SOLUTION: If the perimeter of a rectangle is 260 feet and the length is 30 feet more than the width, what are the dimensions of the rectangle?

Question 1055662: If the perimeter of a rectangle is 260 feet and the length is 30 feet more than the width, what are the dimensions of the rectangle?
Found 2 solutions by ikleyn, addingup:
Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
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If the perimeter of a rectangle is 260 feet and the length is 30 feet more than the width, what are the dimensions of the rectangle?
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Let W be the width, then the length is W+30. W + (W+30) + W + (W+30) = 260, 4W + 60 = 260, 4W = 260-60 ---> 4W = 200 ---> W = = 50. Answer. The wide is 50 ft. The length is 50+30 = 80 ft.

Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
2L+2W = Perimeter
L = W+30 substitute for L in the equation above:
2(W+30)+2W = 260
2W+60+2W = 260
4W = 200
W = 50
and
L = 50+30 = 80
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