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I have the following question in my text book:

Express as a single fraction

$ \frac{3}{x-4} - \frac{2}{(x-4)^2} $

The answer the book gives is this:

$ \frac{3x-14}{(x-4)^2} $

I understand how they get to this through these steps:

$ \frac{3(x-4)-2}{(x-4)^2} = \frac{3x-12-2}{(x-4)^2} $

My question is why can you not cancel the $ (x-4) $ instead like this:

$ \frac{3(x-4)-2}{(x-4)^2} = \frac{3-2}{(x-4)} = \frac{1}{(x-4)} $

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    Can you can cancel $2$ in $\frac{3}{2}-\frac{2}{2^2}$ and write it as $\frac{3-2}{2}=\frac{1}{2}$. Think about it and see the answers and it will be clear.2012-03-28

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in your last line you can't cancel the $(x-4)$! You have $ \frac{3(x-4) - 2}{(x-4)^2} = \frac{3 - \frac 2{x-4}}{x-4} \ne \frac{1}{x-4}. $ Just as in, say $\frac{3+4}9$, you can't cancel the $3$ to obtain $\frac{1+4}{3} = \frac 53$.

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    Even nicer: $\$f$rac{3-$1$}9=\frac{$1$-1}3=0$ :)2012-03-28
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It is never allowed to cancel across a plus or minus in a fractional expression. If it were possible, strange things would happen. Consider your expression when $x = 6$:

$ \frac{3(x-4) - 2}{(x-4)^2} = \frac{3(6-4)-2}{(6-4)^2} = \frac{3(2) - 2}{2^2} = \frac{4}{4} = 1. $ But if you tried canceling the $x-4$ first: $\frac{3(x-4) - 2}{(x-4)^2} \stackrel{?}{=} \frac{3-2}{(x-4)} = \frac{1}{x-4} = \frac{1}{6-4} =\frac{1}{2}. $ Not the same, right?

Hope this helps!