[Risolto] Problemi, sistemi

Traducendo in simboli si ha il sistema

{ x^2 + 2y = x(x - 1) + 1

{ x + y = 3x + 7y

che, riducendo, assume la forma

{ 2y = - x + 1

{ 6y = - 2x

moltiplicando per 3 la prima e confrontando

-3x + 3 = -2x

- x = -3

x = 3

e y = -2/6 * 3 = -1

==========================================================

1° numero $\small =x;$

2° numero $\small = y;$

sistema (con sostituzione):

$\small \begin{Bmatrix} x^2+2y&=&x(x-1)+1\\
x+y&=&3x+7y\\
\end{Bmatrix} \Longrightarrow \begin{Bmatrix} x^2+2y&=&x^2-x+1\\
x-3x&=&7y-y\\
\end{Bmatrix}\Longrightarrow \begin{Bmatrix} \cancel{x^2}+2y\cancel{-x^2}+x&=&1\\
-2x&=&6y\\
\end{Bmatrix}$

$\small \begin{Bmatrix} 2y+x&=&1\\
2x&=&-6y\\
\end{Bmatrix} \Longrightarrow \begin{Bmatrix} x&=&1-2y\\
2(1-2y)&=&-6y\\
\end{Bmatrix}\Longrightarrow \begin{Bmatrix} x&=&1-2y\\
2-4y&=&-6y\\
\end{Bmatrix}$

$\small \begin{Bmatrix} x&=&1-2y\\
-4y+6y&=&-2\\
\end{Bmatrix} \Longrightarrow \begin{Bmatrix} x&=&1-2y\\
2y&=&-2\\
\end{Bmatrix}\Longrightarrow \begin{Bmatrix} x&=&1-2y\\
\dfrac{\cancel2y}{\cancel2}&=&\dfrac{-2}{2}\\
\end{Bmatrix}$

$\small \begin{Bmatrix} x&=&1-2y\\
y&=&-1\\
\end{Bmatrix} \Longrightarrow \begin{Bmatrix} x&=&1-(2·(-1))\\
y&=&-1\\
\end{Bmatrix}\Longrightarrow \begin{Bmatrix} x&=&1-(-2)\\
y&=&-1\\
\end{Bmatrix}$

$\small \begin{Bmatrix} x&=&1+2\\
y&=&-1\\
\end{Bmatrix} \Longrightarrow \begin{Bmatrix} x&=&3\\
y&=&-1\\
\end{Bmatrix}$