If (4x² − 3y²) : (2x² + 5y²) = 12 : 19, find x : y.

Given,

(4x² − 3y²) : (2x² + 5y²) = 12 : 19,

Dividing numerator and denominator both by y² we get,

$\Rightarrow \dfrac{\dfrac{4x^2 − 3y^2}{y^2}}{\dfrac{2x^2 + 5y^2}{y^2}} = \dfrac{12}{19} \\[1em] \Rightarrow \dfrac{4\Big(\dfrac{x}{y}\Big)^2 - 3\dfrac{y^2}{y^2}}{2\Big(\dfrac{x}{y}\Big)^2 + 5\dfrac{y^2}{y^2}} = \dfrac{12}{19} \\[1em] \Rightarrow \dfrac{4\Big(\dfrac{x}{y}\Big)^2 - 3}{2\Big(\dfrac{x}{y}\Big)^2 + 5} = \dfrac{12}{19}$

Let, $\dfrac{x}{y}$ = t

$\Rightarrow \dfrac{4t^2 - 3}{2t^2 + 5} = \dfrac{12}{19} \\[1em] \Rightarrow (4t^2 - 3) \times 19 = (2t^2 + 5) \times 12 \\[1em] \Rightarrow 76t^2 - 57 = 24t^2 + 60 \\[1em] \Rightarrow 76t^2 - 24t^2 = 60 + 57 \\[1em] \Rightarrow 52t^2 = 117 \\[1em] \Rightarrow t^2 = \dfrac{117}{52} \\[1em] \Rightarrow t^2 = \dfrac{9}{4} \\[1em] \Rightarrow t = \sqrt{\dfrac{9}{4}} \\[1em] \Rightarrow t = \dfrac{3}{2}.$

Hence, x : y = 3 : 2.