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$$\dfrac{\qquad\dfrac{5p+10}{p^2-4}\qquad}{\dfrac{3p-6}{(p-2)^2}}$$

Im all confused about this question. Can someone go through it step by step. Can you please list when I can cancel numbers. Thanks.

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you are correct in that you can only cancel whole factors of the numerator and denominator, not parts of a sum or difference –  ricky Jun 14 '11 at 18:41
    
The big trick is to realize that $5p+10 = 5(p+2)$, $p^2-4 = (p-2)(p+2)$, and $3p-6 = 3(p-2)$. Thus, you can cancel out a factor of $p+2$ from the top two terms, and $p-2$ from the bottom two terms, and you get: $$\dfrac{\qquad\dfrac{5}{p-2}\qquad}{\dfrac{3}{p-2}}$$ which obviously can be simplified some more... –  Thomas Andrews Jun 14 '11 at 18:43

2 Answers 2

up vote 10 down vote accepted

Remember that $$\frac{\quad\frac{a}{b}\quad}{\frac{c}{d}} = \frac{a}{b}\div \frac{c}{d} = \frac{ad}{bc}.$$ If you think of the fraction as a "sandwich", with $a$ and $d$ the bread, $b$ and $c$ the ham-and-cheese, then the bread goes on top and the ham and cheese go on the bottom.

Alternatively, dividing by $x$ is the same as multiplying by $\frac{1}{x}$, and $$\frac{1}{\quad\frac{c}{d}\quad} = \frac{d}{c}$$ so $$\frac{\quad\frac{a}{b}\quad}{\frac{c}{d}} = \frac{a}{b}\times\frac{1}{\quad\frac{c}{d}\quad} = \frac{a}{b}\times \frac{d}{c} = \frac{ad}{bc}.$$

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Thanks guys. Im still a bit confused with cancelling. Say I can only cancel if there are only multiplications attached to each other? So I couldent cancel the a in : a + b/ a + c ? –  John Jun 14 '11 at 18:33
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@John: If you mean $(a+b)/(a+c)$, you are indeed correct: the $a$s **do not** cancel. Just do a simple example: take $(2+1)/(2+2)$. I that the same thing as $1/2$? No. –  Arturo Magidin Jun 14 '11 at 18:34

$$\frac{{5p + 10}}{{p^2 - 4}}\bigg/\frac{{3p - 6}}{{(p - 2)^2 }} = \frac{{5p + 10}}{{p^2 - 4}}\frac{{(p - 2)^2 }}{{3p - 6}} = \frac{{5(p + 2)}}{{(p + 2)(p - 2)}}\frac{{(p - 2)(p - 2)}}{{3(p - 2)}} = \frac{5}{3}$$

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Note that this is defined if $p \notin \{-2,2\}$. –  Shai Covo Jun 14 '11 at 18:22
    
Thanks guys. Im still a bit confused with cancelling. Say I can only cancel if there are only multiplications attached to each other? So I couldent cancel the a in : a + b/ a + c ? –  John Jun 14 '11 at 18:33
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@John, indeed, you cannot cancel the $a$ in $(a+b)/(a+c)$. –  Shai Covo Jun 14 '11 at 18:37
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And, on the other hand, $(ab)/(ac)=b/c$. –  Shai Covo Jun 14 '11 at 18:38
    
You can "cancel" (by division) factors, not terms –  The Chaz 2.0 Jun 14 '11 at 18:53

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