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Math Help - [SOLVED] Tricky Fraction Problem

  1. #1
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    [SOLVED] Tricky Fraction Problem

    Question:
    The numerator of a fraction is two less than the denominator. The sum of the fraction and its reciprocal is 25/12. Find the numerator and denominator is a positive integer.
    Answer: 6/8

    Where do I start on this problem? Should there only be one variable?
    Here is what I tried...but it did not work:

    ((x-2)/x) + (x/(x-2)) = 25/2
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  2. #2
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    Figured it out!

    I just didn't take it far enough...my equation was set up right.
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by polt View Post
    Question:
    The numerator of a fraction is two less than the denominator. The sum of the fraction and its reciprocal is 25/12. Find the numerator and denominator is a positive integer.
    Answer: 6/8

    Where do I start on this problem? Should there only be one variable?
    Here is what I tried...but it did not work:

    ((x-2)/x) + (x/(x-2)) = 25/12
    Hi polt,

    \frac{x-2}{x}+\frac{x}{x-2}=\frac{25}{12}

    GCD= 12x(x-2)

    12(x-2)^2+12x^2=25x(x-2)

    12(x^2-4x+4)+12x^2=25x^2-50x

    12x^2-48x+48+12x^2=25x^2-50x

    -x^2+2x+48=0

    x^2-2x+48=0

    (x-8)(x+6)=0

    x=8 \ \ or \ \ x=-6

    Use x = 8 to make it positive

    \frac{x-2}{x}=\frac{6}{8}=\frac{3}{4}
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