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Homework Help: Discrete Math Help

  1. Jan 27, 2005 #1
    Discrete Math Help!!!

    Here is the problem:

    Suppose a, b, and c are integers and x, y, and z are nonzero real numbers that satisfy the following equations:

    [tex]\dfrac{xy}{x+y}=a [/tex] and [tex]\dfrac{xz}{x+z}=b [/tex] and [tex]\dfrac{yz}{y+z}=c [/tex].

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

    I have calculated that [tex]x=\dfrac{-(bz-xz)}{b}[/tex]. I am inclined to answer no, since x, y, or z could be irrational.

    Any help would be appreciated.
  2. jcsd
  3. Jan 27, 2005 #2

    first modify your equation to solve for x explicitly in terms of b and z ( x on one side , b and z terms on the other ).
    then use the other two equations to come up with an expression for z and substitute that into your original equation.
    Your goal is to try to solve for x in terms of a,b,c only... eliminating any z or y

    Post your steps if you run into more trouble
    Last edited: Jan 27, 2005
  4. Jan 27, 2005 #3
    Ok, I solved and got this:

    [tex]x=\dfrac{ay}{y-a}[/tex] and [tex]x=\dfrac{bz}{z-b}[/tex] and [tex]y=\dfrac{cz}{z-c}[/tex].

    I then solved for z and got [tex]z=\dfrac{aby}{(a-b)y+ab}[/tex]. I just cancelled the y's and was left with [tex]z=\dfrac{ab}{(a-b)+ab}[/tex].

    I plugged this in for z in the second equation, and got [tex]z=\dfrac{b \dfrac{ab}{(a-b)+ab}}{\dfrac{ab}{(a-b)+ab}-b}[/tex]. This simplifies to [tex]\dfrac{ab}{b-ab}[/tex].

    If this answer is correct, I would say that yes, x is indeed rational. Am I correct?
    Last edited: Jan 27, 2005
  5. Jan 27, 2005 #4
    Where did the x come from in the third equation (top right) , and where did the y go? That completely changes the problem so the rest is all wrong.

    Also, for
    [tex]z=\dfrac{aby}{(a-b)y+ab}[/tex] you can't cancel the y's, because y is not a factor of the denominator.
    If the denominator was instead [itex](a-b)y + aby[/itex] then you could factor out the y which could become [itex]y((a-b) + ab)[/itex] and the y's could cancel
    (do you see the difference? )

    Another thing: in your result,,, you solved for z (although incorrectly). The equation asks if X is rational... so how does knowing whether or not Z is rational help you?
    based on your equation relating x to z you could use the fact that z is rational to prove that x is rational providing you knew that theorem, but this seems like more trouble than its worth why not just solve for x. /edit

    Your going to have to solve for x in terms of y or z, and then solve for either y or z using the other equations to come up with an equation in terms of x, a, b, and c only

    Last edited: Jan 27, 2005
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