We need to graph the following inequality system:
[tex]\begin{cases}x-y>2 \\ y<-\frac{1}{3}x+1\end{cases}[/tex]Now we need to isolate the y-variable on the left side for the first equation:
[tex]\begin{cases}yNow we have to graph the boundary lines, which are:[tex]\begin{gathered} y=x-2 \\ y=-\frac{1}{3}x+1 \end{gathered}[/tex]We need to points to graph these equations. We will use the points that have x equal to 0 and y = 0.
For the first equation:
[tex]\begin{gathered} y=0-2 \\ y=-2 \end{gathered}[/tex]The first point is (0,-2).
[tex]\begin{gathered} 0=x-2 \\ x=2 \end{gathered}[/tex]The second point is (2, 0).
For the second equation:
[tex]\begin{gathered} y=-\frac{1}{3}\cdot0+1 \\ y=1 \end{gathered}[/tex]The first point (0,1).
[tex]\begin{gathered} 0=-\frac{1}{3}x+1 \\ \frac{1}{3}x+1 \\ x=3 \end{gathered}[/tex]The second point is (3, 0).
Now we can trace both boundary lines:
Finally we can shade the solution set, which is the region that is below both lines, since both have an "<" signal.
