Question 82971
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add or subtract and then put in simplest form 
{{{(5a-12)/(a^2-8a+15)}}} subtracted from {{{(3a-2)/(a^2-8a+15)}}}

One of the chief things here is to correctly interpret the words
"subtracted from".  Many students interpret this incorrectly.

To get "3 subtracted from 7", we START with the 7 and then we subtract
3 from it.  We write "7 - 3", not "3 - 7"!   So in the case of

{{{(5a-12)/(a^2-8a+15)}}} subtracted from {{{(3a-2)/(a^2-8a+15)}}}

We start with the {{{(3a-2)/(a^2-8a+15)}}} and then we subtract
{{{(5a-12)/(a^2-8a+15)}}} from it.

So we begin by writing down the {{{(3a-2)/(a^2-8a+15)}}} then write a 
minus sign and then write down {{{(5a-12)/(a^2-8a+15)}}}.

Be sure not to get that backwards!  This is such a common mistake.
Think about what you START with, write that down FIRST, and then
either add to it or subtract from it as you are told.  So here
we have this.

{{{(3a-2)/(a^2-8a+15)}}} - {{{(5a-12)/(a^2-8a+15)}}} 

Since the denominators are already the same, all we need do is combine 
the numerators over that same denominator, into just one fraction,
using the sign that is between the fractions like this, being sure to
place the numerators in parentheses first:

{{{( (3a-2) - (5a-12))/(a^2-8a+15)}}}

Then we remove the parentheses in the numerator:

{{{(3a-2-5a+12)/(a^2-8a+15)}}}

Now combine the like terms in the numerator:

{{{(-2a+10)/(a^2-8a+15)}}}

Now we factor both numerator and denominator,
that is, take out -2 in the numerator, and factor the
denominator as a trinomial:

{{{(-2(a-5))/((a-5)(a-3))}}}

Now the {{{(a-5)}}}'s cancel and all that's left is

{{{(-2)/((a-3))}}}

or just

{{{-2/(a-3)}}}

Edwin</pre>