Sistemi

$ \begin{cases} \frac{x}{0.1} - \frac{y}{0.\bar{3}} = \frac{1}{0.3} \\ 3x-y = 2 \end{cases} $

$ \begin{cases} 10x - 3y = \frac{10}{3} \\ 3x-y = 2 \end{cases} $

$ \begin{cases} 30x - 9y = 10 \\ 3x-y = 2 \end{cases} $

per riduzione

(10*2° → 2°)

$ \begin{cases} 30x - 9y = 10 \\ 30x-10y = 20 \end{cases} $

sottraiamo la seconda dalla prima

$ y = -10 \; ⇒ \; 3x + 10 = 2 \; ⇒ \; y = -\frac{8}{3} $

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Provo con il metodo sostituzione:

$\small \begin{Bmatrix}
\dfrac{x}{0,1}-\dfrac{y}{0,\overline3}&=&\dfrac{1}{0,3}\\
3x-y&=&2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\dfrac{x}{\frac{1}{10}}-\dfrac{y}{\frac{3-0}{9}}&=&\dfrac{1}{\frac{3}{10}}\\
3x-y&=&2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
10x-\dfrac{y}{\frac{\cancel3^1}{\cancel9_3}}&=&\dfrac{10}{3}\\
3x-y&=&2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
10x-\dfrac{y}{\frac{1}{3}}&=&\dfrac{10}{3}\\
-y&=&2-3x\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
10x-3y&=&\dfrac{10}{3}\\
y&=&3x-2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
10x-3(3x-2)&=&\dfrac{10}{3}\\
y&=&3x-2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
10x-9x+6&=&\dfrac{10}{3}\\
y&=&3x-2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&\dfrac{10}{3}-6\\
y&=&3x-2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&\dfrac{10-18}{3}\\
y&=&3x-2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{8}{3}\\
y&=&3x-2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{8}{3}\\
y&=&\cancel3^1\left(-\dfrac{8}{\cancel3_1}\right)-2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{8}{3}\\
y&=&-8-2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&-\dfrac{8}{3}\\
y&=&-10\\
\end{Bmatrix}$