SOLUTION: The area of a square exceeds the area of a rectangle by 3 inches squared. The width of the rectangle is 3 inches shorter and the length 4 inches longer than the side of the square

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The area of a square exceeds the area of a rectangle by 3 inches squared. The width of the rectangle is 3 inches shorter and the length 4 inches longer than the side of the square      Log On


   

Question 666640: The area of a square exceeds the area of a rectangle by 3 inches squared. The width of the rectangle is 3 inches shorter and the length 4 inches longer than the side of the square. Find the side of the square.
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
A[s]=s^2 of square
A[r]=lw of rectangle
w = s-3
l = s+4
A[s] = A[r] + 3
s^2 = (s-3)(s+4) + 3
s^2 = s^2+s-12 +3
9-s = 0
s = 9 inches