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

May 23, 2015

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

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

$\frac{x + 4}{2 \left(x + 3\right)} + \frac{3}{{x}^{2} - 9}$

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

$\frac{x + 4}{2 \left(x + 3\right)} + \frac{3}{\left(x + 3\right) \left(x - 3\right)}$

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

$\frac{\left(x + 4\right) \left(x - 3\right)}{2 \left(x + 3\right) \left(x - 3\right)} + \frac{6}{2 \left(x + 3\right) \left(x - 3\right)}$

Now simplify.

$\frac{\left(x + 4\right) \left(x - 3\right) + 6}{2 \left(x + 3\right) \left(x - 3\right)} = \frac{{x}^{2} + x - 12 + 6}{2 {x}^{2} - 18} = \frac{{x}^{2} + x - 6}{2 {x}^{2} - 18}$

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

$\frac{{x}^{2} + x - 6}{2 {x}^{2} - 18} = \frac{\left(x + 3\right) \left(x - 2\right)}{2 \left(x + 3\right) \left(x - 3\right)}$

Aha! The $\left(x + 3\right)$s can cancel out

$\frac{\left(x - 2\right)}{2 \left(x - 3\right)} = \frac{x - 2}{2 x - 6}$

May 23, 2015

$\frac{x + 4}{2 x + 6} + \frac{3}{{x}^{2} - 9}$

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

Since $2 x + 6 = 2 \left(x + 3\right)$ and ${x}^{2} - 9 = \left(x + 3\right) \left(x - 3\right)$
the obvious choice for a common denominator is
$2 \left(x + 3\right) \left(x - 3\right)$

$\frac{\left(x + 4\right) \left(x - 3\right) + 3 \left(2\right)}{2 \left(x + 3\right) \left(x - 3\right)}$
$= \frac{{x}^{2} + x - 12 + 6}{2 \left(x + 3\right) \left(x - 3\right)}$
$= \frac{\left(x + 3\right) \left(x - 2\right)}{2 \left(x + 3\right) \left(x - 3\right)}$
$= \frac{x - 2}{2 \left(x - 3\right)}$