Simplifying $\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}$ I realize that is a basic math problem, but I am still having problems with it. 
The expression $$\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}$$ equals one of the following:


*

*$2\sqrt{3}-1$

*$\dfrac{6}{11}$

*$4\sqrt{3}$

*$12$


How do I simplify this?

Hint:
First Rationalize the denominator of first term by multiplying numerator and denominator with denominator's conjugate and then use 
$$(a+b)(a-b)=a^2-b^2$$

$$\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}=\frac{\sqrt{3}}{2\sqrt{3}+1}\cdot\frac{2\sqrt{3}-1}{2\sqrt{3}-1} + \frac{\sqrt{3}}{11}=\frac{6-\sqrt3}{(2\sqrt3)^2-1}+ \frac{\sqrt{3}}{11}=\frac{6-\sqrt3}{11}+ \frac{\sqrt{3}}{11}$$

