Aritmetica

{x + 5·y + z = -6-----> z = -x - 5·y - 6

{3·x + y + 2·z = 9

{x + y + 2·z = 3

Per sostituzione:

{3·x + y + 2·(-x - 5·y - 6) = 9---> x - 9·y = 21----> x = 9·y + 21

{x + y + 2·(-x - 5·y - 6) = 3---> x + 9·y = -15

Per sostituzione:

(9·y + 21) + 9·y = -15---> y = -2

x = 9·(-2) + 21--->x = 3

z = -3 - 5·(-2) - 6---> z = 1

In definitiva: x = 3 ∧ y = -2 ∧ z = 1

Devi risolvere banalmente un sistema di equazioni lineari

\[\begin{cases} x + 5y + z = -6 \\ y + 3x + 2z = 9 \\ x + y + 2z = 3 \end{cases} \iff \begin{cases} x + 5y + z = -6 \\ 2x = 6 \\ x + y + 2z = 3 \end{cases} \iff\]

\[\begin{cases}3 + 5y + z = -6 \\ x = 3 \\ y = -2z \end{cases} \iff \begin{cases} z = 1 \\ x = 3 \\ y = -2 \end{cases}\,.\]

Sapendo che X+5Y+Z=-6 e Y+3X+2Z=9 e x+Y+2Z=3 quale dei seguenti sono i valori delle incognite X e Y e Z?

la risposta è : (3;-2;1).

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$\left\{\begin{matrix}x+5y+z=-6 \\ 3x+y+2z=9 \\ x+y+2z=3\end{matrix}\right\}$

$\left\{\begin{matrix}x=-6-5y-z \\ 3(-6-5y-z)+y+2z=9 \\ -6-5y-z+y+2z=3\end{matrix}\right\}$

$\left\{\begin{matrix}x=-6-5y-z \\ -18-15y-3z+y+2z=9 \\ -4y+z=3+6\end{matrix}\right\}$

$\left\{\begin{matrix}x=-6-5y-z \\ -14y-z=9+18 \\ -4y+z=9\end{matrix}\right\}$

$\left\{\begin{matrix}x=-6-5y-z\\ -14y-z=27\\ z=9+4y\end{matrix}\right\}$

$\left\{\begin{matrix}x=-6-5y-\left(9+4y\right)\\ -14y-\left(9+4y\right)=27\\ z=9+4y\end{matrix}\right\}$

$\left\{\begin{matrix}x=-6-5y-9-4y\\ -14y-9-4y=27\\ z=9+4y\end{matrix}\right\}$

$\left\{\begin{matrix}x=-15-9y\\ -18y=27+9\\ z=9+4y\end{matrix}\right\}$

$\left\{\begin{matrix}x=-15-9y\\ -18y=36\\ z=9+4y\end{matrix}\right\}$

$\left\{\begin{matrix}x=-15-9y\\ -\frac{18y}{-18}=\frac{36}{-18}\\ z=9+4y\end{matrix}\right\}$

$\left\{\begin{matrix}x=-15-9y\\ y=-2\\ z=9+4\cdot\left(-2\right)\end{matrix}\right\}$

$\left\{\begin{matrix}x=-15-9\cdot\left(-2\right)\\ y=-2\\ z=9-8\end{matrix}\right\}$

$\left\{\begin{matrix}x=-15+18\\ y=-2\\ z=1\end{matrix}\right\}$

$\left\{\begin{matrix}x=3\\ y=-2\\ z=1\end{matrix}\right\}$