
Re: simplifying rational expressions
Posted:
Jan 16, 2013 11:53 PM


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. >> >> >> ((bc)/((ab)(ac))) + ((ca)/((bc)(ba))) + ((ab)/((ca)(cb))) + >> (2/(ba))  (2/(ca)) >> >> >> 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 = ab, > y = bc, > z = ca. > >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

