factor 3y^2+20y-7/ 5y +35 / (3y-1)

Answer 1

Assuming the usual carelessness with parentheses,
(3y^2+20y-7)/(5y) + 35/(3y-1)
= ((3y^2+20y-7)(3y-1) + 35(5y)) / ((5y)(3y-1))
= (9y^3+57y^2+134y+7)/(15y^2-5y)
Not much joy there.
However, maybe things got garbled a bit, since
(3y^2+20y-7)/(3y-1) + 35/(5y)
= (3y-1)(y+7)/(3y-1) + (5*7)/(5y)
= (y+7) + 7/y
or maybe something else...