<s>[INST] Fraction involving Surds Fraction involving Surds. Can anyone please show me the working out? 
$$
 \frac{(\sqrt{6}-1)}{\sqrt{3}} + \frac{(\sqrt{6}+2)}{2\sqrt{3}}
$$
I did this and it was incorrect: 
$$
  2\sqrt{3}(\sqrt{6}-1) + \frac{\sqrt{3}(\sqrt{6} + 2)}{\sqrt{3}\times 2\sqrt{3}}
$$
$$
 2\sqrt{18} - 2\sqrt{3} + \sqrt{18} + \frac{2\sqrt{3}}{\sqrt{3}\times 2\sqrt{3}} 
$$
Thanks!
Edit: This is my answer, which I now understand from my math teacher:
$$
 \frac{(\sqrt{6}-1)}{\sqrt{3}} + \frac{(\sqrt{6}+2)}{2\sqrt{3}}
$$
$$
 \frac{2sqrt{3}(\sqrt{6}-1)}{\sqrt{3}} + \frac{sqrt3(\sqrt{6}+2)}{2\sqrt{3}}
$$
$$
2\sqrt{18} - 2\sqrt{3} + sqrt{18} + 2\sqrt{3}\6
 [/INST] $\begin{array}\\
\frac{(\sqrt{6}-1)}{\sqrt{3}} + \frac{(\sqrt{6}+2)}{2\sqrt{3}}
&=\frac{2(\sqrt{6}-1)}{2\sqrt{3}} + \frac{(\sqrt{6}+2)}{2\sqrt{3}}
\qquad\text{put the fractions over a common denominator}\\
&=\frac{2(\sqrt{6}-1)+(\sqrt{6}+2)}{2\sqrt{3}}
\qquad\text{add the numerators}\\
&=\frac{2\sqrt{6}-2+\sqrt{6}+2}{2\sqrt{3}}
\qquad\text{use the distributive law to get rid of parentheses}\\
&=\frac{3\sqrt{6}}{2\sqrt{3}}
\qquad\text{simple algebra}\\
&=\frac{3\sqrt{2\cdot 3}}{2\sqrt{3}}
\qquad\text{setup to get rid of }\sqrt{3}\\
&=\frac{3\sqrt{2}\sqrt{ 3}}{2\sqrt{3}}
\qquad\text{use }\sqrt{xy} = \sqrt{x}\sqrt{y}\\
&=\frac{3\sqrt{2}}{2}
\qquad\text{and its gone}\\
\end{array}
$
</s>