*}. Since the inequality is <, we will shade below the dashed \begin{align*}{\color{blue}\mathbf{blue}}\end{align*} line.
\begin{align*}y \ge 1 \Rightarrow\end{align*} This is a horizontal line through (0, 1). The line will be solid and we shade above the \begin{align*}{\color{red}\mathbf{red}}\end{align*} line.
\begin{align*}x \ge 4 \Rightarrow\end{align*} This is a vertical line through (4, 0). The line will be solid and we will shade \begin{align*}({\color{yellow}\mathbf{yellow}})\end{align*} to the right of the \begin{align*}{\color{green}\mathbf{green}}\end{align*} line.
The solution to this system is the shaded region (triangular) in the center where all three shaded regions overlap. This region can be difficult to see in a graph so it is common practice to erase the shading that is not a part of the solution to make the solution region is more obvious.
Intro Problem Revisit If we graph all three inequalities on the same grid, we find that most of the overlapping area falls in the first quadrant, but a tiny bit of it also falls in the fourth quadrant. Therefore, neither Caren nor Bahir is correct.
Guided Practice
Graph and identify the solutions to the systems.
1. \begin{align*}y & \le \frac{1}{3} x+5\\ y & > \frac{5}{4}x2\end{align*}
2. \begin{align*}4x+y & > 8\\ 3x5y & \le 15\end{align*}
3. \begin{align*}7x+2y & \le 14\\ 3x9y & \ge 18\end{align*}
4. \begin{align*}y & \ge 2x3\\ 2x+y &>8\\ y &>3\end{align*}
Answers
In each of the solutions below, the first inequality in the system is shown in \begin{align*}{\color{blue}\mathbf{blue}}\end{align*} and the second inequality is shown in \begin{align*}{\color{red}\mathbf{red}}\end{align*}. The solution set is the overlapping shaded region in \begin{align*}{\color{magenta}\mathbf{purple}}\end{align*}. When there are three inequalities, only the solution region is shown to eliminate confusion.
1.
The inequalities in this system are both already in slope intercept form so we can graph them using the slope and \begin{align*}y\end{align*}intercept of each line and shade as shown below.
\begin{align*}y \le \frac{1}{3}x+5 \Rightarrow\end{align*} solid line and shade below
\begin{align*}y>\frac{5}{4}x2 \Rightarrow\end{align*} dashed line and shade above
2.
In these inequalities it is easiest to graph using the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} intercepts. Once we have graphed the lines we can use a test point to determine which side should be shaded.
\begin{align*}4x+y>8 \Rightarrow\end{align*} The intercepts are (2, 0) and (0, 8) and the line will be dashed. If we test the point (0, 0), the inequality is not true so we shade on the side of the line that does not contain (0, 0).
\begin{align*}3x5y \le 15 \Rightarrow\end{align*} The intercepts are (5, 0) and (0, 3) and the line will be solid. The test point (0, 0) satisfies the inequality so we shade on the side of the line that includes (0, 0).
3.
Again, it is easiest here to graph using the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} intercepts. Once we have graphed the lines we can use a test point to determine which side should be shaded.
\begin{align*}7x+2y \le 14 \Rightarrow\end{align*} The intercepts are (2, 0) and (0, 7) and the line will be solid. If we test the point (0, 0), the inequality is true so we shade on the side of the line that contains (0, 0).
\begin{align*}3x9y \ge 18 \Rightarrow\end{align*} The intercepts are (6, 0) and (0, 2) and the line will be solid. The test point (0, 0) does not satisfy the inequality so we shade on the side of the line that does not include (0, 0).
4.
Inequality 1 can be graphed using the slope and \begin{align*}y\end{align*}intercept. This line will be solid and the shading will be above this line.
Inequality 2 can be graphed using intercepts. The line will be dashed and we can use a test point to determine that the shaded region will be above this line.
Inequality 3 is a horizontal line. It will be dashed and the shading is above this line.
The intersection of these three regions is shaded in \begin{align*}{\color{magenta}\mathbf{purple}}\end{align*} on the graph.
Explore More
Graph the following systems of linear inequalities.
 .


\begin{align*}y & > \frac{1}{2} x2\\
4x+6y & \le 24\end{align*}

\begin{align*}y & > \frac{1}{2} x2\\
4x+6y & \le 24\end{align*}
 .


\begin{align*}y & >\frac{3}{4}x1\\
y & > 3x+5\end{align*}

\begin{align*}y & >\frac{3}{4}x1\\
y & > 3x+5\end{align*}
 .


\begin{align*}y & \le \frac{2}{3}x+2\\
y & \ge  \frac{5}{3}x1\end{align*}

\begin{align*}y & \le \frac{2}{3}x+2\\
y & \ge  \frac{5}{3}x1\end{align*}
 .


\begin{align*}y & \ge x3\\
y & < \frac{1}{5}x+1\end{align*}

\begin{align*}y & \ge x3\\
y & < \frac{1}{5}x+1\end{align*}
 .


\begin{align*}5x2y &>10\\
y & \le  \frac{1}{3}x+2\end{align*}

\begin{align*}5x2y &>10\\
y & \le  \frac{1}{3}x+2\end{align*}
 .


\begin{align*}y & >  \frac{4}{5}x3\\
y & >x\end{align*}

\begin{align*}y & >  \frac{4}{5}x3\\
y & >x\end{align*}
 .


\begin{align*}y & \le \frac{1}{2}x+4\\
x2y & \le 2\end{align*}

\begin{align*}y & \le \frac{1}{2}x+4\\
x2y & \le 2\end{align*}
 .


\begin{align*}7x3y &>21\\
x4y &<8\end{align*}

\begin{align*}7x3y &>21\\
x4y &<8\end{align*}
 .


\begin{align*}6x+5y &\le 5\\
2x3y & \le 12\end{align*}

\begin{align*}6x+5y &\le 5\\
2x3y & \le 12\end{align*}
 .


\begin{align*}x &<3\\
y & \ge 2x+1\end{align*}

\begin{align*}x &<3\\
y & \ge 2x+1\end{align*}
 .


\begin{align*}y &<2\\
y & \ge 2\end{align*}

\begin{align*}y &<2\\
y & \ge 2\end{align*}
 .


\begin{align*}2xy & \le 4\\
5x+2y &> 10\end{align*}

\begin{align*}2xy & \le 4\\
5x+2y &> 10\end{align*}
 .


\begin{align*}y & \le 2x+4\\
y & \ge 5x+4\\
y &>\frac{1}{2}x1\end{align*}

\begin{align*}y & \le 2x+4\\
y & \ge 5x+4\\
y &>\frac{1}{2}x1\end{align*}
 .


\begin{align*}x+y & \le 3\\
x & \le 3\\
y & < 3\end{align*}

\begin{align*}x+y & \le 3\\
x & \le 3\\
y & < 3\end{align*}
 .


\begin{align*}x & > 2\\
y & > 3\\
2x+y & \le 2\end{align*}

\begin{align*}x & > 2\\
y & > 3\\
2x+y & \le 2\end{align*}
 .


\begin{align*}x & > 2\\
x & \le 4\\
3x+5y & > 15\end{align*}

\begin{align*}x & > 2\\
x & \le 4\\
3x+5y & > 15\end{align*}
 .


\begin{align*}2x+3y &>6\\
5x2y &<10\\
x3y & >3\end{align*}

\begin{align*}2x+3y &>6\\
5x2y &<10\\
x3y & >3\end{align*}
 .


\begin{align*}y & \le x\\
y & \ge x\\
x & < 5\end{align*}

\begin{align*}y & \le x\\
y & \ge x\\
x & < 5\end{align*}