# Need help solving this system of equations

• Apr 4th 2010, 06:28 PM
rainer
Need help solving this system of equations
Hi. I have to solve the following system of equations:

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

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

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

$\displaystyle x+z-c=0$ (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...

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

and then simplifying, I get:

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

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

(solving for lambda in equation three...)

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

and then substituting this in for lambda in equation one:

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

and then setting this equal to equation two and simplifying:

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

$\displaystyle y+z+2x=0$

So using the first method I get $\displaystyle \frac{1}{y+z}-x-z+c=0$, using the second method I get $\displaystyle y+z+2x=0$. 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.)
• Apr 4th 2010, 07:12 PM
Wilmer
Quote:

Originally Posted by rainer
$\displaystyle \frac{1}{y+z}+\lambda=0$ (equation 1)
$\displaystyle \frac{-x}{(y+z)^2}=0$ (equation 2)

STOP! Lambda is obviously = 0
• Apr 4th 2010, 07:38 PM
rainer
Quote:

Originally Posted by Wilmer
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.
• Apr 5th 2010, 05:08 AM
Wilmer
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).
• Apr 6th 2010, 03:49 PM
rainer
Quote:

Originally Posted by Wilmer
Looks like this problem was set up by a MAD math teacher!

Believe it or not I inflicted this problem on myself. $\displaystyle \frac{x}{y+z}$ 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