SOLUTION: The length of a rectangle is 2 less than twice the width. If the length is increased by 1 and the width is increased by 4, the resulting rectangle has an area which is 50 more than
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Question 138693: The length of a rectangle is 2 less than twice the width. If the length is increased by 1 and the width is increased by 4, the resulting rectangle has an area which is 50 more than the area of the original rectangle. Find the dimensions of the original rectangle. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The length of a rectangle is 2 less than twice the width. If the length is increased by 1 and the width is increased by 4, the resulting rectangle has an area which is 50 more than the area of the original rectangle. Find the dimensions of the original rectangle.
:
Let x = width of original rectangle
Then
(2x-2)= length of the original rectangle
Area:
x*(2x-2) = (2x^2 - 2x)
:
(2x-2) + 1 = length of the new rectangle
or simplified:
(2x-1) = the new length
and
(x+4) = the new width
New area:
FOIL (2x-1)(x+4) = (2x^2 + 7x - 4)
:
New area - old area = 50
(2x^2 + 7x - 4) - (2x^2 - 2x) = 50
2x^2 + 7x - 4 - 2x^2 + 2x = 50
2x^2 - 2x^2 + 7x + 2x - 4 = 50
9x = 50 + 4
x =
x = +6
:
Length = 2(6) - 2
Old rectangle dimensions 10 by 6
;
:
Check solutions by finding the difference between the two area
(11*10)-(10*6)= 50
