Solve the following equation:

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

Given, $\dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = 3\dfrac{1}{3}$.

On cross multiplication,

$\Rightarrow \dfrac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \dfrac{10}{3} \\[1em] \Rightarrow 3[x^2 - 4x - x + 4 + x^2 - 2x - 3x + 6] = 10(x^2 - 4x - 2x + 8) \\[1em] \Rightarrow 3(2x^2 - 10x + 10) = 10(x^2 - 6x + 8) \\[1em] \Rightarrow 6x^2 - 30x + 30 = 10x^2 - 60x + 80 \\[1em] \Rightarrow 10x^2 - 6x^2 - 60x + 30x + 80 - 30 = 0 \\[1em] \Rightarrow 4x^2 - 30x + 50 = 0 \\[1em] \Rightarrow 4x^2 - 20x - 10x + 50 = 0 \\[1em] \Rightarrow 4x(x - 5) - 10(x - 5) = 0 \\[1em] \Rightarrow (4x - 10)(x - 5) = 0 \\[1em] \Rightarrow 4x - 10 = 0 \text{ or } x - 5 = 0 \\[1em] \Rightarrow 4x = 10 \text{ or } x = 5 \\[1em] \Rightarrow x = \dfrac{10}{4} \text{ or } x = 5 \\[1em] \Rightarrow x = \dfrac{5}{2} \text{ or } x = 5.$

Hence, roots of the given equations are $\dfrac{5}{2}$, 5.