Is this allowed?

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• Jul 28th 2013, 07:38 AM
uperkurk
Is this allowed?
$\displaystyle \frac{x^2-3x+2}{4x}\times\frac{12x^2}{x^2-2x}\times\frac{x}{x-1}$

$\displaystyle \frac{(x-2)(x-1)}{4x}\times\frac{12x^2}{(x-1)(x+2)}\times\frac{x}{x-1}$

I know there are many things I can do here but am I allowed to cancel out all of the $\displaystyle x-1$ or can I only cancel out the first and the second, or the first and the third occurrence? Because after cancelling and reducing everything I got this:

$\displaystyle \frac{3x(x-2)}{x+2}$

But wolfram alpha tells me the answer is $\displaystyle 3x$ so what did I do wrong or miss?

Thanks
• Jul 28th 2013, 07:45 AM
ChessTal
Re: Is this allowed?
Quote:

Originally Posted by uperkurk
$\displaystyle \frac{x^2-3x+2}{4x}\times\frac{12x^2}{x^2-2x}\times\frac{x}{x-1}$

$\displaystyle \frac{(x-2)(x-1)}{4x}\times\frac{12x^2}{(x-1)(x+2)}\times\frac{x}{x-1}$

I know there are many things I can do here but am I allowed to cancel out all of the $\displaystyle x-1$ or can I only cancel out the first and the second, or the first and the third occurrence? Because after cancelling and reducing everything I got this:

$\displaystyle \frac{3x(x-2)}{x+2}$

But wolfram alpha tells me the answer is $\displaystyle 3x$ so what did I do wrong or miss?

Thanks

You've made one mistake in factoring the denominator of the 2nd fraction....
• Jul 28th 2013, 07:46 AM
BobP
Re: Is this allowed?
Middle fraction,

$\displaystyle x^{2} - 2x$ is not equal to $\displaystyle (x-1)(x+2).$
• Jul 28th 2013, 07:51 AM
uperkurk
Re: Is this allowed?
Thanks I see my mistake.
• Jul 28th 2013, 12:08 PM
uperkurk
Re: Is this allowed?
Just a quick question without making a new topic.

$\displaystyle \frac{9x-9}{(x-3)(x-1)(x+3)} = \frac{9}{(x-3)(x+3)}$

What happened here? The book doesn't explain where the $\displaystyle 9x$ or the $\displaystyle (x-1)$ goes
• Jul 28th 2013, 12:32 PM
Plato
Re: Is this allowed?
Quote:

Originally Posted by uperkurk
Just a quick question without making a new topic.

$\displaystyle \frac{9x-9}{(x-3)(x-1)(x+3)} = \frac{9}{(x-3)(x+3)}$

What happened here? The book doesn't explain where the $\displaystyle 9x$ or the $\displaystyle (x-1)$ goes

$\displaystyle \frac{9x-9}{(x-3)(x-1)(x+3)} = \frac{9(x-1)}{(x-3)(x-1)(x+3)}$
• Jul 28th 2013, 12:38 PM
uperkurk
Re: Is this allowed?
Quote:

Originally Posted by Plato
$\displaystyle \frac{9x-9}{(x-3)(x-1)(x+3)} = \frac{9(x-1)}{(x-3)(x-1)(x+3)}$

omg I'm so stupid (Headbang) I feel so so stupid lol, these stupid mistakes happen when it gets late
• Jul 29th 2013, 11:07 PM
marybalogh
Re: Is this allowed?
Thanks for this thread...