Drew is twice as old as his sister joy.The quotient of their ages five years ago is seven less than Joy's present age. Find out how old they are at present.

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  1. present age of Joy --- x
    present age of Drew --- 2x

    five years ago:
    Joy = x-5
    Drew = 2x-5

    (x-5)(2x-5) = x-7
    2x^2 -15x + 25 = x-7
    2x^2 - 16x + 32 = 0
    x^2 - 8x + 16 = 0
    (x-4)^2 = 0
    x = 4

    Joy is now 4, and Drew is 8

    5 years ago:
    Joy = -1 ??
    Drew = 3
    product of their ages = -3
    7 less than Joy's present age = 4-7 = -3

    notice that "mathematically" the answer is correct, but in reality the question is bogus.

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  2. There are 2 answers though. It could also be 10 and 20. I saw this when I solved it by making the quotient of their ages 5 years ago be equal to Joy's age-7 in one formula.
    multiply both sides by x-5, move everything to one side, and factor the quadratic you get (x-10)(x-4), and 2 answers to the problem are revealed.

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