Question 700462
Not quite sure what you mean here, assume this is problem
{{{2/x}}}+{{{3/(3x)}}}
------------
{{{1/(3x)}}}-{{{1/(2x)}}}
we want to get each pair of fractions over a single denominator
The top one is easy, cancel the 3's
{{{2/x}}}+{{{1/x}}}
------------
{{{1/(3x)}}}-{{{1/(2x)}}} 
which is
{{{(2 + 1)/x}}}
------------
{{{1/(3x)}}}-{{{1/(2x)}}} 
which is
{{{3/x}}}
------------
{{{1/(3x)}}}-{{{1/(2x)}}}
the bottom pair has a common denominator of 6x, so we have:
{{{3/x}}}
--------
{{{(2-3)/(6x)}}}
which is
{{{3/x}}}
---------
{{{-1/(6x)}}}
invert the dividing fraction and multiply, cancel out the x
{{{3/x}}}*{{{(6x)/(-1)}}} = 3*-6 = -18