Sure, here is the translated text:

```
Solve the following system of equations:

$$
\begin{aligned}
& 3^{x+2}=\sqrt[y+1]{9^{8-y}} \\
& 2^{x-2}=\sqrt[y+1]{8^{3-y}}
\end{aligned}
$$
```
Let's transform our system of equations in the following way:

$$
3^{x+2}=3^{\frac{16-2 y}{y+1}}, \quad 2^{x-2}=2^{\frac{9-3 y}{y+1}}
$$

The equalities can only hold if the exponents on both sides are equal:

$$
x+2=\frac{16-2 y}{y+1}, \quad x-2=\frac{9-3 y}{y+1}
$$

or by multiplying each by \(y+1\) (since \(y+1 \neq 0\), as a root with a zero exponent is undefined):

$$
(x+2)(y+1)=16-2 y, \quad(x-2)(y+1)=9-3 y
$$

Subtract the second equation from the first:

$$
\begin{aligned}
4(y+1) & =7+y \\
y & =1
\end{aligned}
$$

from which

$$
x=5
$$