Solve:

$\dfrac{x + 2}{6} - \Big(\dfrac{11 - x}{3} - \dfrac{1}{4}\Big) = \dfrac{3x - 4}{12}$

$\dfrac{x + 2}{6} - \Big(\dfrac{11 - x}{3} - \dfrac{1}{4}\Big) = \dfrac{3x - 4}{12}\\[1em] ⇒ \dfrac{x + 2}{6} - \dfrac{11 - x}{3} + \dfrac{1}{4} = \dfrac{3x - 4}{12}$

Since L.C.M. of denominators 6, 3 and 4 = 12, multiply each term with 12 to get:

$⇒ \dfrac{12(x + 2)}{6} - \dfrac{12(11 - x)}{3} + \dfrac{1 \times 12}{4} = \dfrac{12(3x - 4)}{12}$

⇒ 2(x + 2) - 4(11 - x) + 1 $\times$ 3 = 1(3x - 4)

⇒ (2x + 4) - (44 - 4x) + 3 = 3x - 4

⇒ 2x + 4 - 44 + 4x + 3 = 3x - 4

⇒ 6x - 37 = 3x - 4

⇒ 6x - 3x = 37 - 4

⇒ 3x = 33

⇒ x = $\dfrac{33}{3}$

⇒ x = 11

Hence, the value of x is 11.