Question

If $$(6x^2 - xy) : (2xy - y^2) = 6 : 1$$, find $$x : y$$.

Answer 1

Given equation: $$\frac{6x^2 - xy}{2xy - y^2} = \frac{6}{1}$$ 

Cross-multiply: $$6x^2 - xy = 6(2xy - y^2)$$

$$6x^2 - xy = 12xy - 6y^2$$ 

Rearrange into standard quadratic form: $$6x^2 - 13xy + 6y^2 = 0$$ 

Factor by splitting the middle term ($$-9 \times -4 = 36$$): $$6x^2 - 9xy - 4xy + 6y^2 = 0$$

$$3x(2x - 3y) - 2y(2x - 3y) = 0$$

$$(2x - 3y)(3x - 2y) = 0$$ 

This gives: $$2x - 3y = 0$$

$$\implies 2x = 3y$$

$$\implies \frac{x}{y} = \frac{3}{2}$$ 

or $$3x - 2y = 0$$

$$\implies 3x = 2y$$

$$\implies \frac{x}{y} = \frac{2}{3}$$ 

Therefore, $$x : y = 3 : 2$$ or $$2 : 3$$.