SOLUTION: i am just practicing here. This is my question: The length of a rectangle is 2 ft. more than the width. The perimeter of the rectangle is 20 feet. Find the length. Tell me if th

Algebra ->  Rectangles -> SOLUTION: i am just practicing here. This is my question: The length of a rectangle is 2 ft. more than the width. The perimeter of the rectangle is 20 feet. Find the length. Tell me if th      Log On


   

Question 293259: i am just practicing here. This is my question:
The length of a rectangle is 2 ft. more than the width. The perimeter of the rectangle is 20 feet. Find the length. Tell me if this is the correct way.
P= 2L + 2L
P= 2*(2) + 20
P= 4 + 20
P= 24
L= 6 because 24 divided by 4 = 6(answer)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm not sure how you got that answer (it's the correct answer) since your steps are incorrect.

Let L = length and W = width


Recall that the perimeter of a rectangle is P=2L%2B2W. Since the length is 2 ft more than the width, we can say that L=W%2B2


P=2L%2B2W Start with the given equation.


20=2%28W%2B2%29%2B2W Plug in P=20 and L=W%2B2


20=2W%2B4%2B2W Distribute.


20=4W%2B4 Combine like terms.


20-4=4W Subtract 4 from both sides.


16=4W Combine like terms.


16%2F4=W Divide both sides by 4 to isolate W.


4=W Reduce.


So the width is W=4 ft and the length is L=W%2B2=4%2B2=6 ft