Solve : $\dfrac{x}{3} + \dfrac{3}{6 - x} = \dfrac{2(6 + x)}{15}$; (x ≠ 6)

Given,

$\Rightarrow \dfrac{x}{3} + \dfrac{3}{6 - x} = \dfrac{2(6 + x)}{15} \\[1em] \Rightarrow \dfrac{x(6 - x) + 9}{3(6 - x)} = \dfrac{12 + 2x}{15} \\[1em] \Rightarrow \dfrac{6x - x^2 + 9}{18 - 3x} = \dfrac{12 + 2x}{15} \\[1em] \Rightarrow 15(6x - x^2 + 9) = (12 + 2x)(18 - 3x) \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 36x + 36x - 6x^2 \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 6x^2 \\[1em] \Rightarrow 15x^2 - 6x^2 - 90x + 216 - 135 = 0 \\[1em] \Rightarrow 9x^2 - 90x + 81 = 0 \\[1em] \Rightarrow 9(x^2 - 10x + 9) = 0 \\[1em] \Rightarrow x^2 - 10x + 9 = 0 \\[1em] \Rightarrow x^2 - 9x - x + 9 = 0 \\[1em] \Rightarrow x(x - 9) - 1(x - 9) = 0 \\[1em] \Rightarrow (x - 1)(x - 9) = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = 9.$

Hence, x = 1 or x = 9.