SOLUTION: I need help on this problem that I am simply stuck on. The perimeter of a rectangle is 196 ft. The length of the rectangle is 40 ft. less than twice the width. Find the width an

Algebra ->  Rectangles -> SOLUTION: I need help on this problem that I am simply stuck on. The perimeter of a rectangle is 196 ft. The length of the rectangle is 40 ft. less than twice the width. Find the width an      Log On


   

Question 919411: I need help on this problem that I am simply stuck on.
The perimeter of a rectangle is 196 ft. The length of the rectangle is 40 ft. less than twice the width. Find the width and the length of the rectangle.
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Let length = L, width = W

2(L+W) = 196 <--> L+W = 98

L = 2W - 40

Now you have two equations and two variables. Can you solve for L, W (hint: you can substitute L with 2W-40)?