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Math Help - Need help solving this system of equations

  1. #1
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    Need help solving this system of equations

    Hi. I have to solve the following system of equations:

    \frac{1}{y+z}+\lambda=0 (equation 1)

    \frac{-x}{(y+z)^2}=0 (equation 2)

    \frac{-x}{(y+z)^2}+\lambda=0 (equation 3)

    x+z-c=0 (equation 4)


    I am confused because I see two solutions to this system of equations. First:

    If I solve by setting all of the equations equal to each other...

    \frac{-x}{(y+z)^2}= \frac{-x}{(y+z)^2}+\lambda= \frac{1}{y+z}+\lambda= x+z-c=0

    and then simplifying, I get:

    \frac{1}{y+z}-x-z+c=0

    Then if I solve by first solving for \lambda in one of the equations...

    (solving for lambda in equation three...)

    \lambda=\frac{x}{(y+z)^2}

    and then substituting this in for lambda in equation one:

    \frac{1}{y+z}+\frac{x}{(y+z)^2}=0

    and then setting this equal to equation two and simplifying:

    \frac{1}{y+z}+\frac{x}{(y+z)^2}=-\frac{x}{(y+z)^2}

    y+z+2x=0

    So using the first method I get \frac{1}{y+z}-x-z+c=0, using the second method I get y+z+2x=0. Which one is right? Or do I now have to set them equal?

    Thanks

    (Note: I posted a slightly different version of this question in the calculus forum. I've decided to post it here because it is essentially an algebra question. No need to confuse people with the calculus background to the problem.)
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  2. #2
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    Quote Originally Posted by rainer View Post
    \frac{1}{y+z}+\lambda=0 (equation 1)
    \frac{-x}{(y+z)^2}=0 (equation 2)
    STOP! Lambda is obviously = 0
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  3. #3
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    Quote Originally Posted by Wilmer View Post
    STOP! Lambda is obviously = 0
    I can see that. But I don't see what result that fact leads to. It's probably really obvious to you, can you please help me out? I've second-guessed myself so many times on this problem that I can no longer tell up from down.
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  4. #4
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    Looks like this problem was set up by a MAD math teacher!

    From equations 2 and 3: lambda = 0

    So equation 2 is SAME as equation 3 ; you got only 3 equations

    Equation 1 : 1 / (y + z) = 0 is impossible, since numerator must = 0;
    and 1 is certainly <> 0

    Equation 2: -x / (y + z)^2 = 0 means numerator = 0, so x = 0

    Substitute x=0 in equation 4 to get: c = z

    That's all she wrote (as far as I can see).
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  5. #5
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    Quote Originally Posted by Wilmer View Post
    Looks like this problem was set up by a MAD math teacher!
    Believe it or not I inflicted this problem on myself. \frac{x}{y+z} is Marx's famous formula for the rate of profit (where x=profit, y=constant capital, and z=variable capital). Marx's theory holds that x+z and y will increase absolutely over time, but that x+z will decrease relative to y, thus resulting in a tendency for the rate of profit to fall.

    I was doing a constrained optimization of the formula to see if I could reach a more rigorous formalization. The system of equations are the first order conditions of that process. As you point out, the first equation cannot equal 0. This is what was tripping me up. Big face palm.

    Anyway, in the post I did over in the calculus forum, Opalg pointed out that the issue can be resolved via a much more elementary approach. Here it is in case you're interested:

    http://www.mathhelpforum.com/math-he...ion-again.html
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