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I am familiar with the PEMDAS rule, but how would we simplify something as complex as this:

$$\frac{\frac{3}{4}+2\frac{\sqrt{6}}{5}}{1-\frac{3}{4}(2\frac{\sqrt{6}}{5})}$$

Can we work on the top side first and then move to the bottom? I tried doing that but my result did not match the correct answer that was provided by my book, so either I made a mistake or what I did was outright wrong.

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closed as unclear what you're asking by Rory Daulton, gebruiker, kingW3, Arnaldo, C. Falcon May 21 '17 at 0:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Does your question have anything to do with the partial solution shown at your link, or are you using only the question itself? And what exactly did you try--your explanation is not clear. $\endgroup$ – Rory Daulton May 20 '17 at 10:42
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Yes, you can work the top and then the bottom. But it's probably less work, in these "four-story fraction" problems, to clear the inside denominators. In this case, $20$ is a common denominator of all the little fractions, so multiply top and bottom by $20$ to get

$$\frac{15+8\sqrt{6}}{20 - 6\sqrt{6}}.$$

Often "simplify" includes rationalizing the denominator. If you multiply top and bottom by $20 + 6\sqrt{6}$ you get

$$\frac{15+8\sqrt{6}}{20 - 6\sqrt{6}}\frac{20+6\sqrt{6}}{20+6\sqrt{6}} = \frac{ 300 +160\sqrt{6}+90\sqrt{6} +48\cdot 6}{400-36\cdot 6}$$

$$= \frac{588+250\sqrt{6}}{184} = \frac{294+125\sqrt{6}}{92}. $$

Where, at last step, we divided top and bottom by $2$. That's about as simple as this one gets.

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  • $\begingroup$ You wrote $150\sqrt{6}$ when I think you meant $160\sqrt{6}$. $\endgroup$ – John Wayland Bales May 20 '17 at 18:17
  • $\begingroup$ So the correct answer is $\dfrac{294+125\sqrt{6}}{92}$. $\endgroup$ – John Wayland Bales May 22 '17 at 6:01
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we have $$1-\frac{3}{4}\cdot 2\frac{\sqrt{6}}{5}=1-{3}\frac{\sqrt{6}}{10}=\frac{10-3\sqrt{6}}{10}$$ and $$\frac{3}{4}+\frac{2}{5}\sqrt{6}=\frac{15+8\sqrt{6}}{20}$$

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