Решите систему уравнений: 4x^2+y^2=13; xy=-3.

\[\left\{ \begin{matrix} 4x^{2} + y^{2} = 13\ \\ xy = - 3\ \ \ \ \ \ | \cdot 4 \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} 4x^{2} + y^{2} = 13 \\ 4xy = - 12\ \ \ \ \ \\ \end{matrix}( + ) \right.\ \text{\ \ }\]

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

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

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

\[\left\{ \begin{matrix} y = 1 - 2x\ \ \ \ \ \ \ \ \ \ \ \\ x(1 - 2x) = - 3 \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} y = 1 - 2x\ \ \ \ \ \ \ \ \ \\ x - 2x^{2} + 3 = 0 \\ \end{matrix} \right.\ \]

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

\[D = 1 + 24 = 25\ \ \]

\[x_{1} = \frac{- 1 + 5}{- 4} = - 1,\ \ \]

\[x_{2} = \frac{- 1 - 5}{- 4} = 1,5\]

\[\left\{ \begin{matrix} x = - 1 \\ y = 3\ \ \ \\ \end{matrix}\ \ \ \ \ или\ \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 1,5 \\ y = - 2 \\ \end{matrix} \right.\ \]

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

\[\left\{ \begin{matrix} y = - 1 - 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x( - 1 - 2x) + 3 = 0 \\ \end{matrix}\text{\ \ \ \ } \right.\ \ \]

\[\left\{ \begin{matrix} y = - 1 - 2x\ \ \ \ \ \ \ \\ - x - 2x^{2} + 3 = 0 \\ \end{matrix} \right.\ \]

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

\[D = 1 + 24 = 25\ \]

\[x_{1} = \frac{1 + 5}{- 4} = - 1,5,\ \ \]

\[x_{2} = \frac{1 - 5}{- 4} = 1\]

\[\left\{ \begin{matrix} x = 1\ \ \ \\ y = - 3 \\ \end{matrix}\ \ \ \ \ \ \ \ \ или\ \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - 1,5 \\ y = 2\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:( - 1;3);\ (1,5;\ - 2);\]

\[(1;\ - 3);\ ( - 1,5;\ 2).\]