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

Given, (4x² + xy) : (3xy - y²) = 12 : 5

$\Rightarrow \dfrac{4x^2 + xy}{3xy - y^2} = \dfrac{12}{5} \\[1em] \Rightarrow 5(4x^2 + xy) = 12(3xy - y^2) \\[1em] \Rightarrow 20x^2 + 5xy = 36xy - 12y^2 \\[1em] \Rightarrow 20x^2 - 31xy + 12y^2 = 0 \\[1em] \Rightarrow 20\dfrac{x^2}{y^2} - 31\dfrac{x}{y} + 12 = 0 \\[1em] \Rightarrow 20(\dfrac{x}{y})^2 - 31(\dfrac{x}{y}) + 12 = 0 \\[1em] \Rightarrow 20(\dfrac{x}{y})^2 - 15(\dfrac{x}{y}) - 16(\dfrac{x}{y})+ 12 = 0 \\[1em] \Rightarrow 5(\dfrac{x}{y})\big(4\big(\dfrac{x}{y}\big) - 3\big) - 4\big(4\big(\dfrac{x}{y}\big) - 3\big) \\[1em] \Rightarrow \big(5\big(\dfrac{x}{y}\big) - 4\big) \big(4\big(\dfrac{x}{y}\big) - 3\big) = 0 \\[1em] \Rightarrow \big(5\big(\dfrac{x}{y}\big) - 4\big) = 0 \text{ or } \big(4\big(\dfrac{x}{y}\big) - 3\big) = 0 \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{4}{5} \text{ or } \dfrac{x}{y} = \dfrac{3}{4}.$

We need to find value of (x + 2y) : (2x + y) or $\dfrac{x + 2y}{2x + y}$

Dividing the numerator and denominator by y,

$\Rightarrow \dfrac{\dfrac{x}{y} + 2}{\dfrac{2x}{y} + 1}$

Putting value of $\dfrac{x}{y} = \dfrac{4}{5}$,

$\Rightarrow \dfrac{\dfrac{4}{5} + 2}{\dfrac{8}{5} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{14}{5}}{\dfrac{13}{5}} \\[1em] \Rightarrow \dfrac{14}{13} = 14 : 13. \\[1em]$

Putting value of $\dfrac{x}{y} = \dfrac{3}{4}$,

$\Rightarrow \dfrac{\dfrac{3}{4} + 2}{\dfrac{6}{4} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{11}{4}}{\dfrac{10}{4}} \\[1em] \Rightarrow \dfrac{11}{10} = 11 : 10$

Hence, the value of ratio (x + 2y) : (2x + y) is 14 : 13 or 11 : 10.