Solve the following simultaneous equations:

$\dfrac{x}{7} + \dfrac{y}{3} = 5, \dfrac{x}{2} - \dfrac{y}{9} = 6$

Given,

$\dfrac{x}{7} + \dfrac{y}{3} = 5, \dfrac{x}{2} - \dfrac{y}{9} = 6$

Simplifying,

$\Rightarrow \dfrac{x}{7} + \dfrac{y}{3} = 5 \\[1em] \Rightarrow \dfrac{3x + 7y}{21} = 5 \\[1em] \Rightarrow 3x + 7y = 5 \times 21 \\[1em] \Rightarrow 3x + 7y = 105 \\[1em] \Rightarrow 3x = 105 - 7y \\[1em] \Rightarrow x = \dfrac{105 - 7y}{3} \text{ ….(1)}$

Substituting value of x from equation (1) in $\dfrac{x}{2} - \dfrac{y}{9} = 6$, we get :

$\Rightarrow \dfrac{\dfrac{105 - 7y}{3}}{2} - \dfrac{y}{9} = 6 \\[1em] \Rightarrow \dfrac{(105 - 7y)}{6} - \dfrac{y}{9} = 6 \\[1em] \Rightarrow \dfrac{3(105 - 7y) - 2y}{18} = 6 \\[1em] \Rightarrow \dfrac{315 - 21y - 2y}{18} = 6 \\[1em] \Rightarrow 315 - 23y = 6 \times 18 \\[1em] \Rightarrow 315 - 23y = 108 \\[1em] \Rightarrow 23y = 315 - 108 \\[1em] \Rightarrow 23y = 207 \\[1em] \Rightarrow y = \dfrac{207}{23} = 9.$

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

$\Rightarrow x = \dfrac{105 - 7y}{3} \\[1em] \Rightarrow x = \dfrac{105 - 7(9)}{3} \\[1em] \Rightarrow x = \dfrac{105 - 63}{3} \\[1em] \Rightarrow x = \dfrac{42}{3} \\[1em] \Rightarrow x = 14.$

Hence, x = 14, y = 9.