Extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange Multipliers 
Find the extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange multipliers.

So I set it up:
$$
1 = 2x\lambda_1 + 2\lambda_2 \\
1 = -2y\lambda_1 \\
1 = \lambda_2
$$
Plug in for $\lambda_2$:
$$
1 = 2x\lambda_1 + 2 \\
1 = -2y\lambda_1 \\
$$
So we work with:
$$
1 = 2x\lambda_1 + 2 \\
1 = -2y\lambda_1 \\
1 = x^2 - y^2 \\
1 = 2x + z
$$
After some algebra I got $x = y$ as a solution but that's impossible because of the constraint $1 = x^2 - y^2$. What am I missing?

Note: There is no extremas with the current situation...So I slightly changed one of the 
"constraints" to: 
$x^2 - 2y^2 = 1$, and to show you how to solve if using one parameter $LM$ method.
$x + y + z = x + y + (1 - 2x) = y - x + 1$, subject to: $x^2 - 2y^2 = 1$.
Define $f(x,y) = -x + y + 1$, and $g(x,y) = x^2 - 2y^2$. Then:
$\nabla f = \lambda\cdot \nabla g \to (-1,1) = (2x\lambda,-4y\lambda)$. Thus:
$-1 = 2x\lambda$
$1 = -4y\lambda$. 
Add the above equations: $2\lambda(x - 2y) = 0 \to \lambda = 0$ or $x = 2y$. But $\lambda$ cannot be $0$, so $x = 2y$, and with $x^2 - 2y^2 = 1$ we get:
$4y^2 - 2y^2 = 1 \to 2y^2 = 1 \to y = \pm \dfrac{1}{\sqrt{2}}$. So $x = 2y = \pm \sqrt{2}$.
$f_{min} = f(\sqrt{2},\frac{1}{\sqrt{2}}) = \dfrac{\sqrt{2} - 1}{\sqrt{2}}$, and
$f_{max} = f(-\sqrt{2},-\frac{1}{\sqrt{2}}) = \dfrac{\sqrt{2} + 1}{\sqrt{2}}$
