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How is it possible to solve this system of 2 equations with 4 unknowns?

  1. Sep 14, 2012 #1
    I've always been taught that to solve for 4 unknowns, you need 4 equations. This question seems to suggest that there are in fact solutions. Here's the problem and my work so far. The most I can seem to do is eliminate one variable, I'm not sure how to shake a second one out, let alone all 4.

    Joseph-1.png
     
  2. jcsd
  3. Sep 14, 2012 #2

    jbunniii

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    For linear equations: You need 4 equations in order for there to be a UNIQUE solution. With fewer equations than unknowns, there will be either infinitely many solutions or no solutions, depending on whether the equations are consistent or not.

    However, note that your second equation is not linear in x, y, and z.

    Your work so far is correct, but I don't think it's getting you any closer to a solution.

    I suggest solving one of the equations for one of the variables in terms of the others. Then substitute into the other equation and simplify. Most likely there will be infinitely many solutions, so the goal is to try to characterize the set of solutions in the simplest way possible.
     
  4. Sep 14, 2012 #3

    vela

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    If you have four unknowns, you need four independent equations to find a unique solution. You won't be able to find a unique solution in this case.
     
  5. Sep 14, 2012 #4
    Well, I've found a fairly trivial set of solutions, but are these the only ones?

    Solutions 1: {x = w, z = -y}
    Solution 2: {x = -z, y = w}
    Solution 3: {y = x, z = w}

    The real question is - are these solutions the only ones?
     
  6. Sep 14, 2012 #5

    jbunniii

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    Actually this equation might be on the right track:

    [tex]yz + xz + xy = \frac{xyz}{x + y + z}[/tex]

    Try solving for one of the variables in terms of the others. This will probably involve the quadratic formula. Also be careful that this equation is true only if [itex]x + y + z \neq 0[/itex].
     
  7. Sep 14, 2012 #6
    I think we're off track. I've confirmed using mathematica that


    Untitled-6.png



    which is true for all x + y + z /= 0 and y + z /= 0.

    However, this is a putnam problem, so I know that this isn't the sort of solution they are looking for. There has to be a more elegant solution.
     
  8. Sep 14, 2012 #7

    vela

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    You probably want to take advantage of the symmetry with which x, y, and z appear in the problem.
     
  9. Sep 14, 2012 #8
    Well, I feel like I've already done that with these solutions. I think the union of these three sets would give the complete set of solutions. Isn't this the best we can do with 4 variables and 2 equations?

    2wcjg3p.png
     
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