$\left[\begin{array}{l}\text { Systems of algebraic nonlinear equations }] \\ {[\underline{\text { Cauchy inequality }}}\end{array}\right]$

How many real solutions does the system of two equations with three unknowns have:

$$
\begin{aligned}
& x+y=2 \\
& x y-z^{2}=1 ?
\end{aligned}
$$
$(x-y)^{2}+4 z^{2}=(x+y)^{2}-4\left(xy-z^{2}\right)=4-4=0$. Therefore, $x=y, z=0$. From the first equation, we find $x=y$ $=1$.

## Answer

One solution.