Solve the following equation using the formula :

$\dfrac{2x}{x - 4} + \dfrac{2x - 5}{x - 3} = 8\dfrac{1}{3}$

Given,

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

Comparing $13x^2 - 118x + 240 = 0$ with ax² + bx + c = 0 we get,

$a = 13, b = -118, c = 240.$

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get,

$\Rightarrow x = \dfrac{-(-118) \pm \sqrt{(-118)^2 - 4(13)(240)}}{2(13)} \\[1em] = \dfrac{118 \pm \sqrt{13924 - 12480}}{26} \\[1em] = \dfrac{118 \pm \sqrt{1444}}{26} \\[1em] = \dfrac{118 \pm 38}{26} \\[1em] = \dfrac{118 + 38}{26} \text{ or } \dfrac{118 - 38}{26} \\[1em] = \dfrac{156}{26} \text{ or } \dfrac{80}{26} \\[1em] = 6 \text{ or } 3\dfrac{1}{13}.$

Hence, x = $6, 3\dfrac{1}{13}.$