Solve the following equation by factorisation:

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

Given,

$\Rightarrow \dfrac{4}{x + 2} - \dfrac{1}{x + 3} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow \dfrac{4(x + 3) - (x + 2)}{(x + 2)(x + 3)} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow \dfrac{4x + 12 - x - 2}{x^2 + 3x + 2x + 6} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow \dfrac{3x + 10}{x^2 + 5x + 6} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow (3x + 10)(2x + 1) = 4(x^2 + 5x + 6) \\[1em] \Rightarrow 6x^2 + 3x + 20x + 10 = 4x^2 + 20x + 24 \\[1em] \Rightarrow 6x^2 - 4x^2 + 23x - 20x + 10 -24 = 0 \\[1em] \Rightarrow 2x^2 + 3x - 14 = 0 \\[1em] \Rightarrow 2x^2 + 7x - 4x - 14 = 0 \\[1em] \Rightarrow x(2x + 7) - 2(2x + 7) = 0 \\[1em] \Rightarrow (x - 2)(2x + 7) = 0 \\[1em] \Rightarrow (x - 2) = 0 \text{ or } 2x + 7 = 0 \\[1em] \Rightarrow x = 2 \text{ or } 2x = -7 \\[1em] \Rightarrow x = 2 \text{ or } x = -\dfrac{7}{2}.$

Hence, value of x = 2 or -$\dfrac{7}{2}$.