An algebra problem by A Former Brilliant Member

$\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}} = \dfrac{5}{4}\\ 4\left(x+\dfrac{1}{x}\right) = 5\left(x-\dfrac{1}{x}\right)\\ \dfrac{4}{x} + \dfrac{5}{x} = 5x - 4x\\ x=\dfrac{9}{x}\\ x^2=9\\ \boxed{x= \pm 3}$

$\frac{x+\frac{1}{x}}{x-\frac{1}{x}}=\frac{5}{4}$ $\Rightarrow (x+\frac{1}{x})-(x-\frac{1}{x})=\frac{2}{x}=\frac{5-4}{4} \times (x-\frac{1}{x})=\frac{1}{4} \times (x-\frac{1}{x})$ $\Rightarrow x-\frac{1}{x}=\frac{2}{x}/\frac{1}{4}=\frac{8}{x}$ $\Rightarrow x=\frac{8}{x}+\frac{1}{x}=\frac{9}{x}$ $\Rightarrow x^{2}=9$ $\boxed{x=\pm 3}$