Solve the equation: $\frac{1}{x + 2} + \frac{1}{x + 5} = \frac{1}{x + 4} + \frac{1}{x + 3}$

Created: 1 year ago | Updated: 1 year ago

Updated: 1 year ago

$\frac{1}{x + 2} + \frac{1}{x + 5} = \frac{1}{x + 4} + \frac{1}{x + 3} \Rightarrow \frac{1}{x + 2} - \frac{1}{x + 4} = \frac{1}{x + 3} - \frac{1}{x + 5} \Rightarrow \frac{x + 4 - x - 2}{x^{2} + 4 x + 2 x + 8} = \frac{x + 5 - x - 3}{x^{2} + 5 x + 3 x + 15} \Rightarrow \frac{2}{x^{2} + 6 x + 8} = \frac{2}{x^{2} + 8 x + 15} \Rightarrow \frac{1}{x^{2} + 6 x + 8} = \frac{1}{x^{2} + 8 x + 15} \Rightarrow x^{2} + 8 x + 15 = x^{2} + 6 x + 8 \Rightarrow x^{2} + 8 x - x^{2} - 6 x = 8 - 15 \Rightarrow 2 x = - 7 \Rightarrow x = \frac{- 7}{2} ∴ x = \frac{- 7}{2}$