I'm trying to find extrema for $f(x,y,z)=x+y+z$ with 2 constraints:
$$x^2-y^2=1 \\ 2x+z=1$$
So I'll say that $g(x,y,z)=x^2-y^2$ and $h(x,y,z)=2x+z$. So I'll try to find values of $\lambda$ and $\mu$ when:
$$\nabla f=\lambda . \nabla g + \mu . \nabla h$$
Taking the gradients of the 3 functions, I have: $$\nabla f= \langle 1;1;1\rangle\\ \nabla g=\langle 2x; -2y; 0\rangle \\ \nabla h=\langle 2;0;1\rangle $$
So this leaves me with the following 5 equations (3 from the multiplier definition and 2 from the restrictions):
$$\begin{cases} 1=2x\lambda+2\mu \\ 1=-2y\lambda \\ 1= \mu\\ 1= x^2-y^2\\ 1= 2x+z\\ \end{cases} $$
I started solving the first as $x=\frac{-1}{2 \lambda}$ and then the second as $y=\frac{-1}{2 \lambda}$ (I purposely avoided dividing both terms by $x$ or $y$ because I don't know if any of them are 0).
However, since now $x$ and $y$ have the same value, if I plug it in the fourth equation: $1= x^2-y^2$ I get that $1=0$.
What have I done wrong? And how could I solve an equation for $\lambda$, when I still don't know if $x$ or $y$ are $0$?
I don't think there is a mistake. It is just that the extrema do not exist. Consider:
$x = \sqrt{1 + y^2}$, $z = 1 - 2x$, then $x + y + z = x + y + 1 - 2x = 1 + y - x = 1 + y - \sqrt{1 + y^2}$. Differentiating wrt $y$, and equating to 0 gives $y = \sqrt{1 + y^2}$, which obviously has no solutions.