Решите систему уравнений x^2+6xy+9y^2=16; x-3y=-2.

Ответ:

\[\left\{ \begin{matrix} x² + 6xy + 9y² = 16 \\ x - 3y = - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} (x + 3y)^{2} = 16 \\ x = 3y - 2\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

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

\[\left\{ \begin{matrix} 6y = 6\ \ \ \ \ \ \ \\ x = 3y - 2 \\ \end{matrix} \right.\ \Longleftrightarrow \left\{ \begin{matrix} y = - \frac{1}{3} \\ x = - 3\ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 1 \\ x = 1 \\ \end{matrix} \right.\ \]

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