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October 7th, 2014, 06:51 PM  #1 
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9  Complex fractions
I am asked to simplify $\displaystyle \frac{\frac{1}{a}+1}{\frac{1}{2a}1}$ the answer is listed as $\displaystyle \frac{a}{1a}$ I am not sure how they got the answer. The basic model that I used on previous problems goes like this $\displaystyle \frac{\frac{1}{a}+1}{\frac{1}{2a}1}=\frac{\frac{1+a}{a}}{\frac{12a}{2a}}=\frac{1+a}{a}\div\frac{12a}{2a}=\frac{1+a}{a}\cdot\frac{2a}{12a}=\frac{1+a}{a\div a}\cdot\frac{2a\div a}{12a}=\frac{1+a}{1}\cdot\frac{2}{12a}=\frac{2+2a}{12a}=\frac{2(1+a)}{12a}$ From here, I'm stuck. Help plx Last edited by skipjack; November 22nd, 2014 at 04:30 AM. 
October 7th, 2014, 07:12 PM  #2 
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  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: $\:\dfrac{\dfrac{1+a}{a}}{\dfrac{1a^2}{a^2}} \;=\;\dfrac{1+a}{a} \div \dfrac{1a^2}{a^2} \;=\;\dfrac{1+a}{a} \cdot\dfrac{a^2}{1a^2}$ $\qquad =\;\dfrac{1+a}{a}\cdot\dfrac{a^2}{(1a)(1+a)} \;=\;\dfrac{a}{1a} $ 
October 7th, 2014, 11:07 PM  #3 
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9 
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. 
May 26th, 2015, 04:44 PM  #4 
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9 
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?

May 26th, 2015, 05:07 PM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,773 Thanks: 1427  Quote:
$\displaystyle \frac{\frac{1}{a}+1}{\frac{1}{2a}1} \cdot \frac{2a}{2a} = \frac{2+2a}{12a} = \frac{2(1+a)}{12a}$  

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