Solve the following system of equations:

$$
\begin{gathered}
x+y+\sqrt{x+y-2}=14 \\
\frac{x^{2} y^{2}}{3}-\frac{3 x y}{2}=255
\end{gathered}
$$
Let $x+y=u$ and $xy=v$, then our equations can be written as:

$$
u^{2}-29 u+198=0
$$

$$
2 v^{2}-9 v-1530=0
$$

From these two equations:

$$
u_{1}=18, \quad u_{2}=11, \quad v_{1}=30, \quad v_{2}=-\frac{51}{2}
$$

However, $u_{1}$ would only be a root of equation (1) if the square root were taken with a negative sign, which is why we only consider $u_{2}$. Thus, we still need to solve the following systems of equations:

$$
x+y=11, \quad xy=30 \quad \text { and } \quad x+y=11, \quad xy=-\frac{51}{2}
$$

From these systems of equations:

$$
x_{1}=y_{2}=5, x_{2}=y_{1}=6, x_{3}=y_{4}=\frac{11}{2}-\frac{1}{2} \sqrt{223}, x_{4}=y_{3}=\frac{11}{2}+\frac{1}{2} \sqrt{223}
$$

(Viktor Wáhl, Eger.)

Number of solutions: 44.