Solve :

$\dfrac{2x}{3} - \dfrac{x - 1}{6} + \dfrac{7x - 1}{4} = 2\dfrac{1}{6}$. Hence, find the value of 'a', if $\dfrac{1}{a} + 5x = 8.$

$\dfrac{2x}{3} - \dfrac{x - 1}{6} + \dfrac{7x - 1}{4} = 2\dfrac{1}{6}$

$⇒ \dfrac{2x}{3} - \dfrac{x - 1}{6} + \dfrac{7x - 1}{4} = \dfrac{13}{6}$

Since L.C.M. of denominators 3, 6 and 4 = 12, multiply each term with 12 to get:

$⇒ \dfrac{2x \times 12}{3} - \dfrac{(x - 1) \times 12}{6} + \dfrac{(7x - 1) \times 12}{4} = \dfrac{13 \times 12}{6}$

⇒ 2x $\times$ 4 - (x - 1) $\times$ 2 + (7x - 1) $\times$ 3 = 13 $\times$ 2

⇒ 8x - (2x - 2) + (21x - 3) = 26

⇒ 8x - 2x + 2 + 21x - 3 = 26

⇒ 27x - 1 = 26

⇒ 27x = 26 + 1

⇒ 27x = 27

⇒ x = $\dfrac{27}{27}$

⇒ x = 1

Now, when x = 1

$\dfrac{1}{a} + 5x = 8\\[1em] ⇒ \dfrac{1}{a} + 5 \times 1 = 8\\[1em] ⇒ \dfrac{1}{a} + 5 = 8\\[1em] ⇒ \dfrac{1}{a} = 8 - 5\\[1em] ⇒ \dfrac{1}{a} = 3\\[1em] ⇒ a = \dfrac{1}{3}$

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