$\,\,\,\,\,\,\,\begin{cases}\left(1+\sqrt{2}\right)x+y=\sqrt{2}\\\left(2+\sqrt{2}\right)x-y=1\end{cases}$
$\Leftrightarrow\begin{cases}\left(1+\sqrt{2}\right)x+\left(2+\sqrt{2}\right)x=\sqrt{2}+1\\\left(2+\sqrt{2}\right)x-y=1\end{cases}$
$\Leftrightarrow\begin{cases}\left(2\sqrt{2}+3\right)x=\sqrt{2}+1\\\left(2+\sqrt{2}\right)x-y=1\end{cases}$
$\Leftrightarrow\begin{cases}x=-1+\sqrt{2}\\\left(2+\sqrt{2}\right).\left(-1+\sqrt{2}\right)-y=1\end{cases}$
$\Leftrightarrow\begin{cases}x=-1+\sqrt{2}\\\sqrt{2}-y=1\end{cases}$
$\Leftrightarrow\begin{cases}x=-1+\sqrt{2}\\y=-1+\sqrt{2}\end{cases}$
$\Leftrightarrow x=y=-1+\sqrt{2}$
