Question 203076
1 1
__ - __
4n 2
*OVER*
n 1
__- ___
6 24n 
Assume the problem is:
{{{(1/(4n)) - (1/2)}}}
------------
{{{(n/6)-(1/(24n))}}}
:
Common denominators
{{{(1-2n)/(4n)}}}
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{{{(4n^2-1)/(24n)}}}
invert the dividing fraction and multiply, cancel out 4n, factor the dif of squares
{{{(1-2n)/(4n)}}} * {{{(24n)/(4n^2-1)}}} = (1-2n) * {{{6/(4n^2-1)}}} = {{{6(1-2n)/(4n^2-1)}}} = {{{6(1-2n)/((2n+1)(2n-1))}}}
We can rewrite the numerator, then cancel 2n-1:
{{{(-6(2n-1))/((2n+1)(2n-1))}}} = {{{(-6)/((2n+1))}}} or {{{-6/(2n+1)}}}