SOLUTION: The perimeter of a rectangle is 262 inches. The length exceeds the width by 63 inches. Find the length and the width. I read this question as the equation of: 63(L)+w=262. Am I on

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Question 197543: The perimeter of a rectangle is 262 inches. The length exceeds the width by 63 inches. Find the length and the width. I read this question as the equation of: 63(L)+w=262. Am I on the right track. Please help.
Found 2 solutions by vleith, Alan3354:
Answer by vleith(2983) (Show Source):
You can put this solution on YOUR website!
Look at this, I bet you can figure it out from there.
http://www.algebra.com/algebra/homework/word/age/Age_Word_Problems.faq.question.196407.html
In your case, Length=Width+63 and the perimeter is different. But the solution to problem uses the same process.
Let me know if you still can't get it

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
The perimeter of a rectangle is 262 inches. The length exceeds the width by 63 inches. Find the length and the width.
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P = 2L + 2W = 262
L = W + 63
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262 = 2(W + 63) + 2W
4W + 126 = 262
4W = 136
W = 34
L = 97