Solve: \(\frac{1}{x−3}−\frac{1}{x+5}=\frac{1}{6}\)

Q.Solve: \(\frac{1}{x−3}−\frac{1}{x+5}=\frac{1}{6}\)

\(\frac{1}{x-3} - \frac{1}{x+5} = \frac{1}{6}\) ⇒ \(\frac{(x+5) - (x-3)}{(x-3)(x+5)} = \frac{1}{6}\) ⇒ \(\frac{x + 5 - x + 3}{x^2 + 5x - 3x - 15} = \frac{1}{6}\) ⇒ \(\frac{8}{x^2 + 2x - 15} = \frac{1}{6}\) ⇒ \(x^2 + 2x - 15 = 48\) ⇒ \(x^2 + 2x - 63 = 0\) ⇒ \(x^2 + 9x - 7x - 63 = 0\) ⇒ \(x(x + 9) - 7(x + 9) = 0\) ⇒ \((x + 9)(x - 7) = 0\) That means either \((x + 9) = 0 ⇒ x = -9\) or \((x - 7) = 0 ⇒ x = 7\) ∴ The solutions to the quadratic equation \(\frac{1}{x-3} - \frac{1}{x+5} = \frac{1}{6}\) are \(x = -9\) and \(x = 7\).