Question 475519
add simplify if possible
2y/y^2-9 + y/7-3
:
Assume you mean
{{{(2y)/((y^2-9)) + y/((y-3))}}}
:
We can factor the 1st denominator as the "difference of squares"
{{{(2y)/((y-3)(y+3)) + y/((y-3))}}}
Combine as a single fraction
{{{(2y+y(y+3))/((y-3)(y+3))}}} = {{{(2y+y^2+3y)/((y-3)(y+3))}}} = {{{(y^2+5y)/((y-3)(y+3))}}} = {{{(y(y+5))/((y-3)(y+3))}}}