Solve:

$\dfrac{3x - 2}{3} + \dfrac{2x + 3}{2} = x + \dfrac{7}{6}$

$\dfrac{3x - 2}{3} + \dfrac{2x + 3}{2} = x + \dfrac{7}{6}$

Since L.C.M. of denominators 3 and 2 = 6, multiply each term with 6 to get:

$⇒ \dfrac{6(3x - 2)}{3} + \dfrac{6(2x + 3)}{2} = x \times 6 + \dfrac{7 \times 6}{6}$

⇒ 2(3x - 2) + 3(2x + 3) = 6x + 7 $\times$ 1

⇒ (6x - 4) + (6x + 9) = 6x + 7

⇒ 6x - 4 + 6x + 9 = 6x + 7

⇒ 12x + 5 = 6x + 7

⇒ 12x - 6x = 7 - 5

⇒ 6x = 2

⇒ x = $\dfrac{2}{6}$

⇒ x = $\dfrac{1}{3}$

Hence, the value of x is $\dfrac{1}{3}$.