Solve:

$\dfrac{x + 2}{3} - \dfrac{x + 1}{5} = \dfrac{x - 3}{4} - 1$

$\dfrac{x + 2}{3} - \dfrac{x + 1}{5} = \dfrac{x - 3}{4} - 1\\[1em] ⇒ \dfrac{x + 2}{3} - \dfrac{x + 1}{5} - \dfrac{x - 3}{4} = - 1$

Since L.C.M. of denominators 3, 5 and 4 = 60, multiply each term with 60 to get:

$⇒ \dfrac{60(x + 2)}{3} - \dfrac{60(x + 1)}{5} - \dfrac{60(x - 3)}{4} = - 1 \times 60$

⇒ 20(x + 2) - 12(x + 1) - 15(x - 3) = - 60

⇒ (20x + 40) - (12x + 12) - (15x - 45) = - 60

⇒ 20x + 40 - 12x - 12 - 15x + 45 = - 60

⇒ - 7x + 73 = - 60

⇒ - 7x = - 60 - 73

⇒ - 7x = - 133

⇒ x = $\dfrac{-133}{-7}$

⇒ x = 19

Hence, the value of x is 19.