Question 269957
polynomials that are in fraction form.
{{{(-2)/(y^2-9)}}} + {{{(4y)/(y-3)^2}}} + {{{(6)/(3-y)}}} 
:
Do some factoring, then determine a common denominator
{{{(-2)/((y-3)(y+3))}}} + {{{(4y)/((y-3)(y-3))}}} + {{{(6)/(-1(y-3)))}}}
that -1 with change the sign on the last fraction
{{{(-2)/((y-3)(y+3))}}} + {{{(4y)/((y-3)(y-3))}}} - {{{(6)/((y-3)))}}}
the common denominator: (y-3)(y-3)(y+3)
:
{{{(-2(y-3) + 4y(y+3)-6(y+3)(y-3))/((y-3)(y-3)(y+3))}}}
:
{{{(-2y+6 + 4y^2+12y-6(y^2-9))/((y-3)(y-3)(y+3))}}}
:
{{{((4y^2+10y-6y^2+6+54))/((y-3)(y-3)(y+3))}}} = {{{((-2y^2+10y+60))/((y-3)(y-3)(y+3))}}} = {{{(-2(y^2-5y-30))/((y-3)(y-3)(y+3))}}} = {{{-(2(y^2-5y-30))/((y-3)(y-3)(y+3))}}}