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Topic: simplifying rational expressions
Replies: 12   Last Post: Jan 28, 2013 12:57 AM

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Wasell

Posts: 8
Registered: 9/29/10
Re: simplifying rational expressions
Posted: Jan 16, 2013 3:33 AM
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On Tue, 15 Jan 2013 23:59:53 -0600, in article
<uqfcf8dvhsqr7qjsdi09b8agcahbsr98ah@4ax.com>, stony wrote:
>
> Hi,
>
> Need a little help with this. We are simplifying the following, but
> the solution is pretty lengthy and messy because of the enormous
> number of factors. I was thinking that may be I am missing seeing a
> pattern (some series or something). Is grunt work the only way to
> solve this or is there a pattern that can simplify the whole process?
>
> My daughter was trying to solve this, but ended up with the mess and
> then I got the same mess, but I thought there may be an easy way to
> simplify this that I may be missing.
>
>
> ((b-c)/((a-b)(a-c))) + ((c-a)/((b-c)(b-a))) + ((a-b)/((c-a)(c-b))) +
> (2/(b-a)) - (2/(c-a))
>
>
> of course, I took all the factors in the denominator and then started
> multiplying the numerator with the remaining factors to end up with a
> mess.
>
> Your help is appreciated.
>
> s


Let x = a-b,
y = b-c,
z = c-a.

Then the first three terms become -(x^2 + y^2 + z^2)/(xyz).

Note that x+y = -z,
y+z = -x,
z+x = -y.

I'm sure you can take it from there.

HTH
/Wasell




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