Solve :

$\dfrac{3}{x} - \dfrac{2}{y} = 0 \text{ and } \dfrac{2}{x} + \dfrac{5}{y} = 19$.

Hence, find 'a' if y = ax + 3,

Given, equations :

$\Rightarrow \dfrac{3}{x} - \dfrac{2}{y} = 0$ …….(1)

$\Rightarrow \dfrac{2}{x} + \dfrac{5}{y} = 19$ …….(2)

Multiplying equation (1) by 2, we get :

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

Multiplying equation (2) by 3, we get :

$\Rightarrow 3\Big(\dfrac{2}{x} + \dfrac{5}{y}\Big) = 3 \times 19 \\[1em] \Rightarrow \dfrac{6}{x} + \dfrac{15}{y} = 57 \text{ ………(4)}$

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

$\Rightarrow \Big(\dfrac{6}{x} + \dfrac{15}{y}\Big) - \Big(\dfrac{6}{x} - \dfrac{4}{y}\Big) = 57 - 0 \\[1em] \Rightarrow \dfrac{6}{x} - \dfrac{6}{x} + \dfrac{15}{y} + \dfrac{4}{y} = 57 \\[1em] \Rightarrow \dfrac{19}{y} = 57 \\[1em] \Rightarrow y = \dfrac{19}{57} = \dfrac{1}{3}.$

Substituting value of y in equation (1), we get :

$\Rightarrow \dfrac{3}{x} - \dfrac{2}{\dfrac{1}{3}} = 0 \\[1em] \Rightarrow \dfrac{3}{x} - 6 = 0 \\[1em] \Rightarrow \dfrac{3}{x} = 6 \\[1em] \Rightarrow x = \dfrac{3}{6} = \dfrac{1}{2}.$

Given,

$\Rightarrow y = ax + 3 \\[1em] \Rightarrow \dfrac{1}{3} = a \times \dfrac{1}{2} + 3 \\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{a}{2} + 3 \\[1em] \Rightarrow \dfrac{a}{2} = \dfrac{1}{3} - 3 \\[1em] \Rightarrow \dfrac{a}{2} = \dfrac{1 - 9}{3} \\[1em] \Rightarrow \dfrac{a}{2} = \dfrac{-8}{3} \\[1em] \Rightarrow a = -\dfrac{16}{3} = -5\dfrac{1}{3}.$

Hence, $x = \dfrac{1}{2}, y = \dfrac{1}{3}, a = -5\dfrac{1}{3}$.