Solve:

$\dfrac{6x + 1}{2} + 1 = \dfrac{7x - 3}{3}$

$\dfrac{6x + 1}{2} + 1 = \dfrac{7x - 3}{3}\\[1em] ⇒ \dfrac{6x + 1}{2} - \dfrac{7x - 3}{3} = - 1$

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

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

⇒ 3(6x + 1) - 2(7x - 3) = - 6

⇒ (18x + 3) - (14x - 6) = - 6

⇒ 18x + 3 - 14x + 6 = - 6

⇒ 4x + 9 = - 6

⇒ 4x = - 6 - 9

⇒ 4x = - 15

⇒ x = - $\dfrac{15}{4}$

Hence, the value of x is $-\dfrac{15}{4}$.