The coordinate plane below represents a town. Points A through F are farms in the town.

graph of coordinate plane. Point A is at 2, negative 3. Point B is at negative 3, negative 4. Point C is at negative 4, 2. Point D is at 2, 4. Point E is at 3, 1. Point F is at negative 2, 3.

Part A: Using the graph above, create a system of inequalities that only contains points D and F 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 F 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 − 2. Explain how you can identify farms in which chickens can be raised. (2 points)

Part A: To create a system of inequalities that only contains points D and F in the overlapping shaded regions, we can draw two lines that intersect at D and F, and shade the region in which both points lie.

First, we need to find the equations of the lines that pass through points D and F.

The line passing through points D (2, 4) and F (-2, 3) can be found using the slope-intercept form of a line equation: y = mx + b, where m is the slope and b is the y-intercept.

Slope (m) = (change in y) / (change in x) = (4 - 3) / (2 - (-2)) = 1 / 4

Substituting the slope and the coordinates of point D (2, 4) into the slope-intercept form, we get:
4 = (1/4)(2) + b
4 = 1/2 + b
b = 4 - 1/2
b = 7/2

So, the equation of the line passing through point D is:
y = (1/4)x + 7/2

Similarly, for point F (-2, 3), we can find its line equation as:
3 = (1/4)(-2) + b
3 = -1/2 + b
b = 3 + 1/2
b = 7/2

The equation of the line passing through point F is:
y = (1/4)x + 7/2

Now, to create the system of inequalities, we can write:
y ≤ (1/4)x + 7/2 (equation of the line passing through D)
y ≥ (1/4)x + 7/2 (equation of the line passing through F)

Graphically, the line passing through D is upward sloping (positive slope) and the line passing through F is also upward sloping. Since we want the overlapping shaded region, we will shade the area between these two lines.

Part B: To verify that points D and F are solutions to the system of inequalities created in Part A, we substitute the coordinates of each point into the inequalities and check if the inequalities hold true.

For point D (2, 4), substituting into the inequalities:
4 ≤ (1/4)(2) + 7/2 is true
4 ≥ (1/4)(2) + 7/2 is true

For point F (-2, 3), substituting into the inequalities:
3 ≤ (1/4)(-2) + 7/2 is true
3 ≥ (1/4)(-2) + 7/2 is true

Since all the inequalities hold true, points D and F are indeed solutions to the system of inequalities.

Part C: To identify farms in which chickens can be raised, we need to check if the coordinates of the farms satisfy the inequality y > 2x - 2.

For each farm's coordinates (x, y), substitute them into the inequality and check if it holds true.

If the inequality holds true, then the farm lies in the area where chickens can be raised.

For example, let's check point A (2, -3):
-3 > 2(2) - 2
-3 > 4 - 2
-3 > 2

Since -3 is not greater than 2, point A is not in the area where chickens can be raised.

Repeat this process for each farm's coordinates to identify the farms where chickens can be raised.

Part A: To create a system of inequalities that only contains points D and F in the overlapping shaded regions, we need to determine the equations for the lines that form the boundaries of the shaded region.

Looking at the points D (2, 4) and F (-2, 3), we can see that the line connecting these two points will form one boundary. To find the equation of this line, we can use the slope-intercept form:

m = (y2 - y1) / (x2 - x1)

m = (3 - 4) / (-2 - 2)

m = -1 / -4

m = 1/4

Using the point-slope form of an equation, we have:

y - y1 = m(x - x1)

y - 4 = (1/4)(x - 2)

Multiplying through by 4 to eliminate fractions, we get:

4y - 16 = x - 2

x - 4y = -14

Now, for the second boundary, we can use the line y = 2x - 2. This line will be graphed as a dashed line since the inequality is strict (y > 2x - 2).

To graph these lines on the coordinate grid above, plot the points D and F. Then, draw a line passing through D and F. This line represents the boundary that includes both points. Finally, draw the line y = 2x - 2 as a dashed line on the graph.

Part B: To verify that the points D (2, 4) and F (-2, 3) are solutions to the system of inequalities created in Part A, substitute the coordinates of each point into the inequalities and check if the inequalities hold true.

For point D:
x - 4y = -14

Substituting x = 2 and y = 4, we have:
2 - 4(4) = -14

Simplifying:
2 - 16 = -14

-14 = -14 (True)

For point F:
y > 2x - 2

Substituting x = -2 and y = 3, we have:
3 > 2(-2) - 2

Simplifying:
3 > -4 - 2

3 > -6 (True)

Since both inequalities hold true when the coordinates of points D and F are substituted, we can conclude that D and F are solutions to the system of inequalities created in Part A.

Part C: To identify the farms in which chickens can be raised, we need to determine which farms lie in the area defined by the inequality y > 2x - 2.

For each farm, substitute the coordinates (x, y) into the inequality y > 2x - 2. If the inequality is true for a particular farm, then that farm is suitable for raising chickens. If the inequality is false, then that farm would not be appropriate for raising chickens.

For example, let's take the point A (2, -3):
-3 > 2(2) - 2
-3 > 4 - 2
-3 > 2 (False)

Since the inequality is false for point A, it means that farm A is not suitable for raising chickens.

Repeat this process for each farm in the town to determine which farms satisfy the inequality y > 2x - 2. Farms that satisfy this inequality can be identified as suitable for raising chickens.

A: The line through D and F is

y = x/4 + 7/2
So, since all the other points lie below this line, and above C, you need

2 < y < x/4 + 7/2

C: Just plug in the points and see whether they satisfy y > 2x-2