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October 7th, 2014, 06:51 PM   #1
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Complex fractions

I am asked to simplify $\displaystyle \frac{\frac{1}{a}+1}{\frac{1}{2a}-1}$ the answer is listed as $\displaystyle \frac{a}{1-a}$ I am not sure how they got the answer.


The basic model that I used on previous problems goes like this $\displaystyle \frac{\frac{1}{a}+1}{\frac{1}{2a}-1}=\frac{\frac{1+a}{a}}{\frac{1-2a}{2a}}=\frac{1+a}{a}\div\frac{1-2a}{2a}=\frac{1+a}{a}\cdot\frac{2a}{1-2a}=\frac{1+a}{a\div a}\cdot\frac{2a\div a}{1-2a}=\frac{1+a}{1}\cdot\frac{2}{1-2a}=\frac{2+2a}{1-2a}=\frac{2(1+a)}{1-2a}$


From here, I'm stuck. Help plx

Last edited by skipjack; November 22nd, 2014 at 04:30 AM.
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October 7th, 2014, 07:12 PM   #2
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Hello, Opposite!

Either you copied the problem incorrectly
or the original problem had a typo.

I believe the problem is: $\;\boxed{\text{Simplify: }\:\dfrac{\dfrac{1}{a} + 1}{\dfrac{1}{a^2} - 1}}$

Then we have: $\:\dfrac{\dfrac{1+a}{a}}{\dfrac{1-a^2}{a^2}} \;=\;\dfrac{1+a}{a} \div \dfrac{1-a^2}{a^2} \;=\;\dfrac{1+a}{a} \cdot\dfrac{a^2}{1-a^2}$

$\qquad =\;\dfrac{1+a}{a}\cdot\dfrac{a^2}{(1-a)(1+a)} \;=\;\dfrac{a}{1-a} $

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October 7th, 2014, 11:07 PM   #3
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Oh that makes way more sense.


The actual problem has listed for the bottom denominator a2, I'm now sure they meant a^2

Last edited by Opposite; October 7th, 2014 at 11:15 PM.
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May 26th, 2015, 04:44 PM   #4
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So now I am curious as to whether or not I did my posted problem correctly. If my posted equations as a whole new problem to work on for fun, did I do it correctly?
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May 26th, 2015, 05:07 PM   #5
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Quote:
Originally Posted by Opposite View Post
So now I am curious as to whether or not I did my posted problem correctly. If my posted equations as a whole new problem to work on for fun, did I do it correctly?
looks ok ...

$\displaystyle \frac{\frac{1}{a}+1}{\frac{1}{2a}-1} \cdot \frac{2a}{2a} = \frac{2+2a}{1-2a} = \frac{2(1+a)}{1-2a}$
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