SOLUTION: The width of a rectangle is 2 feet less than 10 times its lenght. If the perimeter of the rectangle is 62 feet find its dimensions.
I began with 2l+2w=P
So.....2L+(2-10L)+(2
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Question 8539: The width of a rectangle is 2 feet less than 10 times its lenght. If the perimeter of the rectangle is 62 feet find its dimensions.
I began with 2l+2w=P
So.....2L+(2-10L)+(2-10L)=62
But that's not correct. How do I set this up to solve?
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
"2 feet less than 10 times its length" means you must start with 10L and then take away 2 feet from this. It should be 10L-2, not 2-10L.
Solve this:
2L + 10L-2 + 10L-2 = 62
22L - 4 = 62
22L = 66
L=3 = Length
Then, 10L-2 = 10(3)-2= 28 = width.
Interesting, the length and the width seem to be reversed. It could be an oversight by the person who made up the problem. I've done it myself.
R^2 at SCC
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