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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

Enzo
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  • Apologies, I wrote the question incorrectly, it is actually: (√6-1)/√3 + (√6+2)/(2√3) – Enzo Feb 20 '16 at 04:58
  • I've edited the text to show you how to write the equations in MathJax. Please edit the expressions in the question using that as a guide. – Biswajit Banerjee Feb 20 '16 at 05:08
  • Will my question not be answered unless I use MathJax? – Enzo Feb 20 '16 at 05:13
  • I wouldn't say that, @Enzo , it's just this community's standard for writing mathematics. And now you have an example to learn from. – pjs36 Feb 20 '16 at 05:29
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    Thanks, I will definitely try to learn MathJax in future questions – Enzo Feb 20 '16 at 05:31
  • Thanks for the edit Biswajit – Enzo Feb 20 '16 at 05:32
  • We want to bring the two expressions to a common denominator. Convenient is $2\sqrt{3}$. So the first term is $\frac{2(\sqrt{6}-1)}{2\sqrt{3}}$. Now we can add the tops freely. There will be a great deal of simplification. – André Nicolas Feb 20 '16 at 05:37
  • @Enzo Actually, this site(http://functionspace.com/equationeditor) might prove helpful for Latex formatting. – S.C.B. Feb 20 '16 at 05:44

1 Answers1

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$\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} $

marty cohen
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