If 4x + y = 7x - 15y, then find the value of $\dfrac{9x + 5y}{9x - 5y}$.

Given,

⇒ 4x + y = 7x - 15y

⇒ 7x - 4x = y + 15y

⇒ 3x = 16y

⇒ x = $\dfrac{16y}{3}$ …….(1)

Substituting value of x from equation (1) in $\dfrac{9x + 5y}{9x - 5y}$, we get :

$\Rightarrow \dfrac{9x + 5y}{9x - 5y} \\[1em] \Rightarrow \dfrac{9 \times \dfrac{16y}{3} + 5y}{9 \times \dfrac{16y}{3} - 5y} \\[1em] \Rightarrow \dfrac{\dfrac{144y}{3} + 5y}{\dfrac{144y}{3} - 5y} \\[1em] \Rightarrow \dfrac{\dfrac{144y + 15y}{3}}{\dfrac{144y - 15y}{3}} \\[1em] \Rightarrow \dfrac{\dfrac{159y}{3}}{\dfrac{129y}{3}} \\[1em] \Rightarrow \dfrac{159}{129} \\[1em] \Rightarrow \dfrac{53}{43} = 1\dfrac{10}{43}.$

Hence, $\dfrac{9x + 5y}{9x - 5y} = 1\dfrac{10}{43}$.