Solve the following pairs of linear equations:

$\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6}$

$\dfrac{2}{x} - \dfrac{1}{y} = 1$

Substituting $\dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b$ in above equations we get,

$2a + \dfrac{2}{3}b = \dfrac{1}{6}$ …….(i)

2a - b = 1 …….(ii)

Subtracting eq. (ii) from (i) we get,

$\Rightarrow 2a + \dfrac{2}{3}b - (2a - b) = \dfrac{1}{6} - 1 \\[1em] \Rightarrow 2a - 2a + \dfrac{2}{3}b + b = \dfrac{1 - 6}{6} \\[1em] \Rightarrow \dfrac{2b + 3b}{3} = -\dfrac{5}{6} \\[1em] \Rightarrow \dfrac{5b}{3} = -\dfrac{5}{6} \\[1em] \Rightarrow b = -\dfrac{5 \times 3}{6 \times 5} \\[1em] \Rightarrow b = -\dfrac{1}{2} \\[1em] \therefore \dfrac{1}{y} = -\dfrac{1}{2} \\[1em] \Rightarrow y = -2.$

Substituting value of b in eq (ii),

$\Rightarrow 2a - \Big(-\dfrac{1}{2}\Big) = 1 \\[1em] \Rightarrow 2a + \dfrac{1}{2} = 1 \\[1em] \Rightarrow 2a = 1 - \dfrac{1}{2} \\[1em] \Rightarrow 2a = \dfrac{1}{2} \\[1em] \Rightarrow a = \dfrac{1}{4} \\[1em] \therefore \dfrac{1}{x} = \dfrac{1}{4} \\[1em] \Rightarrow x = 4.$

Hence, x = 4 and y = -2.