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

Algebra ->  Test -> 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      Log On


   

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) About Me  (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
2w%5E2+%2B+7w+-4+=+2w%5E2+-2w+%2B50

Subtract out the 2w^2:
7w+-4=-2w%2B50+
9w+=+54
w=+6= Original Width
2w-2=+12-2=+10= Original Length

2w-1=12-1=+11 New Width
w%2B4+=+10+ 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