Solve the following simultaneous equations:

$\dfrac{3}{2x} + \dfrac{2}{3y} = 5, \dfrac{5}{x} - \dfrac{3}{y} = 1$

Given,

Equations:

$\dfrac{3}{2x} + \dfrac{2}{3y} = 5 \text{ ….(1) },$

$\dfrac{5}{x} - \dfrac{3}{y} = 1 \text{ ….(2) }$

Multiplying equation (1) by 9, we get:

$\Rightarrow 9\Big(\dfrac{3}{2x} + \dfrac{2}{3y}\Big) = 5 \times 9 \\[1em] \Rightarrow \Big(\dfrac{27}{2x} + \dfrac{6}{y}\Big) = 45 \text{ ……….(3) }$

Multiplying equation (2) by 2, we get:

$\Rightarrow 2\Big(\dfrac{5}{x} - \dfrac{3}{y}\Big) = 1 \times 2 \\[1em] \Rightarrow \Big(\dfrac{10}{x} - \dfrac{6}{y}\Big) = 2 \text{ ………(4) }$

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

$\Rightarrow \Big(\dfrac{27}{2x} + \dfrac{6}{y}\Big) + \Big(\dfrac{10}{x} - \dfrac{6}{y}\Big) = 45 + 2 \\[1em] \Rightarrow \Big(\dfrac{27}{2x} + \dfrac{10}{x}\Big) = 47 \\[1em] \Rightarrow \Big(\dfrac{27 + 20}{2x}\Big) = 47 \\[1em] \Rightarrow \Big(\dfrac{47}{2x}\Big) = 47 \\[1em] \Rightarrow 2x = \dfrac{47}{47} \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow x = \dfrac{1}{2}.$

Substituting value of x in equation (1),

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

Hence, $x = \dfrac{1}{2}, y = \dfrac{1}{3}$.