Prove that $f(x;y;z)=(x+y+z;x-y-2xz)$ can be resolve for $(x,y)=\phi(z)$

Question

Let be $f(x;y;z)=(x+y+z;x-y-2xz)$ a function, prove that it can be resolve for $(x,y)=\phi(z)$ close to $z=0$. Find explicitly $\phi(z)$

I'm very lost for this problem, first I thought it could be resolved withe the Implicit Function Theorem but it's for functions from $\mathbb{R}^{n}\rightarrow \mathbb{R}$.

Answer 1

I presume you want to solve $f=0$. For that purpose solve $$\begin{align} x+y+z&=0\\ x-y-2yz&=0 \end{align}$$ for $x$ and $y$. Substituting $y=-x-z$ in the second equation gives $$x=\frac{z}{2z-2},$$ from which $$y=\frac{z}{2z-2}-z$$ is easily computed.