Sistemi frazionari.

1/y = (x + 1)/(y - y^2)

1/y = (x + 1)/(y·(1 - y))

* y·(1 - y) ≠ 0----> y ≠ 1 ∧ y ≠ 0   ( C.E.)

1 - y = x + 1

Quindi sistema:

{x + y = 0

{x - y = 1

(somma- differenza)

Risolvo ed ottengo: [x = 1/2 ∧ y = - 1/2]

(compatibili con le C.E.)

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$\small \begin{Bmatrix}
\dfrac{1}{y}&=&\dfrac{x+1}{y-y^2} \quad ((mcm=\,y(1-y))\\
x-y&=&1\\
\end{Bmatrix}$ $\quad C.E.: y\not=0\,\land\,\not=1$

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

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

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

sostituisci la "x" nella 1° equazione:

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

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

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

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

sostituisci la "y" nella 2° equazione:

$\small \begin{Bmatrix}
y&=&-\dfrac{1}{2}\\
x&=&1+\left(-\dfrac{1}{2}\right)\\
\end{Bmatrix}$

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

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

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

per cui:

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