The coordinate plane below represents a town. points a through f are farms in the town. graph of coordinate plane. point a is at negative 3, negative 4. point b is at negative 4, 3. point c is at 2, 2. point d is at 1, negative 2. point e is at 5, negative 4. point f is at 3, 4. part a: using the graph above, create a system of inequalities that only contains points a 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 a and e are solutions to the system of inequalities created in part a. (3 points) part c: chickens can only be raised in the area defined by y < −2x + 4. explain how you can identify farms in which chickens can be raised. (2 points)

We have that
A (-3,-4)
B (-4,3)
C (2,2)
D (1,-2)
E (5,-4)
using a graph tool
see the attached figure N 1

Part A: Using the graph above, create a system of inequalities that only contains points A and E in the overlapping shaded regions.

A (-3,-4)  E (5,-4)
y<= -3
y>=-5
is a system of a inequalities that will only contain A and E
to graph it, I draw the constant y = -3 and y=-5  and and I shade the region between both lines
see the attached figure N 2

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

we know that
the system of a inequalities is
y<= -3
y>=-5
the solution is all y real numbers belonging to the interval [-5,-3]

therefore
if points A and E are solutions both points must belong to the interval

points A and E have the same coordinate y=-4
and y=-4 is included in the interval 

therefore
both points are solution

Part C: Chickens can only be raised in the area defined by y < −2x + 4. Explain how you can identify farms in which chickens can be raised

step 1
graph the inequality 
y < −2x + 4
see the attached figure N 3
the farms in which chickens can be raised are the points A, B and D
are those that are included in the shaded part

At a constant it will take one hour to travel 32 mi.at that speed it will take 1/2 hr to travel the remaining 16 mi.it will take 1+1/2 hr to travel the 48mi @ 32mph.so 1 hr = 60 min& 1/2 hr = 30 min1+1/2 hr = 90 min.there are 60s in each 90 × 60 = 3600 seconds.the real question seriously is he why are they driving so slow? ! lol.