Solve the following equation by factorization:

$x + \dfrac{1}{x} = 3\dfrac{1}{3}$, x ≠ 0

Given,

$\Rightarrow x + \dfrac{1}{x} = 3\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{9 + 1}{3} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{10}{3} \\[1em] \Rightarrow 3 \times (x^2 + 1) = 10x \\[1em] \Rightarrow 3x^2 + 3 = 10x \\[1em] \Rightarrow 3x^2 -10x + 3 = 0 \\[1em] \Rightarrow 3x^2 -9x -x + 3 = 0 \\[1em] \Rightarrow 3x(x - 3) - 1(x - 3) = 0 \\[1em] \Rightarrow (x - 3)(3x - 1) = 0.$

⇒ x - 3 = 0 or 3x - 1 = 0      [Using Zero-product rule]

⇒ x = 3 or 3x = 1

⇒ x = 3 or x = $\dfrac{1}{3}$

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