[Risolto] Sistemi

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Metodo riduzione:

$\small \begin{Bmatrix} \dfrac{(x+y)[1-(x-y)]+x^2}{6}+\dfrac{2}{3}-\dfrac{y^2}{6}&=&2-\dfrac{3}{2}+\dfrac{x+y}{4}\\
\dfrac{x+4}{3}&= &\dfrac{x-6}{2} \quad\color{red}(mcm=6)\\
\end{Bmatrix}$

$\small \begin{Bmatrix} \dfrac{(x+y)[1-x+y]+x^2}{6}+\dfrac{2}{3}-\dfrac{y^2}{6}&=&2-\dfrac{3}{2}+\dfrac{x+y}{4}\\
2(y+4)&= &3(x-6)\\
\end{Bmatrix}$

$\small \begin{Bmatrix} \dfrac{x\cancel{-x^2}\cancel{+xy}+y\cancel{-xy}+y^2\cancel{+x^2}}{6}+\dfrac{2}{3}-\dfrac{y^2}{6}&=&2-\dfrac{3}{2}+\dfrac{x+y}{4}\\
2y+8&= &3x-18\\
\end{Bmatrix}$

$\small \begin{Bmatrix} \dfrac{x+y+y^2}{6}+\dfrac{2}{3}-\dfrac{y^2}{6}&=&2-\dfrac{3}{2}+\dfrac{x+y}{4}\quad\color{red}(mcm=12)\\
-3x+2y&= &-18-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix} 2(x+y+y^2)+8-2y^2&=&24-18+3(x+y)\\
-3x+2y&= &-26\\
\end{Bmatrix}$

$\small \begin{Bmatrix} 2x+2y\cancel{+2y^2}+8\cancel{-2y^2} &=& 6+3x+3y\\
3x-2y&= & 26\\
\end{Bmatrix}$

$\small \begin{Bmatrix} 2x+2y-3x-3y&=& 6-8\\
3x-2y&= & 26\\
\end{Bmatrix}$

$\small \begin{Bmatrix} -x-y &=& -2\\
3x-2y&= & 26\\
\end{Bmatrix}$

$\small \begin{Bmatrix} x+y &=& +2\\
3x-2y&= & 26\\
\end{Bmatrix}$

moltiplica la 1° equazione per 2 per eliminare la "y" e  facendo la somma come segue:

$\small \begin{Bmatrix} 2x+2y &=& 4\\
3x-2y&= & 26\\\hline
5x\quad//&=&30
\end{Bmatrix}$

quindi:

$\small 5x=30$

$\small \dfrac{\cancel5x}{\cancel5} = \dfrac{30}{5}$

$\small x= 6$

sostituisci la "x" nella 1° equazione:

$\small 2x+2y = 4$

$\small 2·6+2y = 4$

$\small 12+2y = 4$

$\small 2y = 4-12$

$\small 2y=-8$

$\small \dfrac{\cancel2y}{\cancel2} = \dfrac{-8}{2}$

$\small y= -4$

risultato:

$\small x=6 \land y= -4$

{((x + y)·(1 - (x - y)) + x^2)/6 + 2/3 - y^2/6 = 2 - 3/2 + (x + y)/4

{(y + 4)/3 = (x - 6)/2

-------------------------------------------------------

{((x + y)·(-x + y + 1) + x^2)·2 + 8 - 2·y^2 = 24 - 18 + 3·(x + y)

{2·(y + 4) = 3·(x - 6)

---------------------------------------------------

{(x + y^2 + y)·2 + 8 - 2·y^2 = 24 - 18 + (3·x + 3·y)

{2·y + 8 = 3·x - 18

----------------------------------------------------------

{2·x + 2·y + 8 = 3·x + 3·y + 6

{3·x - 2·y = 26

--------------------------------------------

{x + y = 2   (*2)

{3·x - 2·y = 26

Elimino la y:

{2·x + 2·y = 4

{3·x - 2·y = 26

-------------------(sommo)

5x...........=30----> x = 6

Elimino la x:

{3·x + 3·y = 6

{3·x - 2·y = 26

-----------------(sottraggo)

.........5·y = -20-----> y = -4