Question 107260
{{{x^2/(x-3)-9/(x-3)}}}
here we already have a common denominator so we can add these fractions. 

This becomes
{{{(x^2-9)/(x-3)}}}
{{{x^2-9}}} is a difference of 2 squares so it factorises to (x-3)(x+3)

we now have 
{{{((x-3)(x+3))/(x-3)}}}
the x-3 cancels and we are left with 
x+3