3. (05.06 MC) Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions:
y > 3x + 10
y<- 3/4X-1
Part A: Describe the graph of the system, including shading and the types of lines graphed. Provide a description of the solution area. (6 points)
Part B: Is the point (8, 10) included in the solution area for the system? Justify your answer mathematically. (4 points)
(10 points)

>= means greater than or equal to

<= means less than or equal to

---------------------------------------------

Part A

The graph of y >= -3x+3 will have a solid boundary line and the shading will be above the boundary line.

The boundary line y = -3x+3 has a negative slope so it moves down as you read it from left to right. It goes through the points (0,3) and (1,0)

--------------

The graph of y < (3/2)x - 6 will have a dashed or dotted boundary line. The shading is below the boundary.

The graph y = (3/2)x-6 goes through the two points (0,-6) and (2,-3)

--------------

If you graph both y >= -3x+3 and y < (3/2)x - 6 together, you get what you see in the attached image. This solution shaded region is the result of the overlapping prior shaded regions.  

---------------------------------------------

Part B

Plug (x,y) = (-6,3) into each inequality to see if we get a true inequality or not

For the first inequality we have

y >= -3x+3

3 >= -3*(-6)+3

3 >= 18+3

3 >= 21

which is false. The value 3 is not larger or equal to 21. So right off the bat we know that (-6,3) is NOT a solution. It is NOT in the solution region.

Let's check the other inequality just for the sake of completeness

y < (3/2)x - 6

3 < (3/2)*(-6) - 6

3 < -9 - 6

3 < -15

this is also false. The value -15 is smaller than 3, since it is to the left of 3

We're given more evidence that (-6,3) is NOT in the solution area. It is outside of both shaded areas.  

answer: what

step-by-step explanation:

answer: need picture of problem

step-by-step explanation: