Solve :

$5x + \dfrac{8}{y} = 19$

$3x - \dfrac{4}{y} = 7$

Given, equations :

$\Rightarrow 5x + \dfrac{8}{y} = 19$ ………(1)

$\Rightarrow 3x - \dfrac{4}{y} = 7$ ……..(2)

Multiplying equation (2) by 2, we get :

$\Rightarrow 2\Big(3x - \dfrac{4}{y}\Big) = 2 \times 7 \\[1em] \Rightarrow 6x - \dfrac{8}{y} = 14 \text{ ……..(3)}$

Adding equation (1) and (3), we get :

$\Rightarrow \Big(5x + \dfrac{8}{y}\Big) + \Big(6x - \dfrac{8}{y}\Big) = 19 + 14 \\[1em] \Rightarrow 5x + 6x + \dfrac{8}{y} - \dfrac{8}{y} = 33 \\[1em] \Rightarrow 11x = 33 \\[1em] \Rightarrow x = \dfrac{33}{11} = 3.$

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

$\Rightarrow 5 \times 3 + \dfrac{8}{y} = 19 \\[1em] \Rightarrow 15 + \dfrac{8}{y} = 19 \\[1em] \Rightarrow \dfrac{8}{y} = 19 - 15 \\[1em] \Rightarrow \dfrac{8}{y} = 4 \\[1em] \Rightarrow y = \dfrac{8}{4} = 2.$

Hence, x = 3 and y = 2.