Solve:

$\dfrac{3x - 2}{4} - \dfrac{2x + 3}{3} = \dfrac{2}{3} - x$

$\dfrac{3x - 2}{4} - \dfrac{2x + 3}{3} = \dfrac{2}{3} - x\\[1em] ⇒ \dfrac{3x - 2}{4} - \dfrac{2x + 3}{3} + x = \dfrac{2}{3}\\[1em] ⇒ \dfrac{3x - 2}{4} - \dfrac{2x + 3}{3} + \dfrac{x}{1} = \dfrac{2}{3}\\[1em]$

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

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

⇒ 3(3x - 2) - 4(2x + 3) + x $\times$ 12 = 2 $\times$ 4

⇒ (9x - 6) - (8x + 12) + 12x = 8

⇒ 9x - 6 - 8x - 12 + 12x = 8

⇒ 13x - 18 = 8

⇒ 13x = 8 + 18

⇒ 13x = 26

⇒ x = $\dfrac{26}{13}$

⇒ x = 2

Hence, the value of x is 2.