Question 549121
A SQUARE AND A RECTANGLE HAVE THE SAME AREA. 
THE LENGTH OF THE RECTANGLE IS 8 MORE THAN A SIDE OF THE SQUARE, AND THE WIDTH OF THE RECTANGLEIS 4 LESS THAN A SIDE OF THE SQUARE.
 FIND A SIDE OF THE SQUARE.
:
Let x = the side of the square
then
x^2 = the area of the square
:
(x+8) = the length of the rectangle
(x-4) = the width of the rectangle
then FOIL
(x+8)(x-4) = x^2 - 4x + 8x - 32
x^2 + 4x - 32 = the area of the rectangle
:
Given the areas areas are equal
x^2 + 4x - 32 = x^2
Combine the x^2's on the left
x^2 - x^2 + 4x - 32 = 0
4x = 32
x = 32/4
x = 8 units is the side of the square
:
:
See if this is right
(8+8) * (8-4) = 8^2
16 * 4 = 64; confirms our solution of x=8