Sistema

dalla seconda $ x = \sqrt{2}(y-1) $

che sostituita nella prima

$ 2(y-1) -4y = 2 \; ⇒ \; y = -2$

da cui

$ x = -3\sqrt{2} $

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588)

$\small \begin{Bmatrix}
x\sqrt2-4y&=&2\\
x-\sqrt2·y&=&-\sqrt2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\sqrt2·x-4y&=&2\\
x&=&-\sqrt2+\sqrt2·y\\
\end{Bmatrix}$

sostituisci la "x" della 2° equazione nella 1°:

$\small \begin{Bmatrix}
\sqrt2·(-\sqrt2+\sqrt2·y)-4y&=&2\\
x&=&-\sqrt2+\sqrt2·y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
-2+2y-4y&=&2\\
x&=&-\sqrt2+\sqrt2·y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
-2y&=&2+2\\
x&=&-\sqrt2+\sqrt2·y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
-2y&=&4\\
x&=&-\sqrt2+\sqrt2·y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
2y&=&-4\\
x&=&-\sqrt2+\sqrt2·y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\dfrac{\cancel2y}{\cancel2}&=&\dfrac{-4}{2}\\
x&=&-\sqrt2+\sqrt2·y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&-2\\
x&=&-\sqrt2+\sqrt2·y\\
\end{Bmatrix}$

sostituisci la "y" nella 2°:

$\small \begin{Bmatrix}
y&=&-2\\
x&=&-\sqrt2+\sqrt2·(-2)\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&-2\\
x&=&-\sqrt2+(-2\sqrt2)\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&-2\\
x&=&-\sqrt2-2\sqrt2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&-2\\
x&=&-3\sqrt2\\
\end{Bmatrix}$

$\small x= -3\sqrt2\, \land \, y= -2$