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Is there a way to break the left hand side expression such that it takes the the right hand side form?

$(a+b)/(c+d)=a/c+b/d+k$

Where $k$ is some expression.

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Yes, and that expression would be $(a+b)/(c+d) - a/c - b/d$. Are you looking for something less stupid or more specific? –  Patrick Da Silva Jul 3 '13 at 3:17
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1 Answer

Solve for $k$, as Patrick indicated: $$ \begin{align} k&=\frac{a+b}{c+d}-\frac{a}{c}-\frac{b}{d}\\ &=\frac{cd(a+b)-ad(c+d)-bc(c+d)}{cd(c+d)}\\ &=\frac{acd+bcd-acd-ad^2-bc^2-bcd}{cd(c+d)}\\ &=\frac{-ad^2-bc^2}{cd(c+d)} \end{align} $$ In the words of lots of movie cops over the years, "Move along, folks, there's nothing to see here."

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@jessica: one additional thing to note is that you need $c$ and $d$ non-zero. –  James Jul 3 '13 at 13:21
    
@James: and also $c\ne-d$, else the original expression is undefined. –  Rick Decker Jul 3 '13 at 13:42
    
Eh.... I was wondering if there was an elegant way to divide numbers. For example take 32/65. I was trying to simplify it to such that I take the first digit of both numbers and divide. Getting 30/60=3/6=.5. Now knowing this .5 would be my base number for which the 2&6 on the end of both numbers would be used to balance around the approximated .5. The 2&6 are my remainder term which would either push me to the left or right of .5. –  jessica Jul 3 '13 at 22:24
    
If K is such an ugly expression I guess there really is no elegant technique to dividing numbers with approximation mentally in your head. –  jessica Jul 3 '13 at 22:27
    
Here's an old chestnut related to your problem. Take $64/16$, cancel the 6s, and you get $4/1$ which happens to be the right answer. Too bad it doesn't work in general. –  Rick Decker Jul 4 '13 at 14:05
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