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.)