SOLUTION: The length of a rectangle exceeds its width by 2 feet. If each dimension was increased by 3ft., the area would increased by 51 square feet.. Find the original dimensions.

Question 93071: The length of a rectangle exceeds its width by 2 feet. If each dimension was increased by 3ft., the area would increased by 51 square feet.. Find the original dimensions.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The length of a rectangle exceeds its width by 2 feet. If each dimension was increased by 3 ft., the area would increased by 51 square feet.. Find the original dimensions.
:
Let x = the width
Then
(x+2) = the length
and
Area = x(x+2)
:
If the dimensions are increase by 3 ft, then
(x+3) = new width
(x+5) = new length
and
New Area = (x+3)(x+5)
:
Problem equation
New area - old area = 51 sq/ft
:
(x+3)(x+5) - x(x+2) = 51
x^2 + 8x + 15 - x^2 - 2x = 51
x^2 - x^2 + 8x - 2x + 15 = 51
6x = 51 - 15
6x = 36
x = 36/6
x = 6 ft. width of the original rectangle
6 + 2 = 8 ft, length of the original rectangle
:
:
Check our solution
6 + 3 = 9 ft, width of the new rectangle
6 + 5 = 11 ft, length of the new rectangle
:
11*9 - 8*6 =
99 - 48 = 51

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