Given the following inequalitites:
[tex]\begin{gathered} y\leq\frac{1}{2}x+2 \\ y\ge-2 \\ x\leq-3 \end{gathered}[/tex]We are going to evaluate each point:
A= (-10 , 1)
B= (-3, 7 )
C= ( 5 , 1)
D= (-3 , -5)
The point A:
[tex]\begin{gathered} 1\leqslant\frac{1}{2}(-10)+2=1\leqslant-5+2=1\leqslant-3 \\ \end{gathered}[/tex]As -3 is not greater or equal to 1 the point A does not work.
Point B:
[tex]7\leqslant\frac{1}{2}(-3)+2=7\leqslant-\frac{3}{2}+2=7\leqslant\frac{1}{2}[/tex]1/2 is not greater or equal to seven, the point B does not work.
Point C.
[tex]1\leqslant\frac{1}{2}(5)+2=1\leqslant\frac{5}{2}+2=1\leqslant\frac{9}{2}[/tex]9/2 is greater than 1, now we have to evaluate the other inequalities in the same point.
[tex]\begin{gathered} 1\ge-2 \\ and \\ 1\leq-3 \end{gathered}[/tex]As -3 isn't greater than 1, the point c does not work.
The point D.
[tex]-3\leqslant\frac{1}{2}(-5)+2=-3\leqslant-\frac{1}{2}[/tex]The first one is true, -1/2 is greater than -3.
[tex]-5\ge-2[/tex]The second one is false, the point D does not work.
Point E (0 , 0).
[tex]0\leqslant\frac{1}{2}(0)+2=0\leq2[/tex]The first is true.
Second one:
[tex]0\ge-2[/tex]The second one is true.
Third one.
[tex]0\leq-3[/tex]The point E does not work, because -3 is not greater or equal to -3.
Answer: no point is a solution of the system.
