Решите систему уравнений: y^2-3xy+5x^2=3; 4x^2-3xy=1.

\[(1) + (2):\ \ \ \ \ y^{2} + 6xy - 7x^{2} = 0\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }(x - y)(7x + y) = 0\]

\[\left\{ \begin{matrix} (x - y)(7x + y) = 0 \\ 4x^{2} - 3xy = 1\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[4 \cdot y^{2} - 3y \cdot y = 1\]

\[4y^{2} - 3y^{2} = 1\]

\[y^{2} = 1\]

\[y_{1} = 1 \Longrightarrow x_{1} = 1.\]

\[y_{2} = - 1 \Longrightarrow x_{2} = - 1.\]

\[4x^{2} - 3x( - 7x) = 1\]

\[4x^{2} + 21x^{2} = 1\]

\[25x^{2} = 1\]

\[x^{2} = \frac{1}{25}\]

\[x_{1} = \frac{1}{5} \Longrightarrow \text{\ \ \ \ \ \ }\]

\[\Longrightarrow y_{1} = - 7 \cdot \frac{1}{5} = - \frac{7}{5} = - 1\frac{2}{5}.\]

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

\[\Longrightarrow y_{2} = - 7 \cdot \left( - \frac{1}{5} \right) = \frac{7}{5} = 1\frac{2}{5}.\]

\[Ответ:(1;1),\ ( - 1;\ - 1),\]

\[\ \left( \frac{1}{5};\ - 1\frac{2}{5} \right),\ \left( - \frac{1}{5};1\frac{2}{5} \right).\]