Question 146857
From your explanation, I think this is your problem. 
{{{3/(a+2)+4/(2-a)-6/(a^2-4)}}}
First step is to find a common denominator. 
Let's re-write the equation using the following,
{{{a^2-4=(a+2)(a-2)}}}
Using this identity, we can find the common denominator,
{{{3/(a+2)+4/(2-a)-6/(a^2-4)=3/(a+2)-4/(a-2)-6/((a+2)(a-2))}}}
If we multiply the first term by (a-2)/(a-2)
and the second term by (a+2)/(a+2), then we
will have the common denominator (a+2)(a-2).
{{{3/(a+2)+4/(2-a)-6/(a^2-4)=3/(a+2)-4/(a-2)-6/((a+2)(a-2))}}}
{{{3/(a+2)+4/(2-a)-6/(a^2-4)=3(a-2)/((a+2)(a-2))-4(a+2)/((a+2)(a-2))-6/((a+2)(a-2))}}}
{{{3/(a+2)+4/(2-a)-6/(a^2-4)=(3(a-2)-4(a+2)-6)/((a+2)(a-2))}}}
{{{3/(a+2)+4/(2-a)-6/(a^2-4)=(3a-6-4a-8-6)/((a+2)(a-2))}}}
{{{3/(a+2)+4/(2-a)-6/(a^2-4)=(-a-20)/((a+2)(a-2))}}}
{{{3/(a+2)+4/(2-a)-6/(a^2-4)=-(a+20)/((a+2)(a-2))}}}