Решите систему уравнений способом подстановки 3x+y=1; x^2+y^2+xy=3.

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

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

\[x^{2} + 1 - 6x + 9x^{2} + x - 3x^{2} - 3 = 0\]

\[7x^{2} - 5x - 2 = 0\]

\[D = 25 + 56 = 81\]

\[x_{1} = \frac{5 + 9}{14} = 1;\]

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

\[x_{2} = \frac{5 - 9}{14} = - \frac{4}{14} = - \frac{2}{7};\]

\[y_{2} = 1 - 3 \cdot \left( - \frac{2}{7} \right) = 1 + \frac{6}{7} = 1\frac{6}{7};\]

\[\left\{ \begin{matrix}
x = 1\ \ \ \\
y = - 2 \\
\end{matrix} \right.\ \ \ \ \ и\ \ \ \ \left\{ \begin{matrix}
x = - \frac{2}{7} \\
y = 1\frac{6}{7}\ \\
\end{matrix} \right.\ \]

\[Ответ:(1;\ - 2);\left( - \frac{2}{7};1\frac{6}{7} \right).\]