Question 44527
Hello,
       My answer to the problem No.44551 will clear your doubt.  However, I am reproducing the same here below:  Here {{{(9x^2-16y^2)/xy}}} will be taken as numerator and {{{(3/y)-(4/x)}}} will be taken as denominator and simplified/factorised first.

      {{{((9x^2-16y^2)/xy)/((3/y)-(4/x))}}}  After factorising the numerator and simplifying the denominator, we get,
 
  {{{((3x+4y)(3x-4y)/xy)/((3x-4y)/xy)}}}.  This means the numerator divided by the denominator.  For dividing, we multiply the numerator by the inverse of the denominator.  So, 
 
  {{{((3x+4y)(3x-4y)/xy)*(xy/(3x-4y)))}}} = {{{((3x+4y)(3x-4y)xy)/(xy*(3x-4y))}}} = {{{((3x+4y)cross((3x-4y))cross(xy))/cross((xy))cross((3x-4y))}}}
 
  {{{3x+4y}}}  Answer.  I hope this is clear to you now. 

gsm