## Task B-3.2.

Solve the system of equations in the set of positive real numbers

$$
\begin{gathered}
\sqrt[x-y]{x+y}=2 \sqrt{3} \\
(x+y) \cdot 2^{y-x}=3
\end{gathered}
$$
## First Solution.

We introduce the substitution $x-y=a$ and $x+y=b$.

Then the system transforms into

$$
\sqrt[a]{b}=2 \sqrt{3}, \quad b \cdot 2^{-a}=3
$$

1 point

If we raise the first equation to the power of $a$, we get $b=(2 \sqrt{3})^{a}$. Substituting this into the second equation, we get

$$
\begin{aligned}
(2 \sqrt{3})^{a} 2^{-a} & =3 \\
2^{a}(\sqrt{3})^{a} 2^{-a} & =3 \\
3^{\frac{1}{2} a} & =3
\end{aligned}
$$

From this, we have $a=2$, so $b=(2 \sqrt{3})^{a}$, which means $b=12$.

It remains to determine $x$ and $y$ from the system

$$
x-y=2, \quad x+y=12
$$

Finally, the solution to the original system is $x=7, y=5$.