2. Solve the system of equations

$$
\left\{\begin{array}{r}
\log _{2}(x+y)|+| \log _{2}(x-y) \mid=3 \\
x y=3
\end{array}\right.
$$
Solution. From $x y=3$ we conclude that $x>0$ and $y>0$, or $x<0$ and $y<0$.

Also, $x+y>1$, because $x$ or $y$ is greater than 1, which is derived from the positivity of $x$ and $y$ and from $x y=3$.

From all this, it follows that the given system is equivalent to the system

$$
\left\{\begin{aligned}
\log _{2}(x+y)+\left|\log _{2}(x-y)\right| & =3 \\
x y & =3
\end{aligned}\right.
$$

If $0<1$, it leads to the system

$$
\left\{\begin{array}{l}
x^{2}-y^{2}=8 \\
x y=3
\end{array}\right.
$$

and the solution to this system that satisfies the conditions $x-y>1$ and $x, y>0$, is $x=3, y=1$.