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Rational Number Help

  1. Jul 12, 2004 #1
    Suppose a, b, and c are integers and x, y, and z are nonzero real numbers that satisfy the following equations:
    xy/x+y = a
    xz/x+z = b
    yz/y+z = c

    Is x rational? If so, express it as a ratio of two integers.

    I am pretty sure x is rational, but I don't know how to get the ratio. I am guessing the ratio uses a, b, or c. I tried solving for x, but that got me no where.
  2. jcsd
  3. Jul 12, 2004 #2


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    Now, I have to assume that you mean xy/(x+y), with the brackets. You really should write those brackets in instead of leaving it to those trying to help you having to guess what you mean (technically, without the brackets, the equations become 2y = a, 2z = b = c, and x could be any non-zero, so we have to guess you mean it with the brackets). Now:

    [tex]x = \frac{ya}{y - a},\ y = \frac{zc}{z - c},\ z = \frac{xb}{x - b}[/tex]

    [tex]x = \frac{ya}{y - a},\ y = \frac{\left (\frac{xb}{x - b} \right )c}{\left (\frac{xb}{x - b} \right ) - c}[/tex]

    Take your equation for x, substitute what you have for y, and you have an equation with only x, a, b, and c. Isolate x, and see what you get. I would think you'd get something like [itex]f(a,b,c)/g(a,b,c)[/itex] where the functions f and g are simple ones that take a, b, and c and simply add or multiply or divide them together in different ways. If this is so, and you get a non-zero denominator and numerator, then x is rational.
  4. Jul 12, 2004 #3


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    These equations can be re-written as
    [tex]{1\over x}+{1\over y}={1\over a}[/tex]
    etc. In this form, they are very easy to solve.
  5. Jul 14, 2004 #4
    Krab is right provided that you define [tex] x^{-1} =u [/tex] and the like for y and z and work with those auxiliary variables
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