Решите систему уравнений x-3y=-1; xy+4y=18.

Ответ:

\[\left\{ \begin{matrix} x - 3y = - 1\ \\ xy + 4y = 18 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 3y - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (3y - 1)y + 4y = 18 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[3y^{2} - y + 4y - 18 = 0\]

\[3y^{2} + 3y - 18 = 0\ \ \ \ \ |\ :3\]

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

\[y_{1} + y_{2} = - 1;\ \ \ y_{1} \cdot y_{2} = - 6\]

\[y_{1} = - 3;\ \ y_{2} = 2\]

\[\left\{ \begin{matrix} y = - 3\ \ \\ x = - 10 \\ \end{matrix} \right.\ \ \ \ \ \ \ и\ \ \ \ \ \ \left\{ \begin{matrix} y = 2 \\ x = 5 \\ \end{matrix} \right.\ \]

\[Ответ:( - 10;\ - 3);\ \ (5;2).\]