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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:

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

  • 0
    Perhaps my phraseology is a little misleading. I am attempt to question the method by which I got to the result.2012-09-19

3 Answers 3

2

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 $ad=bc$.

1

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

0

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

with $b, d\neq 0$.