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

1. x = 12, y = 8

2. x = 10, y = 8

3. x = 8, y = 12

4. x = 12, y = 6

Given,

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

Solving first equation,

⇒ $\dfrac{x}{6} + 6 = y$

⇒ $\dfrac{x + 36}{6} = y$

⇒ x + 36 = 6y

⇒ x = 6y - 36     …….(1)

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

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

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

⇒ x = 6y - 36

⇒ x = 6(8) - 36

⇒ x = 48 - 36 = 12.

Hence, option 1 is the correct option.