$\begin{cases}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{12}\\\dfrac{3}{x}+\dfrac{4}{y}=2\end{cases}$
Đặt $\dfrac{1}{x}=a;\dfrac{1}{y}=b$
$⇔ \begin{cases} a+b=\dfrac{7}{12}\\3a+4b=2\end{cases}$
$⇔ \begin{cases}4a+4b=\dfrac{7}{3}\\3a+4b=2\end{cases}$
$⇔ \begin{cases}a=\dfrac{1}{3}\\1+4b=2\end{cases}$
$⇔ \begin{cases}a=\dfrac{1}{3}\\b=\dfrac{1}{4}\end{cases}$
$⇔ \begin{cases}x=3\\y=4\end{cases}$
Vậy $(x;y)=(3;4)$
