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I have the following sum listed in Algebra Demystified:

$\frac{71}{84} - \frac{13}{30x} = \frac{71}{84}\cdot \frac{5x}{5x} - \frac{13}{30x} \cdot \frac{14}{14} = \frac{355x}{420x} - \frac{182}{420x} = \frac{355x-182}{420x}$

However, I previously understood it as follows:

$\frac{a}{b} - \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{d} - \frac{c}{d} \cdot \frac{b}{b}= \frac{ad-bc}{bd}$

In which case, why would

$\frac{71}{84}\cdot \frac{5x}{5x}$

shouldn't it be $\frac{14}{14}$?

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    The are using a common multiple of the denominators that is smaller than the product, obtained by cancelling a common factor $\,g,\,$ i.e. in your abstract example cancel $\,g\,$ from the scalings, i.e.scale by $(d/g)/(d/g)$ and $(b/g)/(b/g).\ $ In the example $\,g= 6 = \gcd(84,13)\ $ You can use *any* common multiple of the denominators (e.g. their least or their product, etc). The goal is to get *any* common denominator, to make the subtraction simple. – 2017-01-02
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    This makes perfect sense, thank you. I've just completed some sums with this in mind and I really understand what's happening now. Thanks for providing the abstract example too. – 2017-01-02
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    We must have $x\neq 0$. $\frac{5x}{5x} = \frac{14}{14} = 1$ – 2017-01-02

1 Answers 1

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No. Because you need to match the denominators. That's it.

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    And because I have the LCD I can multiply the denominator 84 by $\frac{5}{5}$ as I know that 30x * $\frac{14}{14}$ equals the same? – 2017-01-02
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    Yes. Thats true. – 2017-01-02