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Re: simplifying rational expressions
Posted:
Jan 16, 2013 11:53 PM
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Thank you very much. Yes I was able to take it from there and finish
is and find the answer was quite an elegant solution. Your help is
appreciated. Thanks.
p
On Wed, 16 Jan 2013 09:33:19 +0100, Wasell <Wasell@example.invalid>
wrote:
>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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