$\left\{\begin{matrix} \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12}\\ \dfrac{4}{x} + \dfrac{10}{y} = \dfrac{2}{3} \end{matrix}\right.$ (ĐK $x \neq 0; y \neq 0$)
Đặt $u = \dfrac{1}{x}; u = \dfrac{1}{y}$, hệ phương trình trở thành:
$\left\{\begin{matrix} u + v = \dfrac{1}{12}\\ 4u + 10v = \dfrac{2}{3} \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} 4u + 4v = \dfrac{1}{3}\\ 4u + 10v = \dfrac{2}{3} \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} -6v = -\dfrac{1}{3}\\ u + v = \dfrac{1}{12} \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} v = \frac{1}{18}\\ \frac{1}{18} + u = \dfrac{1}{12} \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} u = \frac{1}{36}\\ v = \frac{1}{18} \end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix} \dfrac{1}{x} = \frac{1}{36}\\ \dfrac{1}{y} = \frac{1}{18} \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} x = 36\\ y = 18 \end{matrix}\right.$ (Thỏa mãn ĐK)
