<s>[INST] Solve $\begin{cases} x + y + z = 2 \\ 2xy - z^2 = 4 \\ \end{cases} $ for x, y, z. It came to my mind to rewrite the expression above as 
$$\begin{cases}
x + y = 2 - z \\ 
2xy = (2 - z)^2 + 4z \\
\end{cases}
$$
and see if there any restrictions on the values of the variables occur.
What I can see is that $(2 - z)^2 + 4z \gt 0$ for all $z$ (since $z^2 > -4$) but when I express $y$ in terms of $z$ and $x$ 
$$\begin{cases}
y = 2 - z - x\\ 
x(x - (2 - z)) = (2 - z)^2 + 4z \\
\end{cases}
$$
I see that $x(x - (2 - z)) \gt 0$ not for all $z$. This is all I got for now.
The answer is $(2, 2, -2)$.
 [/INST] $\bf{My\; Solution::}$ Given $$x+y+z = 2$$ and $$2xy-z^2 = 4$$
Now Put $z=2-(x+y)$ into $2xy-z^2=4\Rightarrow 2xy-[2-(x+y)]^2=4$
So we get $$2xy-[4+(x+y)^2-4(x+y)]=4$$
So we get $$2xy-4-x^2-y^2-2xy+4x+4y=4$$
So we get $$x^2+y^2-4x-4y+8=0$$
So we get $$(x-2)^2+(y-2)^2 = 0\;,$$ Which is possible for real no. when 
$(x-2)=0$ and $(y-2) = 0$ , So we get $x=2$ and $y=2$ and $z=-2$
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