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

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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.

the answer is (x^2+2x-3)/(x^2-9)