Answer:
x=2 and y=36
Step-by-step explanation:
you have two equations
1) xy=72 and 2) (x+2)(y-4)=128
solve for one of the variables in equation 2), lets do y
(x+2)(y-4)=128, divide by (x+2) to both sides and add 4 to isolate variable y
therefore, y=(128/(x+2))+4
then plug in y into equation 1)
so x(our new y)=72
x( 128/(x+2) +4 )=72
distribute the x so you will get
(128/(x+2))x +4x=72
we are now solving for x, so isolate the x by subtracting 4x and multiplying by (x+2) to get
128x=(x+2)(72-4x)
multiply by distribution and combine like terms
-4x^2 -64x +144 =0
solve for x by quadratic formula, you'll get x=2 or x=-18
substitute back into equation 1), (2)(y)=72, solve for y to get 36
then you can check by substituting back into equation 2)
(2+2)(36-4)=128
(4)(32)=128, 128=128
