## Algebra and Trigonometry 10th Edition

Solve : $\frac{1}{x-3} + \frac{1}{x+3} = \frac{10}{x^{2}-9}$ $\frac{x+3}{x^{2}-9} + \frac{x-3}{x^{2}-9} = \frac{10}{x^{2}-9}$ as (x+3)(x-3) = ${x^{2}-9}$ x+3 + x-3 = 10 2x = 10 x = 5