# How do you combine (4-3x)/ (16-x^2) + 3/( x-4)?

May 18, 2015

You need to find the lowest common denominator among these two functions. We can try to factor the first function's denominator in order to ease things for us, as shown:

$16 - {x}^{2} = 0$
$16 = {x}^{2}$
$\sqrt{16} = x$
$x = \pm 4$, which means $x - 4 = 0$ and $x + 4 = 0$.

However, the factors $\left(x - 4\right) \left(x + 4\right)$ when multiplied, give us the opposite of the first function's denominator:

$\left(x - 4\right) \left(x + 4\right) = {x}^{2} - 16$

We just need to multiply it by $\left(- 1\right)$ to validate our equality.

So, we can rewrite the whole sum as

$\frac{4 - 3 x}{\left(- 1\right) \left(x + 4\right) \left(x - 4\right)} + \frac{3}{x - 4}$

Now, our lowest common denominator is $\left(x + 4\right) \left(x - 4\right)$. Adopting it as l.c.d., then we have:

$\frac{\left(4 - 3 x\right) + 3 \left(- x - 4\right)}{\left(- 1\right) \left(x + 4\right) \left(x - 4\right)}$=$\frac{4 - 6 x - 8}{\left(- 1\right) \left(x + 4\right) \left(x - 4\right)}$

Distributing our factors in the denominator, we have the shortest answer as follows:

$\textcolor{g r e e n}{\frac{- 6 x - 8}{16 - {x}^{2}}}$