Sistemi

{(1/2 - 1/3)^(-1)·x + (1/2 + 1/3)^(-1)·y = 6

{0.1·x + 0.2·y = 0.4

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{(1/6)^(-1)·x + (5/6)^(-1)·y = 6

{1/10·x + 1/5·y = 2/5

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{6·x + 6/5·y = 6

{x + 2·y = 4

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per sostituzione: x = 4 - 2·y

6·(4 - 2·y) + 6/5·y = 6

24 - 54·y/5 = 6

54·y/5 = 18----> y = 5/3

x = 4 - 2·(5/3)----> x = 2/3

soluzione sistema: [x = 2/3 ∧ y = 5/3]

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$\small \begin{Bmatrix}
\left(\dfrac{1}{2}-\dfrac{1}{3}\right)^{-1}x+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)^{-1}y&=&6\\
0,1x+0,2y&=&0,4\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\left(\dfrac{3-2}{6}\right)^{-1}x+\left(\dfrac{3+2}{6}\right)^{-1}y&=&6\\
\dfrac{1}{10}x+\dfrac{1}{5}y&=&\dfrac{2}{5}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\left(\dfrac{1}{6}\right)^{-1}x+\left(\dfrac{5}{6}\right)^{-1}y&=&6\\
x+2y&=&4\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
6x+\dfrac{6}{5}y&=&6\\
x+2y&=&4\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
30x+6y&=&30\\
x+2y&=&4\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
5x+y&=&5\\
x+2y&=&4\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&5-5x\\
x+2(5-5x)&=&4\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&5-5x\\
x+10-10x&=&4\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&5-5x\\
-9x&=&4-10\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&5-5x\\
-9x&=&-6\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&5-5x\\
9x&=&6\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&5-5x\\
\dfrac{\cancel9x}{\cancel9}&=&\dfrac{\cancel6^2}{\cancel9_3}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&5-5x\\
x&=&\dfrac{2}{3}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&5-5·\dfrac{2}{3}\\
x&=&\dfrac{2}{3}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&5-\dfrac{10}{3}\\
x&=&\dfrac{2}{3}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&\dfrac{15-10}{3}\\
x&=&\dfrac{2}{3}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
y&=&\dfrac{5}{3}\\
x&=&\dfrac{2}{3}\\
\end{Bmatrix}$

quindi: $\small x=\dfrac{2}{3} \land y=\dfrac{5}{3}$