Question

If $\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}$, then find the value of $x:y$

Answer 1

$\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}$
Applying componendo and dividendo
$\frac{(x^2+y^2)+(x^2-y^2)}{(x^2+y^2)-(x^2-y^2)}=\frac{17+8}{17-8}$
$\Rightarrow \frac{2x^2}{2y^2}=\frac{25}{9}\Rightarrow \frac{x^2}{y^2}=\frac{25}{9}$
$\Rightarrow \frac{x}{y}=\frac{5}{3}$
$\Rightarrow x:y=5:3$.