SOLUTION: The length of a rectangle is 2in. more than twice its width. If the perimeter of the rectangle is 52in., find the width of the rectangle. I need help with this one please, than

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 2in. more than twice its width. If the perimeter of the rectangle is 52in., find the width of the rectangle. I need help with this one please, than      Log On


   

Question 70374: The length of a rectangle is 2in. more than twice its width. If the perimeter of the rectangle is 52in., find the width of the rectangle.
I need help with this one please, thank you
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 2in. more than twice its width. If the perimeter of the rectangle is 52in., find the width of the rectangle.
The formula for the perimeter of a rectangle is highlight%28P=2L%2B2W%29
Let width (W) be: x
Then the length (L) is: 2x+2
P=52
Problem to solve:
52=2(2x+2)+2(x)
52=4x+4+2x
52=(4+2)x+4
52=6x+4
52-4=6x+4-4
48=6x
48/6=6x/6
8=x
The width (W) is: x=8 in.
Sanity check:
If the width is 8 in., then the length is 2(8)+2=16+2=18 in
Would a rectangle that is 8 in wide and 18 in long have a perimeter of 52 in?
52?=2(18)+2(8)
52?=36+16
52=52! We seem sane.
Happy Calculating!!!!