If $\dfrac{4x + 3y}{4x - 3y} = \dfrac{7}{4}$, use the properties to find the value of $\dfrac{2x^2 - 11y^2}{2x^2 + 11y^2}$.

Given,

Equation : $\dfrac{4x + 3y}{4x - 3y} = \dfrac{7}{4}$

Applying componendo and dividendo we get :

$\Rightarrow \dfrac{4x + 3y + (4x - 3y)}{4x + 3y - (4x - 3y)} = \dfrac{7 + 4}{7 - 4} \\[1em] \Rightarrow \dfrac{8x}{6y} = \dfrac{11}{3} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{6 \times 11}{3 \times 8} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{11}{4}$

Squaring both sides,

$\Rightarrow \Big(\dfrac{x}{y}\Big)^2 = \dfrac{121}{16} \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{121}{16}$

Multiplying both sides by $\dfrac{2}{11}$, we get :

$\Rightarrow \dfrac{2x^2}{11y^2} = \dfrac{121}{16} \times \dfrac{2}{11} \\[1em] \Rightarrow \dfrac{2x^2}{11y^2} = \dfrac{11}{8}$

Applying componendo and dividendo we get :

$\Rightarrow \dfrac{2x^2 + 11y^2}{2x^2 - 11y^2} = \dfrac{11 + 8}{11 - 8} \\[1em] \Rightarrow \dfrac{2x^2 + 11y^2}{2x^2 - 11y^2} = \dfrac{19}{3} \\[1em] \Rightarrow \dfrac{2x^2 - 11y^2}{2x^2 + 11y^2} = \dfrac{3}{19}.$

Hence, $\dfrac{2x^2 - 11y^2}{2x^2 + 11y^2} = \dfrac{3}{19}$.