416*. Solve the system of equations:

$$
\left\{\begin{array}{l}
2 \sqrt{2 x+3 y}+\sqrt{5-x-y}=7 \\
3 \sqrt{5-x-y}-\sqrt{2 x+y-3}=1
\end{array}\right.
$$
$\triangle$ Let's introduce three new variables:

$$
\sqrt{2 x+3 y}=z, \quad \sqrt{5-x-y}=t, \quad \sqrt{2 x+y-3}=u
$$

where $z, t, u$ are non-negative. This results in a system of five rational equations with five unknowns:

$$
\left\{\begin{array}{l}
2 x+3 y=z^{2} \\
5-x-y=t^{2} \\
2 x+y-3=u^{2} \\
2 x+t=7 \\
3 t-u=1
\end{array}\right.
$$

From the first and third equations of this system, we express $x$ and $y$ in terms of $z$ and $u$:

$$
x=\frac{1}{4}\left(3 u^{2}-z^{2}+9\right), \quad y=\frac{1}{2}\left(z^{2}-u^{2}-3\right)
$$

Substitute these expressions into the second equation:

$5-\frac{1}{4}\left(3 u^{2}-z^{2}+9\right)-\frac{1}{2}\left(z^{2}-u^{2}-3\right)=t^{2}$,

$20-3 u^{2}+z^{2}-9-2 z^{2}+2 u^{2}+6=4 t^{2}$,

$z^{2}+4 t^{2}+u^{2}=17$

From the system of equations

$$
z^{2}+4 t^{2}+u^{2}=17, \quad 2 z+t=7, \quad 3 t-u=1
$$

we find $z, t, u$. Then we compute $x$ and $y$. Complete the solution on your own.

Answer: $(3 ; 1)$.