Solve the following simultaneous equations:

$\dfrac{x}{6} + 6 = y, \dfrac{3x}{4} = 1 + y$

Simplifying, equation : $\dfrac{x}{6} + 6 = y$

$\Rightarrow \dfrac{x}{6} + 6 = y \\[1em] \Rightarrow \dfrac{x + 36}{6} = y \\[1em] \Rightarrow x + 36 = 6y \\[1em] \Rightarrow x = 6y - 36 \text{ ….(1)}$

Substituting value of x from equation (1) in $\dfrac{3x}{4} = 1 + y$, we get :

$\Rightarrow \dfrac{3(6y - 36)}{4} = 1 + y \\[1em] \Rightarrow \dfrac{18y - 108}{4} = 1 + y \\[1em] \Rightarrow 18y - 108 = 4(1 + y) \\[1em] \Rightarrow 18y - 108 = (4 + 4y) \\[1em] \Rightarrow 18y - 4y = 4 + 108 \\[1em] \Rightarrow 14y = 112 \\[1em] \Rightarrow y = \dfrac{112}{14} \\[1em] \Rightarrow y = 8.$

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

⇒ x = 6y - 36

⇒ x = 6(8) - 36

⇒ x = 48 - 36

⇒ x = 12.

Hence, x = 12, y = 8.