Решите систему уравнений 2x+y=7; x^2-xy=6.

Ответ:

\[\left\{ \begin{matrix} 2x + y = 7\ \\ x^{2} - xy = 6 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = 7 - 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - x(7 - 2x) = 6 \\ \end{matrix}\ \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} y = 7 - 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x² - 7x + 2x^{2} = 6 \\ \end{matrix}\ \right.\ \Longrightarrow \left\{ \begin{matrix} y = 7 - 2x\ \ \ \ \ \ \ \ \ \ \ \\ 3x² - 7x - 6 = 0 \\ \end{matrix} \right.\ \]

\[3x^{2} - 7x - 6 = 0\]

\[D = 49 + 72 = 121\]

\[x = \frac{7 + 11}{6} = 3\]

\[x = \frac{7 - 11}{6} = - \frac{2}{3}\]

\[\left\{ \begin{matrix} x = 3\ \ \ \ \ \ \ \ \\ y = 7 - 6 \\ \end{matrix} \right.\ \Leftrightarrow \left\{ \begin{matrix} x = - \frac{2}{3}\text{\ \ \ \ } \\ y = 7 + \frac{4}{3} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 3 \\ y = 1 \\ \end{matrix} \right.\ \Leftrightarrow \left\{ \begin{matrix} x = - \frac{2}{3} \\ y = 8\frac{1}{3}\ \\ \end{matrix} \right.\ \]

\[Ответ:(3;1);\ \ \left( - \frac{2}{3};8\frac{1}{3} \right).\]