Answer
$\frac{8x+1}{4x-1}; x \ne 0,\frac{1}{4}$
Work Step by Step
$\frac{8+\frac{1}{x}}{4-\frac{1}{x}} = \frac{\frac{8x+1}{x}}{\frac{4x-1}{x}} ; x\ne 0, \frac{1}{4}$
$= \frac{8x+1}{x} \times \frac{x}{4x-1};x\ne 0, \frac{1}{4}$
$= \frac{8x+1}{4x-1} ; x\ne 0, \frac{1}{4}$
