Solve the following system of equations:

$$
\begin{aligned}
& x^{2}-4 \sqrt{3 x-2}+6=y \\
& y^{2}-4 \sqrt{3 y-2}+6=x
\end{aligned}
$$
I. solution. The equations make sense when $x \geq \frac{2}{3}$ and $y \geq \frac{2}{3}$.

Add the equations and rearrange to 0:

$$
x^{2}+y^{2}-4 \sqrt{3 x-2}-4 \sqrt{3 y-2}+6+6-x-y=0
$$

We assume that the expression can be transformed into a sum of complete squares. In this case, the expressions $2 \cdot 2 \sqrt{3 x-2}$ and $2 \cdot 2 \sqrt{3 y-2}$ play the role of the double products.

Complete the left side with the necessary expressions:

$$
\begin{gathered}
x^{2}+y^{2}+[(3 x-2)-2 \cdot 2 \sqrt{3 x-2}+4]+ \\
+[(3 y-2)-2 \cdot 2 \sqrt{3 y-2}+4]+4+4-4 x-4 y=0
\end{gathered}
$$

Rearranging:

$$
x^{2}-4 x+4+y^{2}-4 y+4+(\sqrt{3 x-2}-2)^{2}+(\sqrt{3 y-2}-2)^{2}=0
$$

Our assumption is confirmed, the left side of the equation is indeed a sum of squares:

$$
(x-2)^{2}+(y-2)^{2}+(\sqrt{3 x-2}-2)^{2}+(\sqrt{3 y-2}-2)^{2}=0
$$

Each term on the left side is greater than or equal to 0, but their sum is 0. This is only possible if each term is 0.

If $(x-2)^{2}=0$ and $(y-2)^{2}=0$, then $x=2$ and $y=2$. For these values, the other two expressions are also 0, and our conditions are satisfied.

Thus, the only solution to the system of equations is $x=y=2$.

II. solution. Suppose that $3 x-2 \geq 0$ and $3 y-2 \geq 0$ hold. Then we can apply the arithmetic and geometric mean inequality to the following expressions:

$$
\begin{aligned}
& 4 \sqrt{3 x-2}=2 \cdot \sqrt{4(3 x-2)} \leq 2 \cdot \frac{(3 x-2)+4}{2}=3 x+2 \\
& 4 \sqrt{3 y-2}=2 \cdot \sqrt{4(3 y-2)} \leq 2 \cdot \frac{(3 y-2)+4}{2}=3 y+2
\end{aligned}
$$

With this, we can estimate the left side of our original equations:

$$
\begin{aligned}
& y=x^{2}-4 \sqrt{3 x-2}+6 \geq x^{2}-3 x-2+6=x^{2}-3 x+4 \\
& x=y^{2}-4 \sqrt{3 y-2}+6 \geq y^{2}-3 y-2+6=y^{2}-3 y+4
\end{aligned}
$$

On the other hand, for every value of $x$, $x \leq x^{2}-3 x+4$, since subtracting $x$ gives

$$
0 \leq x^{2}-4 x+4=(x-2)^{2}
$$

(equality holds at $x=2$). Similarly, for every $y$, $y \leq y^{2}-3 y+4$, and equality holds at $y=2$.

Comparing the four inequalities:

$$
x \leq x^{2}-3 x+4 \leq y \leq y^{2}+3 y+4 \leq x
$$

It is clear that equality must hold in each estimate, which only happens at $x=y=2$; this value also satisfies the system of equations.