Pls give full explanation asap - class is litterally in an hour if 9x+2y^2−3z^2=132 and 9y+2y^2−3z^2=−12, then x−y = (a) 3 (b) 8 (c) 16 (d) 120 (e) 144

(C) 16

Step-by-step explanation:

Subtract the second equation from the first equation.

9x + 2y^2 - 3z^2 = 132 minus 9y + 2y^2 - 3z^2 = -12

                            9x + 2y^2 - 3z^2 = 132

                            -(9y + 2y^2 - 3z^2 = -12 )  

This will change all the signs of the second equation.

9x + 2y^2 - 3z^2 = 132

-9y - 2y^2 + 3z^2 = +12

You have left 9x - 9y = 144.

Factor out the 9 so you have 9(x - y) = 144

Divide both sides by 9

x - y = 16

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3