Question 119186
{{{2/(a^2-2a)-3/(a^2-a-2)}}}
Whenever you subtract fractions, you need to have the same denominators. 
Let's look at each denominator first. 
{{{D[1]=a^2-2a=a(a-2)}}}
{{{D[2]=a^2-a-2=(a+1)(a-2)}}}
So if you multiply the first denominator by (a+1) and then second denominator by (a), they will have a common denominator of a(a+1)(a-2). 
Remember whatever you multiply the denominator by, you also must multiply the numerator by.
{{{2/(a^2-2a)-3/(a^2-a-2)=2/(a(a-2))-3/((a+1)(a-2))}}}
{{{2/(a^2-2a)-3/(a^2-a-2)=2*highlight(a+1)/(a*highlight((a+1))(a-2))-3*highlight(a)/(highlight(a)(a+1)(a-2))}}}
{{{2/(a^2-2a)-3/(a^2-a-2)=(2(a+1)-3a)/(a(a+1)(a-2))}}}
{{{2/(a^2-2a)-3/(a^2-a-2)=(2a+2-3a)/(a(a+1)(a-2))}}}
{{{2/(a^2-2a)-3/(a^2-a-2)=(2-a)/(a(a+1)(a-2))}}}
{{{2/(a^2-2a)-3/(a^2-a-2)=(-(a-2))/(a(a+1)(a-2))}}}
{{{2/(a^2-2a)-3/(a^2-a-2)=(-cross((a-2)))/(a(a+1)cross((a-2)))}}}
{{{2/(a^2-2a)-3/(a^2-a-2)=(-1)/(a(a+1))}}}
Remember this relation only makes sense when a does not equal 0,-1,or 2.