-2x + y = +2
3x + y = -1
Sistem de 2 ecuatii si 2 necunoscute.
Pentru rezolvare se poate folosi metoda substitutiei sau metoda reducerii.
[tex]\displaystyle\\
\text{Metoda substitutiei:}\\\\
\begin{cases}
-2x + y = +2~~~~~\Longrightarrow~~\boxed{\bf y=2x+2}~~\text{\bf substitutie} \\
3x + y = -1\\
\end{cases}\\\\
\text{Inlocuim pe y in a 2-a ecuatie. }\\\\
3x + 2x+2 = -1\\
3x + 2x = -1-2\\
5x=-3\\\\
\boxed{\bf x= -\frac{3}{5}} \\\\
\text{Ne intoarcem la substitutie pentru a-l afla pe y.}\\
y=2x+2\\\\
y=2\cdot \frac{-3}{5}+2 = \frac{2(-3)}{5}+\frac{10}{5} =\frac{-6}{5}+\frac{10}{5} = \frac{10-6}{5} \\\\
\boxed{\bf y=\frac{4}{5}}[/tex]
[tex]\displaystyle\\ \text{Metoda reducerii:}\\\\
\begin{cases}
-2x + y = +2~~~~\Big| \cdot 3\\
3x + y = -1~~~~~\Big| \cdot 2\\
\end{cases}\\\\
\text{Inmultim ecuatia 1 cu 3 si ecuatia 2 cu 2. }\\\\
\begin{cases}
-6x + 3y = 6\\
~~6x + 2y = -2\\
\end{cases}\\
\text{-----------------------Adunam ecuatiile si se va reduce necunoscuta x.}\\
~~~~~~/ ~~~~5y = 4\\\\
\boxed{\bf y = \frac{4}{5} }[/tex]
[tex]\displaystyle\\
\text{Rescriem una din ecuatii.}\\
3x + y = -1\\
x = \frac{-y-1}{3}=-\frac{y+1}{3}=-\frac{\dfrac{4}{5} +1}{3}-\frac{\dfrac{9}{5}}{3}=- \frac{9}{15} \\\\
\boxed{\bf x = - \frac{3}{5}}[/tex]
