If x and y both are positive and (2x² - 5y²) : xy = 1 : 3, find x : y.

Given,

$\Rightarrow \dfrac{2x^2 - 5y^2}{xy} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{3(2x^2 - 5y^2)}{xy} = 1 \\[1em] \Rightarrow \dfrac{6x^2 - 15y^2}{xy} = 1 \\[1em] \Rightarrow \dfrac{6x}{y} - \dfrac{15y}{x} = 1$

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

$\Rightarrow 6t - \dfrac{15}{t} = 1 \\[1em] \Rightarrow \dfrac{6t^2 - 15}{t} = 1 \\[1em] \Rightarrow 6t^2 - 15 = t \\[1em] \Rightarrow 6t^2 - t - 15 = 0 \\[1em] \Rightarrow 6t^2 - 10t + 9t - 15 = 0 \\[1em] \Rightarrow 2t(3t - 5) + 3(3t - 5) = 0 \\[1em] \Rightarrow (2t + 3)(3t - 5) = 0 \\[1em] \Rightarrow 2t + 3 = 0 \text{ or } 3t - 5 = 0 \\[1em] \Rightarrow t = -\dfrac{3}{2} \text{ or } t = \dfrac{5}{3}.$

Since, x and y are positive.

∴ t ≠ $-\dfrac{3}{2}$.

∴ t = $\dfrac{x}{y} = \dfrac{5}{3}$

⇒ x : y = 5 : 3.

Hence, x : y = 5 : 3.