Question

If $$(4x^2 - 3y^2) : (2x^2 + 5y^2) = 12 : 19$$, find $$x : y$$.

Answer 1

Given: $$\frac{4x^2 - 3y^2}{2x^2 + 5y^2} = \frac{12}{19}$$ 

Cross-multiply: $$19(4x^2 - 3y^2) = 12(2x^2 + 5y^2)$$

$$76x^2 - 57y^2 = 24x^2 + 60y^2$$ 

Group like terms: $$76x^2 - 24x^2 = 60y^2 + 57y^2$$

$$52x^2 = 117y^2$$

Divide both sides by 13: $$4x^2 = 9y^2$$

$$\frac{x^2}{y^2} = \frac{9}{4}$$ 

Taking the positive square root: $$\frac{x}{y} = \frac{3}{2}$$ 

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