There are many systems of equation that will satisfy the requirement for Part A.
an example is y≤(1/4)x-3 and y≥(-1/2)x-6
y≥(-1/2)x-6 goes through the point (0,-6) and (-2, -5), the shaded area is above the line. all the points fall in the shaded area, but
y≤(1/4)x-3 goes through the points (0,-3) and (4,-2), the shaded area is below the line, only A and E are in the shaded area.
only A and E satisfy both inequality, in the overlapping shaded area.
Part B. to verify, put the coordinates of A (-3,-4) and E(5,-4) in both inequalities to see if they will make the inequalities true.
for y≤(1/4)x-3: -4≤(1/4)(-3)-3
-4≤-3&3/4 This is valid.
For y≥(-1/2)x-6: -4≥(-1/2)(-3)-6
-4≥-4&1/3 this is valid as well. So Yes, A satisfies both inequalities.
Do the same for point E (5,-4)
Part C: the line y<-2x+4 is a dotted line going through (0,4) and (-2,0)
the shaded area is below the line
farms A, B, and D are in this shaded area.
Answer:
Look at the explanation.
Step-by-step explanation:
Start y=-2x-3 at the Y -3 and X 0, then go up two and to the right one, like a staircase. For y=-x+3 start on Y 3 and X 0 and go down 1 x and 1 y like a downstair staircase with 1 unit down.
Answer:
This equation is in standard form, so to make graphing easier you need to put it in y=mx+b form or slope, intercept form.
Step-by-step explanation:
I have attached a photo of what the graph would look like.
However, in the future you should know how to turn this into a slope intercept equation.
-2x+7y=-21
+2x +2x
7y=2x-21
divide the entire equation by 7
y= 2/7x-3
That is a full equation in slope intercept form.
Hopefully this helps!
