Task 12/79 All positive numbers $x$ and $y$ are to be determined that satisfy the following equations:

$$
x^{x+y}=y^{6} \quad ; \quad y^{x+y}=x^{12} y^{6}
$$
By multiplying the two equations, we obtain

$$
(x y)^{x+y}=(x y)^{12}
$$

This equation is certainly satisfied if a) $x+y=12$ or b) $x y=1$. (For $x+y \neq 12$ and $x y \neq 1$, there is obviously no solution.)

a) Substituting $x+y=12$ into the first equation, we get $y=x^{2}$. From $x+y=12$, it follows that the quadratic equation in $x$ is $x^{2}+x-12=0$ with the solutions $x_{1}=3, x_{2}=-4$. Therefore, the pair $(x ; y)=(3 ; 9)$ is a solution.

b) From $x y=1$, it follows that $y=\frac{1}{x}$, and thus from the first equation

$$
x^{x+\frac{1}{x}}=x^{-6}
$$

Since $x+\frac{1}{x}>0$, the only solution that remains is $(x ; y)=(1 ; 1)$. The pairs $(3 ; 9)$ and $(1 ; 1)$ are solutions of the system, as confirmed by the verification.