SOLUTION: the length of a rectangle is 8 in more than twice its width. if the perimeter of the rectangle is 58 in, find the width of the rectangle.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: the length of a rectangle is 8 in more than twice its width. if the perimeter of the rectangle is 58 in, find the width of the rectangle.      Log On


   



Question 82110: the length of a rectangle is 8 in more than twice its width. if the perimeter of the rectangle is 58 in, find the width of the rectangle.
Answer by weepingwillowcats(10) About Me  (Show Source):
You can put this solution on YOUR website!
In order to solve this, you need to set up an equation.
First we'll find the length.
The length is going to be twice the width, so l = 2w.
Then it also says it's eight more than twice the width, so l = 2w+8.
Now, set up your equation for the perimeter (add all the sides).
2w + 8 + 2w + 8 + w + w = 58
Combine like-terms.
6w + 16 = 58
Subtract 16 from both sides.
6w = 42
Divide by six on both sides.
w = 7
To check your answer, simply plug the answer back into the equation.
Also, if you have any further questions, don't hesitate to ask.
Good luck!
-weepingwillowcats