Решите систему уравнений x+3y=5; 4y+xy=6.

\[\left\{ \begin{matrix} x + 3y = 5\ \ \ \\ 4y + xy = 6 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 5 - 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y + y(5 - 3y) = 6 \\ \end{matrix} \right.\ \]

\[4y + 5y - 3y^{2} - 6 = 0\]

\[- 3y^{2} + 9y - 6 = 0\ \ \ \ |\ :( - 3)\]

\[y^{2} - 3y + 2 = 0\]

\[y_{1} + y_{2} = 3;\ \ \ y_{1} \cdot y_{2} = 2\]

\[y_{1} = 2;\ \ \ y_{2} = 1\]

\[\left\{ \begin{matrix} y = 2\ \ \ \\ x = - 1 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = 1 \\ x = 2 \\ \end{matrix} \right.\ \]

\[Ответ:( - 1;2);\ \ (2;1).\]