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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 $\frac{14}{14}$?

Hemmed
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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. – Bill Dubuque Jan 02 '17 at 15:23
  • 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. – Hemmed Jan 02 '17 at 15:26
  • We must have $x\neq 0$. $\frac{5x}{5x} = \frac{14}{14} = 1$ – amWhy Jan 02 '17 at 17:21

1 Answers1

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

SchrodingersCat
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