$${\text{If }}\,\sqrt {1 + \frac{x}{9}} = \frac{{13}}{3}{\text{,}}$$    then the value of x is?

$$\eqalign{ & \sqrt {1 + \frac{x}{9}} = \frac{{13}}{3} \cr & {\text{By option }} \cr & {\text{Put }}x = 160 \cr & \sqrt {1 + \frac{{160}}{9}} = \frac{{13}}{3} \cr & \Rightarrow \sqrt {\frac{{169}}{9}} = \frac{{13}}{3} \cr & \Rightarrow \frac{{13}}{3} = \frac{{13}}{3} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Squaring both sides}} \cr & {\left( {\sqrt {1 + \frac{{x}}{9}} } \right)^2} = {\left( {\frac{{13}}{3}} \right)^2} \cr & \Rightarrow 1 + \frac{x}{9} = \frac{{169}}{9} \cr & \Rightarrow \frac{{9 + x}}{9} = \frac{{169}}{9} \cr & \Rightarrow 9 + x = 169 \cr & \Rightarrow \boxed{x = 160} \cr} $$