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

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justlooking for someone else

Posts: 77
Registered: 12/10/04
Re: simplifying rational expressions
Posted: Jan 17, 2013 12:10 AM
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I did heed your advice and not look at your solution at all because
Jussi (2nd reply on this thread) had already pointed out the mistake
in my observation. As soon as I saw his point, we solved the problem
and got the answer. I then read your reply to see that we had arrived
at the same solution. Thank you for your generous help and caution.
We grunt it out all the time, but missed a key observation that would
have alleviated the situation. Thank you for your time. Appreciated.

s



On Wed, 16 Jan 2013 08:30:26 -0500, Stan Brown
<the_stan_brown@fastmail.fm> wrote:

>On Tue, 15 Jan 2013 23:59:53 -0600, 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

>
>It would be nice if you had showed us your steps. You do recognize,
>I hope, that a-b and b-a are just the same thing with a factor of -1
>pulled out? So if you're treating (a-b) and (b-a) as different
>factors in finding your common denominator, you're doing much more
>work than you need to. In fact there are only three factors, a-b, a-
>c, b-c, so your common denominator will be (a-b)(a-c)(b-c).
>
>PLEASE, with that hint, solve it on your own, and only ten look at
>what I've done.
>
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>First step: write all the binomial factors in the same form:
>
>(b-c)/[(a-b)(a-c)] + (-1)(a-c)/[(b-c)(-1)(a-b)]
> + (a-b)/[(-1)(a-c)(-1)(b-c)] + 2/[(-1)(a-b)] - 2/[(-1)(a-c)]
>
>(b-c)/[(a-b)(a-c)] + (a-c)/[(a-b)(b-c)
> + (a-b)/[(a-c)(b-c)] + (-2)/(a-b) + 2/(a-c)
>
>(b-c)^2/[(a-b)(a-c)(b-c)] + (a-c)^2/[a-b)(a-c)(b-c)]
> + (a-b)^2/[(a-b)(a-c)(b-c)]
> + (-2)(a-c)(b-c)/[(a-b)(a-c)(b-c)]
> + 2(a-b)(b-c)/[(a-b)(a-c)(b-c)]
>
>Temporarily disregarding the common denominator, you have a numerator
>of
>
> b^2 -2bc + C^2
> +a^2 -2ac + c^2
> +a^2 -2ab + b^2
> -2ab +2ac +2bc -2c^2
> +2ab -2b^2 -2ac +2bc
>
>Which is
>
>2a^2 - 2ab - 2ac + 2bc
>= 2a(a-b) -2c(a-b)
>= 2(a-b)(a-c)
>
>Now restoring the denominator, the whole fraction is
>
>2(a-b)(a-c)/[(a-b)(a-c)(b-c)] = 2/(b-c)




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