Solve the following equation by factorization:

$\dfrac{5}{(2x + 1)} + \dfrac{6}{(x + 1)} = 3$

Given,

$\Rightarrow \dfrac{5}{(2x + 1)} + \dfrac{6}{(x + 1)} = 3 \\[1em] \Rightarrow \dfrac{5(x + 1) + 6(2x + 1)}{(2x + 1)(x + 1)} = 3 \\[1em] \Rightarrow \dfrac{5x + 5 + 12x + 6}{(2x + 1)(x + 1)} = 3 \\[1em] \Rightarrow \dfrac{17x + 11}{(2x^2 + 2x + x + 1)} = 3 \\[1em] \Rightarrow \dfrac{17x + 11}{(2x^2 + 3x + 1)} = 3 \\[1em] \Rightarrow 17x + 11 = 3(2x^2 + 3x + 1) \\[1em] \Rightarrow 17x + 11 = 6x^2 + 9x + 3 \\[1em] \Rightarrow 6x^2 + 9x - 17x + 3 - 11 = 0 \\[1em] \Rightarrow 6x^2 - 8x - 8 = 0 \\[1em] \Rightarrow 2(3x^2 - 4x - 4) = 0 \\[1em] \Rightarrow 3x^2 - 4x - 4 = 0 \\[1em] \Rightarrow 3x^2 - 6x + 2x - 4 = 0 \\[1em] \Rightarrow 3x(x - 2) + 2(x - 2) = 0 \\[1em] \Rightarrow (3x + 2)(x - 2) = 0 \\[1em] \Rightarrow (3x + 2) \text{ or } (x - 2) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 3x = -2 \text{ or } x = 2 \\[1em] \Rightarrow x = \dfrac{-2}{3} \text{ or } x = 2.$

Hence, $x = \Big{2, \dfrac{-2}{3}\Big}$.