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?
Graphing Systems of Linear Inequalities
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.
Let's graph and identify the solutions to the following systems.
\begin{align*}y &> 2x3\\
y & \le 4x+1\end{align*}
Since both of these inequalities are given in slope intercept form, we can use the \begin{align*}y\end{align*}
\begin{align*}y & \ge \frac{2}{3}x+2\\
y & \le \frac{2}{3}x5\end{align*}
Since inequality 1 has “\begin{align*}y \ge\end{align*}
\begin{align*}3xy & <6\\
8x+5y & \le 40\end{align*}
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*}
For \begin{align*}8x+5y \le 40\end{align*}
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*}
For the second inequality, the symbol is \begin{align*}\le\end{align*}
Again, Inequality 1 is graphed in blue and inequality 2 is graphed in red. The overlap of the shaded regions (purple shading) represents the solution.
Now, let's graph the following system of linear inequalities:
\begin{align*}y & < \frac{2}{3}x+3\\
y & \ge 1\\
x & \ge 4\end{align*}
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*}
\begin{align*}y \ge 1 \Rightarrow\end{align*}
\begin{align*}x \ge 4 \Rightarrow\end{align*}
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.
Examples
Example 1
Earlier, you were asked who was correct in their answer to which quadrant(s) the solution falls within.
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.
Graph and identify the solutions to the following systems.
In each of the solutions below, the first inequality in the system is shown in blue and the second inequality is shown in red. The solution set is the overlapping shaded region in purple. When there are three inequalities, only the solution region is shown to eliminate confusion.
Example 2
\begin{align*}y & \le \frac{1}{3} x+5\\
y & > \frac{5}{4}x2\end{align*}yy≤13x+5>54x−2
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*}
\begin{align*}y \le \frac{1}{3}x+5 \Rightarrow\end{align*}
\begin{align*}y>\frac{5}{4}x2 \Rightarrow\end{align*}
Example 3
\begin{align*}4x+y & > 8\\
3x5y & \le 15\end{align*}4x+y3x−5y>8≤15
In these inequalities it is easiest to graph using the \begin{align*}x\end{align*}
\begin{align*}4x+y>8 \Rightarrow\end{align*}
\begin{align*}3x5y \le 15 \Rightarrow\end{align*}
Example 4
\begin{align*}7x+2y & \le 14\\
3x9y & \ge 18\end{align*}
Again, it is easiest here to graph using the \begin{align*}x\end{align*}
\begin{align*}7x+2y \le 14 \Rightarrow\end{align*}
\begin{align*}3x9y \ge 18 \Rightarrow\end{align*}
Example 5
\begin{align*}y & \ge 2x3\\
2x+y &>8\\
y &>3\end{align*}y2x+yy≥2x−3>−8>−3
Inequality 1 can be graphed using the slope and \begin{align*}y\end{align*}
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 purple on the graph.
Review
Graph the following systems of linear inequalities.
 .

 \begin{align*}y & > \frac{1}{2} x2\\
4x+6y & \le 24\end{align*}
y4x+6y>12x−2≤24
 \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 & \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*}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 & \le \frac{1}{2}x+4\\ x2y & \le 2\end{align*}
 .

 \begin{align*}7x3y &>21\\ x4y &<8\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*}y &<2\\ y & \ge 2\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*}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\\ x & \le 4\\ 3x+5y & > 15\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*}
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 3.11.