Sistemi frazionari

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$\small \begin{Bmatrix}
\dfrac{x-1}{y+1}&=&\dfrac{x}{y}\\
x-y+1&=&0\\
\end{Bmatrix}$ $\small [C.E.: y\not=0\, \land\not=-1] \quad[1°eq.\,mcm= y(y+1)]$

$\small \begin{Bmatrix}
y(x-1)&=&x(y+1)\\
x-y&=&-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
xy-y&=&xy+x\\
x-y&=&-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\cancel{xy}-y\cancel{-xy}-x&=&0\\
x-y&=&-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
-y-x&=&0\\
x&=&y-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y+x&=&0\\
x&=&y-1\\
\end{Bmatrix}$

sostituisci la "x" nella 1° equazione:

$\small \begin{Bmatrix}
y+y-1&=&0\\
x&=&y-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
2y&=&1\\
x&=&y-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\dfrac{\cancel2y}{\cancel2}&=&\dfrac{1}{2}\\
x&=&y-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&\dfrac{1}{2}\\
x&=&y-1\\
\end{Bmatrix}$

sostituisci la "y" nella 2° equazione:

$\small \begin{Bmatrix}
y&=&\dfrac{1}{2}\\
x&=&\dfrac{1}{2}-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&\dfrac{1}{2}\\
x&=&\dfrac{1-2}{2}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&\dfrac{1}{2}\\
x&=&-\dfrac{1}{2}\\
\end{Bmatrix}$

quindi:

$\small x=-\dfrac{1}{2}\;\land\;y=\dfrac{1}{2}$