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## 2x / x-3 + 4x-6 / x^2 - 9

The 2x is above the x-3

Added to 4x - 6 over x^2 - 9

2x/(x-3) + (4x-6)/(x2-9)

The first thing you want to do is factor the denominator in the second expression.  x2-9 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) = a2-b2.

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.

(2x2+6x+4x-6) / ((x+3)(x-3))

Combine like terms in the numerator.

(2x2+10x-6) / ((x+3)(x-3))

Factor a 2 out of the numerator.

2(x2+5x-3) / ((x+3)(x-3))

You can't factor x2+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.