SOLUTION: the area of a square exceeds the area of a rectangle is 13squared. if the area of the square is y squared and the area of rectangle is (y+1)(y-3). find y.

Question 615177: the area of a square exceeds the area of a rectangle is 13squared. if the area of the square is y squared and the area of rectangle is (y+1)(y-3). find y.

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
the area of a square exceeds the area of a rectangle is 13 squared.
if the area of the square is y squared and the area of rectangle is (y+1)(y-3). find y.
:
Square area - rectangle area = 13^2
y^2 - (y+1)(y-3) = 169
FOIL
y^2 - (y^2 - 2y - 3) = 169
y^2 - y^2 + 2y + 3 = 169; removing the brackets changes the signs
2y = 169 - 3
2y = 166
y = 166/2
y = 83
:
:
See if that checks out
83^2 - (84*80) =
6889 - 6720 = 169 which is 13^2

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