<PRE><B>Question 103909</B>
suppose that the width of a rectangle is 5 inches shorter than the length and that the perimeter of the rectangle is 50.
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Let length be L ;then width = L-5
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set up an equation for the perimeter involving only L, the length of the rectangle.
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Perimeter = 2L + 2(L-5) = 4L-10
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solve this equation algebraically to find the length of the rectangle. find the width as well.
50 = 4L-10
4L = 60
L = 15 inches (this is the length)
L-5 =10 inches (this is the width)
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Cheers,
Stan H.
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