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

Part A;
There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.

Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve.

An example of such system of equation is
x < 0
y > 0
The system of equation above represent all the points in the second quadrant of the coordinate system.
The area above the x-axis and to the left of the y-axis is shaded.

Part B:
It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true.

Substituting D(-4, 2) into the system we have:
-4 < 0
2 > 0
as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities.

Also, substituting E(-1, 5) into the system we have:
-1 < 0
5 > 0
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.

Part C:
Given that chicken can only be raised in the area defined by y > 3x - 4.

To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area.

For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true

For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false

For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false

For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true

For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true

For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true

Therefore, the farms in which chicken can be raised are the farms at point A, D, E and F.

20 percentage, you can also do it on a calculator.