SOLUTION: The length of a rectangle exceeds its width by 6 units. If each dimension were increased by 3 units. The area would be increased by 57 square units. Find the dimensions of the rect

Question 1066555: The length of a rectangle exceeds its width by 6 units. If each dimension were increased by 3 units. The area would be increased by 57 square units. Find the dimensions of the rectangle.
Found 3 solutions by Boreal, MathTherapy, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
l=w+6
area=w(w+6)=w^2+6w
(w+3)(w+9)-(w^2+6w)=57
w^2+12w+18-w^2-6w=57
6w+18=57
6w=39
w=6.5 units
l=9.5 units, area=61.75 u^2
make them 9.5*12.5 and the area is 118.75 u^2, 57 more
6.5 x 9.5
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

The length of a rectangle exceeds its width by 6 units. If each dimension were increased by 3 units. The area would be increased by 57 square units. Find the dimensions of the rectangle.

Answer by ikleyn(52855)   (Show Source): You can put this solution on YOUR website!
.

l = w+6 area = w(w+6) = w^2+6w (w+3)(w+9)-(w^2+6w)=57 w^2+12w+27-w^2-6w = 57 6w+27 = 57 6w = 30 w = 5 units l = 11 units, area = 55 u^2

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