7.285. $\left\{\begin{array}{l}x^{x^{2}-y^{2}-16}=1, \\ x-y=2\end{array} \quad(x>0)\right.$.

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7.285. $\left\{\begin{array}{l}x^{x^{2}-y^{2}-16}=1, \\ x-y=2\end{array} \quad(x>0)\right.$.
## Solution.

The first equation is equivalent to two equations: $x=1$ or $x^{2}-y^{2}-16=0$ for $x>0$. Then the system of equations is equivalent to two systems:

1) $\left\{\begin{array}{l}x=1, \\ x-y=2\end{array} \Leftrightarrow\left\{\begin{array}{l}x=1 \\ y=-1\end{array}\right.\right.$
2) $\left\{\begin{array}{l}x^{2}-y^{2}-16=0, \\ x-y=2\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{2}=5 \\ y_{2}=3\end{array}\right.\right.$

Answer: $\quad(1 ;-1),(5 ; 3)$.