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 143685: 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 rapaljer(4671) (Show Source): You can put this solution on YOUR website!
Let w= width of the original rectangle.
2w-2 = length of the original rectangle.
Original area = Width* Length=w(2w-2)
A=2w^2-2w
NEW RECTANGLE:
New length = 2w-2 + 1 = 2w-1
New width = w+4
New area = (w+4)*(2w-1)
A=2w^2 +7w -4
NEW AREA = Original Area + 50
Subtract out the 2w^2:
= Original Width
= Original Length
New Width
New Length (Seems these may be backwards!! This is the problem of the author!! It's NOT your problem!!
Check: Original area = 6*10 = 60 square units
New area = 11*10= 110 square units
It checks, since the new area is 50 more than the original area!!
R^2
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