Solve :

$\dfrac{4 - 3x}{5} + \dfrac{7 - x}{3} + 4\dfrac{1}{3} = 0$. Hence, find the value of 'p', if 3p - 2x + 1 = 0.

$\dfrac{4 - 3x}{5} + \dfrac{7 - x}{3} + 4\dfrac{1}{3} = 0\\[1em] ⇒ \dfrac{4 - 3x}{5} + \dfrac{7 - x}{3} + \dfrac{13}{3} = 0\\[1em]$

Since L.C.M. of denominators 5 and 3 = 15, multiply each term with 15 to get:

$⇒ \dfrac{(4 - 3x) \times 15}{5} + \dfrac{(7 - x) \times 15}{3} + \dfrac{13 \times 15}{3} = 0 \times 15$

⇒ (4 - 3x) $\times$ 3 + (7 - x) $\times$ 5 + 13 $\times$ 5 = 0

⇒ (12 - 9x) + (35 - 5x) + 65 = 0

⇒ 12 - 9x + 35 - 5x + 65 = 0

⇒ 112 - 14x = 0

⇒ 14x = 112

⇒ x = $\dfrac{112}{14}$

⇒ x = 8

Now, when x = 8

3p - 2x + 1 = 0

⇒ 3p - 2 $\times$ 8 + 1 = 0

⇒ 3p - 16 + 1 = 0

⇒ 3p - 15 = 0

⇒ 3p = 15

⇒ p = $\dfrac{15}{3}$

⇒ p = 5

Hence, the value of x is 8 and p is 5.