Question

$1) \frac{1}{3x-y} - \frac{1}{3x+y}$

Answer 1

$\frac{1}{3x-y}- \frac{1}{3x+y}= \frac{1(3x+y)}{9x^{2}-y^{2}}- \frac{1(3x-y)}{9x^{2}+y^{2}}= \frac{3x+y-3x+y}{9x^{2}-y^{2}} = \frac{2y}{9x^{2}-y^{2}}$