Finding extrema using Lagrange multiplier (confusion)

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Use Lagrange Multiplier to locate the maximum and minimum points and values of $F(x, y, z) = 2x^2 − 2y + z^2$ subject to the constraint $x^2 + y^2 + z^2 = 1.$

Using the lagrange multiplier, I obtain 4 equations:

$4x\:=\:\lambda 2x,$

$\:-2\:=\:\lambda \:2y,$

$\:2z\:=\:\lambda \:2z,$

$\:x^{2\:}+\:y^2=1$

Now dividing equation $1$ and $3$ by $2$:

$\frac{4x}{-2}\:=\:\frac{\lambda 2x}{\:\lambda \:\:2y},$

$\:\frac{\:2z}{-2}\:=\:\frac{\lambda \:2z}{\:\lambda \:\:2y},$

In means that:

$x\:=\:0 $ or $\:y\:=\:-\frac{1}{2},$ $\:z\:=\:0$ or $\:y\:=\:-1$

Now here is my problem, I am not too sure how to interpret these values. Do I just pick a combination of these values and sub them into equation 4 or the constraint equation and find the other variable ?

For example:

the combinations are:

$x\:=\:0 $ and $\:y\:=\:-\frac{1}{2},$ and find $z$

$x = 0$ and $\:y\:=\:-1$ and find $z$

$x\:=\:0 $ and $z = 0$ and find $y$

$z\:=\:0$ or $\:y\:=\:-1$ and find $x$

$z\:=\:0 $ and $\:y\:=\:-\frac{1}{2},$and find $x$

But apparently this whole is not the correct "match"? How do I find the correct ones? Thanks.

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The set of equations to be solved is $$ \begin{cases}2x(2-\lambda) =0 \\ \lambda y = -1\\2z(1-\lambda)=0\\x^2+y^2+z^2 = 1 \end{cases} $$

Starting with the first equation, we have that $x = 0 \vee \lambda =2$, so, two cases:

CASE 1: $\lambda =2$.

The remaining variables are computed from $$ \begin{cases} 2 y = -1\\ -2z = 0\\ x^2+y^2+z^2 = 1\end{cases} $$ This yields the solutions $\left(\pm\frac{\sqrt{3}}{2}, -\frac 12, 0 \right)$

CASE 2: $x = 0$

The remaining variables must be computed from $$ \begin{cases}\lambda y =-1 \\ 2z(1-\lambda)=0\\y^2+z^2=1 \end{cases} $$

Looking at the second equation in this system, we see two sub-cases:

CASE 2.1: $z = 0$ (and $x=0$)

The last equation becomes $y^2 = 0$ and the solutions that arise from this case are $(0,\pm 1, 0)$

CASE 2.2: $\lambda = 1$ (and $x=0$)

Here $y=-1$ from the first equation and $z=0$ from the last. This yields the solution $(0,-1,0)$