SOLUTION: Please Help The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle.

Algebra ->  Linear-equations -> SOLUTION: Please Help The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle.      Log On


   



Question 73831This question is from textbook Beginning Algebra 1A
: Please Help
The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle.
This question is from textbook Beginning Algebra 1A

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
THE FORMULA FOR THE PERIMETER IS P=2L+2W WHERE L=LENGTH, W=WIDTH
LENGTH=2W+2
52=2(2W+2)+2W
52=4W+4+2W
52-4=6W
48=6W
W=48/6
W=8 IN. IS THE WIDTH.
THEN THE LENGTH IS:
L=2*8+2
L=16+2
L=18 IN. IS THE LENGTH.
PROOF
52=2*8+2*18
52=16+36
52=52