Question 30705
7/w-5+ 6/w-3 - 3w/w^2-8w+15 
= 7/w-5+ 6/w-3 - 3w/(w-5)(w-3)  
(factoring w^2-8w+15)
(the sum is -8 and product is 15 and hence the quantities arse -5 and -3) 
=[7X(w-3)+6(w-5)-3w]/(w-5)(w-3)
=[7w-21+6w-30-3w]/(w-5)(w-3)
=[(7w+6w-3w)-21-30]/(w-5)(w-3)
=(10w-51)/(w-5)(w-3)