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Mathematics
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Suppose that the width of a certain rectangle is 8 inches less than its length. The area is numerically 7 less than twice the perimeter. Find the length and width of the rectangle
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you know that w = l - 8 and that l*w = 2*2*(l + w) - 7 plugging in the first equation into the second, you get l^2 - 8*l = 8*l - 23 or l^2 - 16*l + 39 = 0 which factors into (l - 13)(l-3) = 0 or l = 3,13. Therefore l = 13, w = 5
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