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December 8th, 2014, 08:02 AM   #1
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algebra to invert fractional equation?

so.. if a/b=c/d then b/a=d/c because you can use each side of the equation do divide one by to invert. how do i represent this as an explicit step in place of just flipping the fractions? would
a/b=c/d
1/(a/b)=1/(c/d)
b/a=d/c
be legit algebra? this seems wrong somehow.
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December 8th, 2014, 08:18 AM   #2
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Yep. Doing "1 over" is a valid mathematics operation on both sides of an equation provided you don't have any zeroes.

It's okay to do
$\displaystyle x = -y$
$\displaystyle \frac{1}{x} = -\frac{1}{y}$

(if x and y are non-zero), but not

$\displaystyle x+y = 0$
$\displaystyle \frac{1}{x+y} = \frac{1}{0}$

because 1 divided by 0 is undefined. Also, you need to pay attention to the fact that in the first case presented above, x and y must be non-zero. Otherwise you're doing the same thing as the second case.

As for your specific example, another way of showing that they are the same is to multiply both sides by $\displaystyle \frac{bd}{ac}$

$\displaystyle \frac{a}{b} = \frac{c}{d}$
$\displaystyle \frac{a}{b} \times \frac{bd}{ac}= \frac{c}{d} \times \frac{bd}{ac}$
$\displaystyle \frac{\cancel{a}}{\cancel{b}} \times \frac{\cancel{b}d}{\cancel{a}c}= \frac{\cancel{c}}{\cancel{d}} \times \frac{b\cancel{d}}{a\cancel{c}}$
$\displaystyle \frac{d}{c} = \frac{b}{a}$

which is okay if a, b, c and d are non-zero.
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December 8th, 2014, 10:32 AM   #3
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Another way of simplifying 1/(a/b) with a*b is nonzero, is as follows:

$\displaystyle \frac{1}{\frac{a}{b}} = \frac{1}{\frac{a}{b}} \cdot 1 = \frac{1}{\frac{a}{b}} \cdot \frac{b}{b} = \frac{b}{\frac{a}{b} \cdot b} = \frac{b}{\frac{ab}{b}} = \frac{b}{ \frac{b}{b} \cdot a} = \frac{b}{1 \cdot a} = \frac{b}{a}$

where you might like to leave out some steps.
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