Part A.
We could take, for example:
[tex]\begin{cases}x\geq-3\\y\geq3\end{cases}[/tex]
For x ≥ -3 we have vertical line (x-intercept = -3) and shaded region on the right side of the line.
For y ≥ 3 we have horizontal line (y-intercept = 3) and shaded region above that line.
Part B.
Simply, substitute x and y coordinates of D and E to the system of inequalities from part A, and check if its is true, what we get.
[tex]D=(-2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}-2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}[/tex]
so point D is a solution,
[tex]E=(2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}[/tex]
and point E is also a solution to our system of inequalities.
Part C.
There are two ways we could do that. First is the same as in part B. We substitute x and y coordinates of each school and check if inequality is true or false:
[tex]A=(-5,5)\\\\y\ \textless \ 3x-3\quad\implies\quad5\ \textless \ 3\cdot(-5)-3\quad\implies\quad5\ \textless \ -18\quad\text{False}\\\\\\
B=(-4,-2)\\\\y\ \textless \ 3x-3\quad\implies\quad-2\ \textless \ 3\cdot(-4)-3\quad\implies\quad-2\ \textless \ -15\quad\text{False}\\\\\\
C=(2,1)\\\\y\ \textless \ 3x-3\quad\implies\quad1\ \textless \ 3\cdot2-3\quad\implies\quad1\ \textless \ 3\quad\text{True}\\\\\\
D=(-2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot(-2)-3\quad\implies\quad4\ \textless \ -9\quad\text{False}\\\\\\
E=(2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot2-3\quad\implies\quad4\ \textless \ 3\quad\text{False}\\\\\\[/tex]
[tex]F=(3,-4)\\\\y\ \textless \ 3x-3\quad\implies\quad-4\ \textless \ 3\cdot3-3\quad\implies\quad-4\ \textless \ 6\quad\text{True}\\\\\\[/tex]
So Timothy can attend schools C and F.
We can also draw a graph of that inequality (pic 2).
