
Re: simplifying rational expressions
Posted:
Jan 16, 2013 8:30 AM


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. > > > ((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
It would be nice if you had showed us your steps. You do recognize, I hope, that ab and ba are just the same thing with a factor of 1 pulled out? So if you're treating (ab) and (ba) 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, ab, a c, bc, so your common denominator will be (ab)(ac)(bc).
PLEASE, with that hint, solve it on your own, and only ten look at what I've done.
First step: write all the binomial factors in the same form:
(bc)/[(ab)(ac)] + (1)(ac)/[(bc)(1)(ab)] + (ab)/[(1)(ac)(1)(bc)] + 2/[(1)(ab)]  2/[(1)(ac)]
(bc)/[(ab)(ac)] + (ac)/[(ab)(bc) + (ab)/[(ac)(bc)] + (2)/(ab) + 2/(ac)
(bc)^2/[(ab)(ac)(bc)] + (ac)^2/[ab)(ac)(bc)] + (ab)^2/[(ab)(ac)(bc)] + (2)(ac)(bc)/[(ab)(ac)(bc)] + 2(ab)(bc)/[(ab)(ac)(bc)]
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(ab) 2c(ab) = 2(ab)(ac)
Now restoring the denominator, the whole fraction is
2(ab)(ac)/[(ab)(ac)(bc)] = 2/(bc)  Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Shikata ga nai...

