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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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That's not the way you manipulate fractions! – Andrea Mori Mar 28 '12 at 12:34
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. – Kirthi Raman Mar 28 '12 at 12:50
up vote 1 down vote accepted

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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OK, I get it, thank you. – bot_bot Mar 28 '12 at 12:35
Even nicer: $\frac{3-1}9=\frac{1-1}3=0$ :) – Andrea Mori Mar 28 '12 at 12:36

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!

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