Substitution is blind - 2

Method 1

$\begin{aligned} x + \frac 1x & = 7-4\sqrt 3 + \frac 1{7-4\sqrt 3} \\ & = 7-4\sqrt 3 + \frac {7+4\sqrt 3}{(7-4\sqrt 3)(7+4\sqrt 3)} \\ & = 7-4\sqrt 3 + 7+4\sqrt 3 \\ & = \boxed{14} \end{aligned}$

Method 2

$\begin{aligned} x & = 7 - 4\sqrt 3 & \small \color{#3D99F6} \text{Squaring both sides} \\ x^2 & = 97 - 56\sqrt 3 \\ x^2 & = 14(7-4\sqrt 3) - 1 \\ x^2 & = 14x - 1 & \small \color{#3D99F6} \text{Dividing both sides by }x \\ x & = 14 - \frac 1x & \small \color{#3D99F6} \text{Rearranging} \\ \implies x + \frac 1x & = \boxed{14} \end{aligned}$

Notice that $(7-4\sqrt3)(7+4\sqrt3)=7^{2}-4^{2}3=1$ , therefore $\frac{1}{x}=7+4\sqrt3$ .

$x=7-4\sqrt{3}=\sqrt{49}-\sqrt{48}$

$\implies \dfrac{1}{x} = \sqrt{49}+\sqrt{48}$

$\implies x+ \dfrac{1}{x} =\sqrt{49}-\sqrt{48}+\sqrt{49}+\sqrt{48}=2\sqrt{49} = 2\times 7 =\textcolor{#20A900}{\boxed{14}}$