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$ \begin{cases} \frac{x}{2} + \frac{y}{5} =1 \\ \frac{x}{10}+\frac{2y}{10} = -1 \end{cases} $

$ \begin{cases} 5x + 2y =10 \\ x+2y = -10 \end{cases} $

per riduzione (1°-2° → 1°)

$ \begin{cases} 4x  =20 \\ x+2y = -10 \end{cases} $

dalla prima $ x = 5 $

dalla seconda $ y = -\frac{15}{2} $ 



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$\small \begin{Bmatrix}
2^{-1}x+5^{-1}y&=&1\\
0,1x+0,2y&=&-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\dfrac{1}{2}x+\dfrac{1}{5}y & = & 1\\
\dfrac{1}{10}x+\dfrac{\cancel{2}^1}{\cancel{10}_5}y & = &-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
5x+2y & = & 10\\
\dfrac{1}{10}x+\dfrac{1}{5}y & = &-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
5x+2y & = & 10\\
x+2y & = &-10\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
5(-10-2y)+2y & = & 10\\
x & = &-10-2y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
-50-10y+2y & = & 10\\
x & = &-10-2y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
-8y & = & 10+50\\
x & = &-10-2y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
-8y & = & 60\\
x & = &-10-2y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
8y & = & -60\\
x & = &-10-2y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\dfrac{\cancel8y}{\cancel8} & = & \dfrac{-\cancel{60}^{15}}{\cancel8_2}\\
x & = &-10-2y\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y & = & -\dfrac{15}{2}\\
x & = &-10-\cancel2·\left(-\dfrac{15}{\cancel2}\right)\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y & = & -\dfrac{15}{2}\\
x & = &-10+15\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y & = & -\dfrac{15}{2}\\
x & = &5\\
\end{Bmatrix}$

$\small x= 5\,\land\,y= -\dfrac{15}{2}.$



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