Question

\(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{2}{y}=3\\\dfrac{4}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{2}{y}=3\\\dfrac{8}{x}+\dfrac{2}{y}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-5}{x}=5\\\dfrac{3}{x}+\dfrac{2}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\\dfrac{2}{y}=6\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(-1;\dfrac{1}{3}\right)\)

Answer 2

Đặt 1/x = u ; 1/y = v 

Ta có hệ \(\left\{{}\begin{matrix}3u+2v=3\\4u+v=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=-1\\v=3\end{matrix}\right.\)

Theo cách đặt x = -1 ; y = 1/3