Решите систему уравнений способом подстановки 4x-y=2; x^2+y^2-xy=3.

Ответ:

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

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

\[(*):\]

\[x^{2} + 16x^{2} - 16x + 4 - 4x^{2} + 2x - 3 = 0\]

\[13x^{2} - 14x + 1 = 0\]

\[D = 49 - 13 = 36\]

\[x_{1} = \frac{7 + 6}{13} = 1;\ \ x_{2} = \frac{7 - 6}{13} = \frac{1}{13}\]

\[y_{2} = 4 \cdot \frac{1}{13} - 2 = \frac{4}{13} - 2 = - 1\frac{9}{13}\]

\[y_{1} = 4 \cdot 1 - 2 = 2\]

\[\left\{ \begin{matrix} x = \frac{1}{13}\text{\ \ \ \ } \\ y = 1\frac{9}{13} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 1 \\ y = 2 \\ \end{matrix} \right.\ \]

\[Ответ:\left( \frac{1}{13};1\frac{9}{13} \right);(1;\ 2).\]