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

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) (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 . Since the length is 2 ft more than the width, we can say that

Start with the given equation.

Plug in and

Distribute.

Combine like terms.

Subtract 4 from both sides.

Combine like terms.

Divide both sides by 4 to isolate W.

Reduce.

So the width is ft and the length is ft
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