Solve the following equation by factorization:

$\dfrac{2x}{x - 4} + \dfrac{2x - 5}{x - 3} = \dfrac{25}{3}$

Given,

$\Rightarrow \dfrac{2x}{x - 4} + \dfrac{2x - 5}{x - 3} = \dfrac{25}{3} \\[1em] \Rightarrow \dfrac{2x(x - 3) + (2x - 5)(x - 4)}{(x - 4)(x - 3)} = \dfrac{25}{3} \\[1em] \Rightarrow \dfrac{2x^2 - 6x + 2x^2 - 8x - 5x + 20}{x^2 - 3x - 4x + 12} = \dfrac{25}{3} \\[1em] \Rightarrow \dfrac{4x^2 - 19x + 20}{x^2 - 7x + 12} = \dfrac{25}{3} \\[1em] \Rightarrow 3(4x^2 - 19x + 20) = 25(x^2 - 7x + 12) \\[1em] \Rightarrow 12x^2 - 57x + 60 = 25x^2 - 175x + 300 \\[1em] \Rightarrow 25x^2 - 175x + 300 - (12x^2 - 57x + 60) = 0 \\[1em] \Rightarrow 25x^2 - 175x + 300 - 12x^2 + 57x - 60 = 0 \\[1em] \Rightarrow 13x^2 - 118x + 240 = 0 \\[1em] \Rightarrow 13x^2 - 78x - 40x + 240 = 0 \\[1em] \Rightarrow 13x(x - 6) - 40(x - 6) = 0 \\[1em] \Rightarrow (13x - 40)(x - 6) = 0 \\[1em] \Rightarrow (13x - 40) \text{ or } (x - 6) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 13x = 40 \text{ or } x = 6 \\[1em] \Rightarrow x = \dfrac{40}{13} \text{ or } x = 6.$

Hence, $x = \Big{6, \dfrac{40}{13}\Big}$.