1. Find all triples of real numbers $(u ; v ; w)$, satisfying the system of equations:

$$
\left\{\begin{array}{l}
u+v \cdot w=12 \\
v+w \cdot u=12 \\
w+u \cdot v=12
\end{array}\right.
$$
Answer: $(\mathbf{3} ; \mathbf{3} ; \mathbf{3}),(-\mathbf{4} ;-\mathbf{4} ;-\mathbf{4}),(\mathbf{1} ; \mathbf{1} ; \mathbf{1 1}),(\mathbf{1} ; \mathbf{1 1} ; \mathbf{1}),(\mathbf{1 1} ; \mathbf{1} ; \mathbf{1})$.

Solution: Subtract the first equation from the second

$$
\begin{gathered}
(v-u)-w(v-u)=0 \\
(v-u)(1-w)=0
\end{gathered}
$$

1) $v=u \Rightarrow\left\{\begin{array}{l}v+w \cdot v=12, \\ w+v^{2}=12 .\end{array} \Rightarrow(w-v)-v(w-v)=0 \Rightarrow(w-v)(1-v)=0\right.$

From which it follows that $v=u=w$ or $v=u=1$.

If $v=u=w$, then $\quad v=u=w=3$ or $v=u=w=-4$.

If $v=u=1, \quad$ then $\quad w=11$.
2) $w=1 \Rightarrow\left\{\begin{array}{l}v+u=12, \\ u \cdot v=11 .\end{array} \Rightarrow\left[\begin{array}{l}v=11, u=1 \\ v=1, u=11 .\end{array}\right.\right.$