Solve the following simultaneous equations:

$4x + \dfrac{6}{y} = 15, 3x - \dfrac{4}{y} = 7$

Given,

Equations:

$\Rightarrow 4x + \dfrac{6}{y} = 15 \text{ ….(1)} \\[1em] \Rightarrow 3x - \dfrac{4}{y} = 7 \text{ ….(2)}$

Multiplying equation (1) by 4, we get :

$\Rightarrow 4\Big(4x + \dfrac{6}{y}\Big) = 15 \times 4 \\[1em] \Rightarrow 4\Big(4x + \dfrac{6}{y}\Big) = 15 \times 4 \\[1em] \Rightarrow 16x + \dfrac{24}{y} = 60\text{ ….(3)}$

Multiplying equation (2) by 6, we get :

$\Rightarrow 6\Big(3x - \dfrac{4}{y}\Big) = 7 \times 6 \\[1em] \Rightarrow 18x - \dfrac{24}{y} = 42\text{ ….(4)}$

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

$\Rightarrow \Big(16x + \dfrac{24}{y}\Big) + \Big(18x - \dfrac{24}{y}\Big) = 60 + 42 \\[1em] \Rightarrow 18x + \dfrac{24}{y} + 16x - \dfrac{24}{y} = 102 \\[1em] \Rightarrow 34x = 102 \\[1em] \Rightarrow x = \dfrac{102}{34} \\[1em] \Rightarrow x = 3.$

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

$\Rightarrow 4(3) + \dfrac{6}{y} = 15 \\[1em] \Rightarrow 12 + \dfrac{6}{y} = 15 \\[1em] \Rightarrow \dfrac{6}{y} = 15 - 12 \\[1em] \Rightarrow \dfrac{6}{y} = 3 \\[1em] \Rightarrow y = \dfrac{6}{3} \\[1em] \Rightarrow y = 2.$

Hence, x = 3, y = 2.