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Why is the following not possible?

$$\frac{2x-1}{2x}\neq4x-2$$

And the following method not correct?

$$\bigg(\frac{2x-1}{2x} + \frac{1}{1}\bigg)-1\equiv\frac{2x-1}{2x}$$

Cross multiplying:

$$\big(1[2x-1]+1[2x]\big)-1$$

With the result:

$$2x-1+2x-1=4x-2$$

My question, essentially is, is the following possible:

$$\frac{a}{b}+\frac{c}{d}\equiv ad+bc?$$

I know it is true for the following:

$$\frac{a}{b}=\frac{c}{d}\equiv ad=bc$$

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What? Of course $\frac{2x-1}{2x}\neq 4x-2$ in general, just consider $x=1$. I'm confused why you say it is "not possible". –  Alex Becker Sep 19 '12 at 19:36
    
Perhaps my phraseology is a little misleading. I am attempt to question the method by which I got to the result. –  falcontoast Sep 19 '12 at 20:05

3 Answers 3

No -- the correct rule for adding fractions is $$ \frac ab + \frac cd = \frac{ad+bc}{bd} $$ You're missing the $bd$ denominator.

The rule for subtraction is similar $$ \frac ab - \frac cd = \frac{ad-bc}{bd} $$ so $\frac ab=\frac bc$ if and only if $\frac{ad-bc}{bd}=0$. But a quotient is zero exactly when its numerator is, so $\frac{ad-bc}{bc}=0$ if and only if $ad-bc=0$ or in other words $ad=bc$.

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The mistake is that you don't do anything to the -1. You have $\frac{(1[2x-1]+1[2x])}{2x}-1$.

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In general, $$\frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{bd} \neq ad+bc$$

with $b, d\neq 0$.

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