50 points and

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 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: william can only attend a school in his designated zone. william's zone is defined by y < −x − 1. explain how you can identify the schools that william is allowed to attend. (2 points)

Part A--

In our graph, point A is at (2, -3) and point E is at (3, 1). Since we want to include these and only these points, the procedure is to first find a line through A and E, then take what's to the right as the TRUE space. What's to the left is the FALSE space.

To find the through A and E, we need the slope, m, first.

m = y₂ - y₁ / x₂- x₁

Choose x₁ as 3, y₁ as 1, x₂ as 2, y₂ as -3.

m =

m =

The slope of the line is 4. Using x = 3, y = 1 and m = 4, we can find the equation of the line and report it in slope-intercept form.

y = mx + b

1 = 3 * 4 + b

1 = 12 + b

-11 = b

With a slope of 4 and y-intercept of -11, the equation of the line is y = 4x -11.

Now we make the inequality. Choose one symbol and a test value. If the tests are true, keep it. If false, switch it. Anything to the right of A and E is what we want to be true; anything to the left is false. Let's choose a point to the right - any will work, but (5,0) will be ours. We want to make the inequality true when x = 5, y = 0.

y ≥ 4x - 11 (use ≥ to include the points)

0 ≥ 20 - 11

0 ≥ 9 FALSE

Try switching the inequality.

y ≤ 4x - 11

0 ≤ 20 -11

0 ≤ 9 TRUE

To check, choose a point to the left (where it's false). (0,0) works well.

y ≤ 4x - 11

0 < 0 -11

0 ≤ -11 FALSE

So the line we want is y ≤ 4x -11

To create a second equation, let's divide the city in two diagonally (like cutting bread). Let's go from bottom left to top right, and the equation for that line is

y = x (slope is 1, y-intercept is 0). Also, you can pick any horizontal or vertical line that works)

Let's use (5.0) as our test point yet again. We want it to true, so we want x to be bigger than y. Flip it around, and y is smaller than x.

A working system for the school choice is:

y ≤ 4x -11

y ≤ x

Part B--

We created the second equation to contain points A and E, but we can test each point in the two inequalities to see if they are true.

Point A is (2, -3) and Point E is (3, 1)

y ≤ x

Testing Point A: -3 ≤ 2 TRUE

Testing Point E: 1 ≤ 3 TRUE

y ≤ 4x -11

Testing Point A:

-3 ≤ 4(2) - 11

-3 ≤ 8 -11

-3 ≤ -3 TRUE

Testing Point E

1 ≤ 4(3) - 11

1 ≤ 12 -11

1 ≤ 1 TRUE

Both points work pairwise in both equations.

Part C--

We are now told that William can only go to schools where the equation y < - x - 1 is TRUE. Let's graph our line. Luckily, the line is already in slope-intercept form. The y-intercept is -1, so (0-1) is on the graph. A slope of 1 means that if we go down one unit and to the right one unit (negative y, positive x), that gives us a second point. So (1, -2) is on the graph. We need to find where the equation is true, so let's test x = 0, y = 0 and see which side to shade.

y < -x - 1

0 < 0 -1

0 < -1 TRUE

Anything above the line will make the equation true. Since it is < and not ≤, the points on the line are not included. Anything above the line is included. You could either (1) shade above the line (but not the line inclusive - use a broken line) and find the schools in the shaded area or (2) test the six points one by one to see if make the inequality y < - x - 1 true or false.

2nd one in the first row is correct

stethere all not eveanp-by-step explanation: