Question 22911
First of all, congratulations on getting all the parentheses in the right place in typing this problem.  If you put 3 sets of set braces before and 3 after this, then your question will be expressed in perfect form for a complex fraction:

 {{{ ((x)/(x-4)-(2)/(x+3))/((x^2-4x)/(2x+6))}}}


Let's just "unstack" the problem and write it like this:

{{{ ((x)/(x-4)-(2)/(x+3))}}} divided by {{{((x^2-4x)/(2x+6))}}}


On the left side, you need to find a common denominator, which is (x-4)(x+3).  Multiply the first fraction by {{{(x+3)/(x+3)}}}, and the second fraction by {{{(x-4)/(X-4)}}}.  It should look like this:

{{{ (((x)/(x-4))*((x+3)/(x+3))-((2)/(x+3))*((x-4)/(x-4)))}}} divided by {{{((x^2-4x)/(2x+6))}}}


Now, put down the common denominator for the first fraction; combine numerators to form the first numerator, and invert and factor the second fraction:

{{{ (x^2+3x -2x+8)/((x-4)(x+3))}}} times {{{(2(x+3))/ (x(x-4))  }}}


Combine like terms in the first numerator, divide out the (x+3) factors in the first denominator and the second numerator:

{{{ (x^2+x +8)/(x-4)}}} times {{{2/ (x(x-4))  }}}


This brings (I hope!!) the final answer:

{{{ (2*(x^2+x +8))/(x*(x-4)^2)}}} 


R^2 at SCC


P.S.  That was ugly!!  Somebody let me know if there might be an error in all of that!!