Решите систему уравнений: 4xy-y=-40; 5x-4xy=27.

\[\left\{ \begin{matrix} 4xy - y = - 40 \\ 5x - 4xy = 27\ \ \\ \end{matrix} \right.\ ( + )\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 5x - y = - 13\ \ \ \\ 4xy - y = - 40 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = 5x + 13\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4x(5x + 13) - 5x - 13 = - 40 \\ \end{matrix} \right.\ \]

\[20x^{2} + 52x - 5x - 13 + 40 = 0\]

\[20x^{2} + 47x + 27 = 0\]

\[D = 2209 - 2160 = 49\]

\[x_{1} = \frac{- 47 + 7}{40} = - 1,\ \ \]

\[x_{2} = \frac{- 47 - 7}{40} = - 1\frac{7}{20}\]

\[\left\{ \begin{matrix} x = - 1 \\ y = 8\ \ \ \\ \end{matrix} \right.\ \ \]

\[ИЛИ:\ \]

\[\left\{ \begin{matrix} x = - 1\frac{7}{20}\text{\ \ \ \ \ \ \ } \\ y = - \frac{27}{4} + 13 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} x = - 1\frac{7}{20}\text{\ \ \ \ \ \ \ } \\ y = \frac{- 27 + 52}{4} \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} x = - 1\frac{7}{20} \\ y = \frac{25}{4}\text{\ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = - 1,35 \\ y = 6,25\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:( - 1;8);( - 1,35;\ 6,25).\]