If (6x² − xy) : (2xy − y²) = 6 : 1, find x : y.

Given,

(6x² − xy) : (2xy − y²) = 6 : 1,

Dividing numerator and denominator by y², we get :

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

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

$\Rightarrow \dfrac{6t^2 - t}{2t - 1} = 6 \\[1em] \Rightarrow 6t^2 - t = (2t - 1) \times 6 \\[1em] \Rightarrow 6t^2 - t = 12t - 6 \\[1em] \Rightarrow 6t^2 - t - 12t + 6 = 0 \\[1em] \Rightarrow 6t^2 - 13t + 6 = 0 \\[1em] \Rightarrow 6t^2 - 4t - 9t + 6 = 0 \\[1em] \Rightarrow 2t(3t - 2) - 3(3t - 2) = 0 \\[1em] \Rightarrow (2t - 3)(3t - 2) = 0 \\[1em] \Rightarrow (2t - 3) = 0 \text{ or }(3t - 2) = 0 \text{ [Using zero - product rule] } \\[1em] \Rightarrow 2t = 3 \text{ or } 3t = 2 \\[1em] \Rightarrow t = \dfrac{3}{2} \text{ or } t = \dfrac{2}{3} .$

Thus,

$\dfrac{x}{y} = \dfrac{3}{2} \text{ or } \dfrac{x}{y} = \dfrac{2}{3}$.

Hence, x : y = 3 : 2 or 2 : 3.