If a : b = 9 : 10, find the value of

(i) $\dfrac{5a + 3b}{5a - 3b}$

(ii) $\dfrac{2a^2 - 3b^2}{2a^2 + 3b^2}.$

(i) Given,

$\dfrac{a}{b} = \dfrac{9}{10} \\[0.5em] \Rightarrow a = \dfrac{9b}{10}$

Putting $a = \dfrac{9b}{10}$ in $\dfrac{5a + 3b}{5a - 3b}$, we get,

$\dfrac{5 \times \dfrac{9b}{10} + 3b}{5 \times \dfrac{9b}{10} - 3b} \\[1em] = \dfrac{\dfrac{9b}{2} + 3b}{\dfrac{9b}{2} - 3b} \\[1em] = \dfrac{\dfrac{9b + 6b}{2}}{\dfrac{9b - 6b}{2}} \\[1em] = \dfrac{15b}{3b} \\[1em] = 5.$

Hence, the value of $\dfrac{5a + 3b}{5a - 3b}$ = 5.

(ii) Given,

$\dfrac{a}{b} = \dfrac{9}{10} \\[0.5em] \Rightarrow a = \dfrac{9b}{10}$

Putting $a = \dfrac{9b}{10}$ in $\dfrac{2a^2 - 3b^2}{2a^2 + 3b^2}$, we get,

$\dfrac{2 \times \Big(\dfrac{9b}{10}\Big)^2 - 3b^2}{2 \times \Big(\dfrac{9b}{10}\Big)^2 + 3b^2} \\[1em] = \dfrac{2 \times \dfrac{81b^2}{100} - 3b^2}{2 \times \dfrac{81b^2}{100} + 3b^2} \\[1em] = \dfrac{\dfrac{81b^2 - 150b^2}{50}}{\dfrac{81b^2 + 150b^2}{50}} \\[1em] = -\dfrac{69b^2}{231b^2} \\[1em] = -\dfrac{23}{77}.$

Hence, the value of $\dfrac{2a^2 - 3b^2}{2a^2 + 3b^2} = -\dfrac{23}{77}$.