372. Solve the system of equations:

$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=1 \\
\frac{1}{3-x}+\frac{1}{3-y}=2
\end{array}\right.
$$
$\triangle$ Transform each equation of the system into an algebraic form:

$y+x=x y, \quad 6-x-y=18-6 x-6 y+2 x y$;

$x y=x+y, \quad 2 x y=5 x+5 y-12$.

We obtain a symmetric system (see § 8, point 8.3). Therefore, introduce new variables

$$
u=x+y, \quad v=x y
$$

This results in a system of linear equations

$$
v=u, \quad 2 v=5 u-12
$$

From this system, we find $u$ and $v$. Knowing $u$ and $v$, we will compute $x$ and $y$. Answer: $(2 ; 2)$.