Solve (question no. 2-22) for x :

$\dfrac{4(y + 2)}{5} = 7 + \dfrac{5y}{13}$

$\dfrac{4(y + 2)}{5} = 7 + \dfrac{5y}{13}\\[1em] ⇒ \dfrac{4y + 8}{5} = 7 + \dfrac{5y}{13}\\[1em] ⇒ \dfrac{4y}{5} + \dfrac{8}{5} = 7 + \dfrac{5y}{13}\\[1em] ⇒ \dfrac{4y}{5} - \dfrac{5y}{13} = 7 - \dfrac{8}{5}\\[1em]$

Since L.C.M. of denominators 5 and 13 = 65, multiply each term by 65 to get:

$⇒ \dfrac{4y \times 65}{5} - \dfrac{5y \times 65}{13} = 7 \times 65 - \dfrac{8 \times 65}{5}$

⇒ 4y $\times$ 13 - 5y $\times$ 5 = 455 - 8 $\times$ 13

⇒ 52y - 25y = 455 - 104

⇒ 27y = 351

⇒ y = $\dfrac{351}{27}$

⇒ y = 13

Hence, the value of y is 13.