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Topic: simplifying rational expressions

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Stan Brown

Posts: 1,326
Registered: 12/6/04
Re: simplifying rational expressions
Posted: Jan 16, 2013 8:30 AM

On Tue, 15 Jan 2013 23:59:53 -0600, stony wrote:
> Hi,
> the solution is pretty lengthy and messy because of the enormous
> simplify this that I may be missing.

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.

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

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)
Shikata ga nai...


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