Biết nghiệm của hệ phương trình \(\left\{ {\begin{array}{*{20}{c}}{\dfrac{1}{{3x}} + \dfrac{1}{{3y}} = \dfrac{1}{4}}\\{\dfrac{5}{{6x}} + \dfrac{1}{y} = \dfrac{2}{3}}\end{array}} \right.\)là $\left( {x;y} \right)$. Tính $x - 3y$

Điều kiện: $x \ne 0;y \ne 0$

Ta có \(\left\{ {\begin{array}{*{20}{c}}{\dfrac{1}{{3x}} + \dfrac{1}{{3y}} = \dfrac{1}{4}}\\{\dfrac{5}{{6x}} + \dfrac{1}{y} = \dfrac{2}{3}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\dfrac{1}{3}.\dfrac{1}{x} + \dfrac{1}{3}.\dfrac{1}{y} = \dfrac{1}{4}}\\{\dfrac{5}{6}.\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{3}}\end{array}} \right.\)

Đặt $\dfrac{1}{x} = a;\dfrac{1}{y} = b$ khi đó ta có hệ phương trình

 $\left\{ \begin{array}{l}\dfrac{1}{3}a + \dfrac{1}{3}b = \dfrac{1}{4}\\\dfrac{5}{6}a + b = \dfrac{2}{3}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{1}{3}a + \dfrac{1}{3}b = \dfrac{1}{4}\\b = \dfrac{2}{3} - \dfrac{5}{6}a\end{array} \right.$\( \Leftrightarrow \left\{ \begin{array}{l}b = \dfrac{2}{3} - \dfrac{5}{6}a\\\dfrac{1}{3}a + \dfrac{1}{3}b = \dfrac{1}{4}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = \dfrac{2}{3} - \dfrac{5}{6}a\\\dfrac{1}{3}a + \dfrac{1}{3}\left( {\dfrac{2}{3} - \dfrac{5}{6}a} \right) = \dfrac{1}{4}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}b = \dfrac{2}{3} - \dfrac{5}{6}a\\\dfrac{1}{3}a + \dfrac{2}{9} - \dfrac{5}{{18}}b = \dfrac{1}{4}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = \dfrac{2}{3} - \dfrac{5}{6}a\\\dfrac{1}{{18}}b = \dfrac{1}{{36}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = \dfrac{1}{2}\\a = \dfrac{1}{4}\end{array} \right.\)

Thay lại cách đặt ta được

$\left\{ \begin{array}{l}\dfrac{1}{x} = \dfrac{1}{4}\\\dfrac{1}{y} = \dfrac{1}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 4\\y = 2\end{array} \right.$ (Thỏa mãn điều kiện)

Khi đó $x - 3y = 4 - 3.2 =  - 2$.