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$\begingroup$

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

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 difference2011-06-14

2 Answers 2

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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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    @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.2011-06-14
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$\frac{{5p + 10}}{{p^2 - 4}}\bigg/\frac{{3p - 6}}{{(p - 2)^2 }} = \frac{{5p + 10}}{{p^2 - 4}}\frac{{(p - 2)^2 }}{{3p - 6}} = \frac{{5(p + 2)}}{{(p + 2)(p - 2)}}\frac{{(p - 2)(p - 2)}}{{3(p - 2)}} = \frac{5}{3}$

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    To cancel additive terms: $(a+b) - (a+c) = b-c$2011-06-14