Question

Please help me!! 100 points!!

The coordinate plane below represents a city. Points A through F are schools in the city.


Part A: Using the graph above, create a system of inequalities that only contain points D and E in the overlapping shaded regions. Explain how the lines will be graphed and shaded on the coordinate grid above. (5 points)

Part B: Explain how to verify that the points D and E are solutions to the system of inequalities created in Part A. (3 points)

Part C: Timothy can only attend a school in his designated zone. Timothy's zone is defined by y < 3x − 3. Explain how you can identify the schools that Timothy is allowed to attend. (2 points)

Answer 1

Hello!

x => -3

y => 3

Answer 2

Part A.

We could take, for example:


`\begin{cases}x\geq-3\\y\geq3\end{cases}`

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.


`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}`

so point D is a solution,


`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}`

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:


`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}\\\\\\`


`F=(3,-4)\\\\y\ \textless \ 3x-3\quad\implies\quad-4\ \textless \ 3\cdot3-3\quad\implies\quad-4\ \textless \ 6\quad\text{True}\\\\\\`

So Timothy can attend schools C and F.

We can also draw a graph of that inequality (pic 2).