The 2x is above the x-3
Added to 4x - 6 over x^2 - 9
The 2x is above the x-3
Added to 4x - 6 over x^2 - 9
2x/(x-3) + (4x-6)/(x^{2}-9)
The first thing you want to do is factor the denominator in the second expression. x^{2}-9 is can be factored to (x-3)(x+3) -- this is a special factoring case, the difference of squares, that you should learn to recognize because you'll be using it a lot! (a+b)(a-b) = a^{2}-b^{2}.
2x/(x-3) + (4x-6)/((x+3)(x-3))
Now you need a common denominator so you can add the two fractions. (x+3)(x-3) will be the common denominator. The first expression should be multiplied by (x+3)/(x+3); the second expression already has the correct denominator.
(2x(x+3))/((x+3)(x-3)) + (4x-6)/((x+3)(x-3))
Now, since the fractions have a common denominator, you can add the numerators.
(2x(x+3) + (4x-6)) / ((x+3)(x-3))
Next, distribute the 2x to the x+3.
(2x^{2}+6x+4x-6) / ((x+3)(x-3))
Combine like terms in the numerator.
(2x^{2}+10x-6) / ((x+3)(x-3))
Factor a 2 out of the numerator.
2(x^{2}+5x-3) / ((x+3)(x-3))
You can't factor x^{2}+5x-3, so this is your final answer. Often, in problems like this, you would be able to factor the numerator and one of the factors would end up canceling with the denominator.