If 15(2x² - y²) = 7xy, find x : y; if x and y both are positive.

Given,

$\Rightarrow 15(2x^2 - y^2) = 7xy \\[1em] \Rightarrow 30x^2 - 15y^2 = 7xy \\[1em] \Rightarrow \dfrac{30x^2 - 15y^2}{xy} = \dfrac{7xy}{xy} \\[1em] \Rightarrow 30\dfrac{x}{y} - 15\dfrac{y}{x} = 7$

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

$\Rightarrow 30t - 15\dfrac{1}{t} = 7 \\[1em] \Rightarrow \dfrac{30t^2 - 15}{t} = 7 \\[1em] \Rightarrow 30t^2 - 15 = 7t \\[1em] \Rightarrow 30t^2 - 7t - 15 = 0 \\[1em] \Rightarrow 30t^2 - 25t + 18t - 15 = 0 \\[1em] \Rightarrow 5t(6t - 5) + 3(6t - 5) = 0 \\[1em] \Rightarrow (5t + 3)(6t - 5) = 0 \\[1em] \Rightarrow 5t + 3 = 0 \text{ or } 6t - 5 = 0 \\[1em] \Rightarrow t = -\dfrac{3}{5} \text{ or } t = \dfrac{5}{6}.$

Since, x and y both are positive,

∴ t ≠ $-\dfrac{3}{5}$.

Hence, x : y = 5 : 6.