Q.Solve: \[ \frac{1}{x} - \frac{1}{3} = \frac{1}{x + 2} - \frac{1}{5} \]

\[ \frac{1}{x} - \frac{1}{3} = \frac{1}{x + 2} - \frac{1}{5} \] Or, \[ \frac{1}{x} - \frac{1}{x + 2} = \frac{1}{3} - \frac{1}{5} \] \[ \frac{x + 2 - x}{x(x + 2)} = \frac{5 - 3}{15} \Rightarrow \frac{2}{x^2 + 2x} = \frac{2}{15} \Rightarrow \frac{1}{x^2 + 2x} = \frac{1}{15} \Rightarrow x^2 + 2x = 15 \Rightarrow x^2 + 2x - 15 = 0 \] Now factorizing: \[ x^2 + 5x - 3x - 15 = 0 \Rightarrow x(x + 5) - 3(x + 5) = 0 \Rightarrow (x + 5)(x - 3) = 0 \] So either: \[ x + 5 = 0 \Rightarrow x = -5 \text{or} \quad x - 3 = 0 \Rightarrow x = 3 \] ∴ The required solutions are \(x = -5\) and \(x = 3\)(Answer).
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