Question 238203
Hello,

This is as far as I could do it.  I hope I'm not mising further steps.

{{{(y+3)/(y+3)+7/(3y+9)-2/(y^2+3y)}}}
now factor the second and third denominators
{{{(y+3)/(y+3)+7/3(y+3)-2/y(y+3)}}}
now factor the common denominator
{{{(1/(y+3))((y+3)+7/3-2/y)}}}