Solve :

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

Let $\Big(\dfrac{3x + 1}{x + 1}\Big) = a$ …….(i)

$\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) + \Big(\dfrac{x + 1}{3x + 1}\Big) = \dfrac{5}{2} \\[1em] \Rightarrow a + \dfrac{1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{a^2 + 1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow 2(a^2 + 1) = 5a \\[1em] \Rightarrow 2a^2 + 2 - 5a = 0 \\[1em] \Rightarrow 2a^2 - 5a + 2 = 0 \\[1em] \Rightarrow 2a^2 - 4a - a + 2 = 0 \\[1em] \Rightarrow 2a(a - 2) - 1(a - 2) = 0 \\[1em] \Rightarrow (2a - 1)(a - 2) = 0 \\[1em] \Rightarrow 2a - 1 = 0 \text{ or } a - 2 = 0 \\[1em] \Rightarrow 2a = 1 \text{ or } a = 2 \\[1em] \Rightarrow a = \dfrac{1}{2} \text{ or } a = 2.$

Substituting value of a = $\dfrac{1}{2}$ in (i) we get,

$\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) = \dfrac{1}{2} \\[1em] \Rightarrow 2(3x + 1) = x + 1 \\[1em] \Rightarrow 6x + 2 = x + 1 \\[1em] \Rightarrow 6x - x = 1 - 2 \\[1em] \Rightarrow 5x = -1 \\[1em] \Rightarrow x = -\dfrac{1}{5}$

Substituting value of a = 2 in (i) we get,

$\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) = 2 \\[1em] \Rightarrow 3x + 1 = 2(x + 1) \\[1em] \Rightarrow 3x + 1 = 2x + 2 \\[1em] \Rightarrow 3x - 2x = 2 - 1 \\[1em] \Rightarrow x = 1.$

Hence, x = 1, $-\dfrac{1}{5}$.