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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

in my observation. As soon as I saw his point, we solved the problem
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)