Solve $\dfrac{1}{x - 3} + \dfrac{1}{x - 4} = \dfrac{1}{x - 2} + \dfrac{1}{x - 5}$:

Pairing the terms as $\frac{1}{x-3}-\frac{1}{x-2}=\frac{1}{x-5}-\frac{1}{x-4}$ makes each side a single fraction with equal numerators, forcing $(x-3)(x-2)=(x-5)(x-4)$. Expanding and cancelling the $x^2$ terms gives $-5x+6=-9x+20$, so $x=\dfrac{7}{2}$.