Решите систему уравнений: x^2+xy-3y=-1; 4x-y=3.

Ответ:

\[\left\{ \begin{matrix} x^{2} + xy - 3y = - 1 \\ 4x - y = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[5x² - 15x + 10 = 0\ \ \ |\ :5\]

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

\[x_{1} + x_{2} = 3,\ \ \ x_{1} \cdot x_{2} = 2\]

\[x_{1} = 1,\ \ x_{2} = 2\]

\[\left\{ \begin{matrix} x = 1 \\ y = 1 \\ \end{matrix} \right.\ \ \ \ \ \ или\ \ \ \ \left\{ \begin{matrix} x = 2 \\ y = 5 \\ \end{matrix} \right.\ \]

\[Ответ:(1;1);\ \ (2;5).\]