Question 119455
{{{((w+3)/4w)/((w-3)/2w)}}}


Remember the rule for dividing one fraction by another:  Invert the denominator fraction and multiply:


{{{(p/q)/(r/s)=(p/q)(s/r)}}} for all real p, q, r, and s, so


{{{((w+3)/4w)/((w-3)/2w)=((w+3)/4w)(2w/(w-3))}}}


Now you just multiply the numerators and the denominators the same way you would multiply any other two fractions:


{{{((w+3)/4w)(2w/(w-3))=(2w(w+3))/(4w(w-3))}}}


Since there is a factor of 2w in both the numerator and denominator, we can eliminate that, so:


{{{(2w(w+3))/(4w(w-3))=(w+3)/(2(w-3))}}}


Since there are no more common factors in the numerator and denominator, we are done except for distributing the 2 in the denominator across the (w - 3) to remove the parentheses, thus:


{{{(w+3)/(2w-6)}}}


Let's check our work:


We know that if {{{p/q=r}}} then {{{rq=p}}}, so if {{{((w+3)/4w)/((w-3)/2w)=(w+3)/(2w-6)}}}, then {{{((w-3)/2w)((w+3)/(2w-6))=(w+3)/4w}}} must be true.


{{{((w-3)/2w)((w+3)/(2w-6))=((w-3)(w+3))/2w(2w-6))}}}


Now factor a 2 out of the binomial in the denominator resulting in:


{{{((w-3)(w+3))/4w(w-3)=(w+3)/4w}}} once you eliminate the common factor of (w - 3), and that verifies the answer.

 
Hope that helps.
John