Giải hệ phương trình sau \(\left\{ \begin{array}{l}\dfrac{6}{x} - \dfrac{4}{y} = - 4\\\dfrac{3}{x} + \dfrac{8}{y} = 3\end{array} \right.\)

* Hướng dẫn giải

Điều kiện : \(x \ne 0;\,\,y \ne 0\).

Đặt \(\left\{ \begin{array}{l}u = \dfrac{1}{x}\\v = \dfrac{1}{y}\end{array} \right.\), khi đó hệ phương trình trở thành

\(\begin{array}{l}\left\{ \begin{array}{l}6u - 4v = - 4\\3u + 8v = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3u - 2v = - 2\\3u + 8v = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}10v = 5\\3u + 8v = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}v = \dfrac{1}{2}\\3u + 4 = 3\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}v = \dfrac{1}{2}\\3u = - 1\end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l}v = \dfrac{1}{2}\\u = \dfrac{{ - 1}}{3}\end{array} \right.\end{array}\)

\(\begin{array}{l}v = \dfrac{1}{2} \Rightarrow \dfrac{1}{y} = \dfrac{1}{2}\\ \Leftrightarrow y = 2\,\,\,\,\,\,\,\,\,\,\left( {tm} \right)\\u = \dfrac{{ - 1}}{3} \Rightarrow \dfrac{1}{x} = \dfrac{{ - 1}}{3} \\\Leftrightarrow x = - 3\,\,\,\left( {tm} \right)\end{array}\)