Решите систему уравнений: 4x^2+y^2=29; y-2x=3.

\[\left\{ \begin{matrix} 4x² + y² = 29\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y - 2x = 3 \Longrightarrow y = 3 + 2x \\ \end{matrix} \right.\ \]

\[4x^{2} + (3 + 2x)^{2} = 29\]

\[4x^{2} + 4x^{2} + 12x + 9 = 29\]

\[8x^{2} + 12x - 20 = 0\ \ \ \ \ |\ :4\]

\[2x^{2} + 3x - 5 = 0\]

\[D = 9 + 40 = 49\]

\[x_{1} = \frac{- 3 + 7}{4} = 1;\ \ \ \ \ \ \ \ \]

\[\ x_{2} = \frac{- 3 - 7}{4} = - \frac{10}{4} = - 2,5\]

\[y_{1} = 3 + 2 \cdot 1 = 5;\ \ \ \ \ \ \ \]

\[y_{2} = 3 + 2 \cdot ( - 2,5) = 3 - 5 =\]

\[= - 2\]

\[Ответ:\ \ (1;5);\ \ ( - 2,5;\ - 2).\]