If 7x - 15y = 4x + y, find the value of x : y. Hence, use componendo and dividendo to find the values of :

(i) $\dfrac{9x + 5y}{9x - 5y}$

(ii) $\dfrac{3x^2 + 2y^2}{3x^2 - 2y^2}$

7x - 15y = 4x + y

⇒ 7x - 4x = y + 15y

⇒ 3x = 16y

⇒ $\dfrac{x}{y} = \dfrac{16}{3}$

(i) $\dfrac{9x + 5y}{9x - 5y}$

$\phantom{\Rightarrow} \dfrac{x}{y} = \dfrac{16}{3} \\[1em] \dfrac{9x}{5y} = \dfrac{9 \times 16}{5 \times 3} \\[1em] \dfrac{9x}{5y} = \dfrac{144}{15}$

Applying componendo and dividendo:

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

Hence, $\dfrac{9x + 5y}{9x - 5y} = \dfrac{53}{43}$.

(ii) $\dfrac{3x^2 + 2y^2}{3x^2 - 2y^2}$

$\Rightarrow \dfrac{x}{y} = \dfrac{16}{3} \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{256}{9} \\[1em] \Rightarrow \dfrac{3x^2}{2y^2} = \dfrac{3 \times 256}{2 \times 9} \\[1em] \Rightarrow \dfrac{3x^2}{2y^2} = \dfrac{768}{18}$

Applying componendo and dividendo:

$\Rightarrow \dfrac{3x^2 + 2y^2}{3x^2 - 2y^2} = \dfrac{768 + 18}{768 - 18} \\[1em] \Rightarrow \dfrac{3x^2 + 2y^2}{3x^2 - 2y^2} = \dfrac{786}{750} \\[1em] \Rightarrow \dfrac{3x^2 + 2y^2}{3x^2 - 2y^2} = \dfrac{131}{125}.$

Hence, $\dfrac{3x^2 + 2y^2}{3x^2 - 2y^2} = \dfrac{131}{125}$.