Solve the following simultaneous equations:

$\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6},\;\dfrac{3}{x} + \dfrac{2}{y} = 0$

Given,

Equations:

$\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6} \text{ ….(1) }$,

$\dfrac{3}{x} + \dfrac{2}{y} = 0 \text{ ….(2) }$

Multiplying equation (1) by 3, we get:

$\Rightarrow 3\Big(\dfrac{2}{x} + \dfrac{2}{3y}\Big) = \dfrac{1}{6} \times 3 \\[1em] \Rightarrow \Big(\dfrac{6}{x} + \dfrac{2}{y}\Big) = \dfrac{1}{2} \text{ ….(3) }$

Multiplying equation (2) by 2, we get:

$\Rightarrow 2\Big(\dfrac{3}{x} + \dfrac{2}{y} \Big) = 0 \times 2 \\[1em] \Rightarrow \Big(\dfrac{6}{x} + \dfrac{4}{y}\Big) = 0 \text{ ….(4) }$

Subtracting equation (3) from (4), we get:

$\Rightarrow \Big(\dfrac{6}{x} + \dfrac{4}{y}\Big) - \Big(\dfrac{6}{x} + \dfrac{2}{y}\Big) = 0 - \dfrac{1}{2} \\[1em] \Rightarrow \Big(\dfrac{6}{x} + \dfrac{4}{y} - \dfrac{6}{x} - \dfrac{2}{y}\Big) = -\dfrac{1}{2} \\[1em] \Rightarrow \Big(\dfrac{4}{y} - \dfrac{2}{y}\Big) = -\dfrac{1}{2} \\[1em] \Rightarrow \Big(\dfrac{2}{y}\Big) = -\dfrac{1}{2} \\[1em] \Rightarrow y = \dfrac{2 \times 2}{-1} \\[1em] \Rightarrow y = -4.$

Substituting value of y in equation (1),

$\Rightarrow \dfrac{2}{x} + \dfrac{2}{3(-4)} = \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{2}{x} - \dfrac{1}{6} = \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{2}{x} = \dfrac{1}{6} + \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{2}{x} = \dfrac{2}{6} \\[1em] \Rightarrow \dfrac{2}{x} = \dfrac{1}{3} \\[1em] \Rightarrow x = 2 \times 3 \\[1em] \Rightarrow x = 6.$

Hence, x = 6, y = -4.