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
$\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} $