Решите систему уравнений: x/y+y/x=10/3; x-y=6.

Ответ:

\[3 \cdot (6 + y)^{2} + 3y \cdot y =\]

\[= 10y(6 + y)\]

\[3 \cdot \left( 36 + 12y + y^{2} \right) + 3y^{2} =\]

\[= 60y + 10y^{2}\]

\[108 + 36y + 3y^{2} + 3y^{2} =\]

\[= 60y + 10y^{2}\]

\[4y^{2} + 24y - 108 = 0\ \ \ \ \ |\ :4\]

\[y^{2} + 6y - 27 = 0\]

\[D_{1} = 9 + 27 = 36\]

\[y_{1} = - 3 + 6 = 3;\ \ \]

\[y_{2} = - 3 - 6 = - 9.\]

\[\left\{ \begin{matrix} y = 3 \\ x = 9 \\ \end{matrix} \right.\ \ \ \ \ \ или\ \ \ \left\{ \begin{matrix} y = - 9 \\ x = - 3 \\ \end{matrix} \right.\ \]

\(Ответ:(9;3);( - 3; - 9).\)