SOLUTION: Find the length and width of a rectangle when the length is 5 feet longer than the width. The perimeter of the rectangle is 50 feet.
Here is what I have:
2l+2w=P
2(x+5)+2
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-> SOLUTION: Find the length and width of a rectangle when the length is 5 feet longer than the width. The perimeter of the rectangle is 50 feet.
Here is what I have:
2l+2w=P
2(x+5)+2
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Question 250295: Find the length and width of a rectangle when the length is 5 feet longer than the width. The perimeter of the rectangle is 50 feet.
Here is what I have:
2l+2w=P
2(x+5)+2(x)=50
2x+10+2x=50
4x+10=50
subtract 10 from both sides
4x=40
divide then x=10
Is this correct? Answer by richwmiller(17219) (Show Source):
