Use componendo and dividendo to find the value of x, when :

Given,

$\frac{x^2 + 3x}{3x^2 + 1} = \frac{14}{13}$

Applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{x^2 + 3x + 3x^2 + 1}{x^2 + 3x - (3x^2 + 1)} = \frac{14 + 13}{14 - 13} \\[1em] \Rightarrow \frac{x^2 + 3x + 3x^2 + 1}{x^2 + 3x - 3x^2 - 1}= \frac{27}{1} \\[1em] \Rightarrow \frac{(x + 1)^3}{(x - 1)^3} = \frac{3^3}{1^3} \\[1em] \Rightarrow \Big(\frac{x + 1}{x - 1}\Big)^3 = \Big(\frac{3}{1}\Big)^3 \\[1em] \Rightarrow \frac{x + 1}{x - 1} = 3 \\[1em] \Rightarrow x + 1 = 3(x - 1) \\[1em] \Rightarrow x + 1 = 3x - 3 \\[1em] \Rightarrow 3x - x = 1 + 3 \\[1em] \Rightarrow 2x = 4 \\[1em] \Rightarrow x = \frac{4}{2} = 2.\end{aligned}$$

Hence, x = 2.