Giải thích các bước giải:
Ta có :
$\begin{cases}2(y-2)+3x=4xy-8x\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{cases}$
$\to \begin{cases}2(y-2)+3x=4x(y-2)\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{cases}$
$\to \begin{cases}\dfrac{1}{x}+\dfrac{3}{2(y-2)}=2\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{cases}$
$\to \begin{cases}\dfrac{3}{2(y-2)}=2-\dfrac{1}{x}\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{cases}$
$\to \begin{cases}\dfrac{1}{y-2}=\dfrac 43-\dfrac{2}{3x}\\\dfrac{4}{x}-(\dfrac 43-\dfrac{2}{3x})=1\end{cases}$
$\to \begin{cases}y=3\\x=2\end{cases}$
