SOLUTION: The length of a rectangle is 5 more than twice its width. If the width is decreased by 3 and the length by 1, the perimeter becomes four-fifths of the original. Find the original d

Question 218159: The length of a rectangle is 5 more than twice its width. If the width is decreased by 3 and the length by 1, the perimeter becomes four-fifths of the original. Find the original dimensions of the rectangle.
Please send me the solution as soon as possible. Thanks MATH Masters~
Answer by alicealc(293) (Show Source): You can put this solution on YOUR website!
length = l
width = w

l = 5 + 2*w
2*(w - 3) + 2*(l - 1) = 4/5 * (2*l + 2*w)

substitute the first equation into the second equation:
2*(w - 3) + 2*(5 + 2w - 1) = 4/5 * (2*(5 + 2*w) + 2*w)
2w - 6 + 10 + 4w - 2 = 4/5 * (10 + 4w + 2w)
6w + 2 = 40/5 + 16/5 w + 8/5 w
6w + 2 = 8 + 24/5 w
6w - 24/5 w = 8 - 2
30/5 w - 24/5 w = 6
6/5 w = 6
w = 6 * 5/6 = 30/6 = 5
l = 5 + 2*w = 5 + 2*5 = 5 + 10 = 15

so, the original dimension of the rectangle is:
length = 15, and width = 5
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