Question 107277
we need to factorse what we can first.

3b-6=3(b-2)

4b+8=4(b+2)

s0 we have:

{{{2/(3(b-2))+3/(4(b+2))}}}

we need a common  denominator when we add fractions so we mulitply the first fraction by 4(b+2) and the second fraction by 3(b-2)

we get 
{{{2*4*(b+2)/(3(b-2)*4*(b+2))+(3*3*(b-2))/(4*(b+2)*3*(b-2))}}}

then we can add so we have

{{{(8(b+2)+9(b-2))/(12*(b+2)*(b-2))}}}

expand and simplify the numerator but leave the denominator in the current form