Step 1. Solve both inequalities for [tex]y[/tex]:
[tex]3x+y\ \textgreater \ -3[/tex]
[tex]y\ \textgreater \ -3x-3[/tex]
[tex]x+2y\ \textless \ 4[/tex]
[tex]2y\ \textless \ -x+4[/tex]
[tex]y\ \textless \ - \frac{1}{2} x+2[/tex]
Step 2. To check a point in the solution of the given system of inequalities, look for the intercepts of the lines [tex]-3x-3[/tex] and [tex]- \frac{1}{2} x+2[/tex]:
[tex]y=-3x-3[/tex] (1)
[tex]y=- \frac{1}{2} x+2[/tex] (2)
Replace (1) in (2):
[tex]-3x-3=- \frac{1}{2} x+2[/tex]
Solve for [tex]x[/tex]:
[tex] \frac{5}{2} x=-5[/tex]
[tex]x=-2[/tex] (3)
Replace (3) in (1):
[tex]y=-3x-3[/tex]
[tex]y=-3(-2)-3[/tex]
[tex]y=6-3[/tex]
[tex]y=3[/tex]
We can conclude that the point (-2,3) is in the solution of the system if inequalities; also any point inside the dark shaded area of the graph of the system of inequalities is also a solution of the system.
