Question

How do you add `(x + 4)/(2x +6) +3/(x^2-9)`?

Answer 1

It's all about using factoring to "whittle down" these guys.

First, 2x + 6 is 2 times x + 3.

`(x+4)/(2(x+3)) + 3/(x^2-9)`

Also, `x^2 - 9` is the difference of squares, so it can be factored into (a+b) and (a-b).

`(x+4)/(2(x+3))+3/((x+3)(x-3))`

The common denominator is 2(x+3)(x-3)

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

Now simplify.

`((x+4)(x-3) + 6)/(2(x+3)(x-3)) = (x^2 + x - 12 + 6)/(2x^2 - 18) = (x^2 + x - 6)/(2x^2-18)`

This might be the final answer, but we need to check for any common factors in the numerator and denominator.

`(x^2+x-6)/(2x^2-18) = ((x+3)(x-2))/(2(x+3)(x-3))`

Aha! The `(x+3)`s can cancel out

`((x-2))/(2(x-3)) = (x-2)/(2x-6)`

Final Answer

Answer 2

`(x+4)/(2x+6) +3/(x^2-9)`

As with simple fractions, the first step is to create a common denominator.

Since `2x+6 = 2(x+3)` and `x^2-9=(x+3)(x-3)`
the obvious choice for a common denominator is
`2(x+3)(x-3)`

and our addition becomes
`((x+4)(x-3) + 3(2))/(2(x+3)(x-3))`

`=(x^2 +x -12 +6)/(2(x+3)(x-3))`

`=((x+3)(x-2))/(2(x+3)(x-3))`

`=(x-2)/(2(x-3))`