# How do you evaluate 3/(16-a^2) - 5/(12+3a)?

Jul 23, 2015

$\frac{5 a - 11}{3 \left(4 + a\right) \left(4 - a\right)}$

#### Explanation:

Assuming you want to evaluate to a single expression:

Factoring the denominators we have:
$\frac{3}{16 - {a}^{2}} - \frac{5}{12 + 3 a} = \frac{3}{\left(4 + a\right) \left(4 - a\right)} - \frac{5}{3 \left(4 + a\right)}$

we can then make the denominators identical:
$\frac{3}{\left(4 + a\right) \left(4 - a\right)} - \frac{5}{3 \left(4 + a\right)} =$

$\frac{3}{\left(4 + a\right) \left(4 - a\right)} - \frac{\frac{5}{3} \left(4 - a\right)}{\left(4 + a\right) \left(4 - a\right)} =$

$\frac{3 - \frac{5}{3} \left(4 - a\right)}{\left(4 + a\right) \left(4 - a\right)} = \frac{5 a - 11}{3 \left(4 + a\right) \left(4 - a\right)}$

Jul 23, 2015

I tried factorizing and using a common denominator:

#### Explanation:

You can write it as:
$\frac{3}{\left(4 + a\right) \left(4 - a\right)} - \frac{5}{3 \left(4 + a\right)} =$
$= \frac{9 - 5 \left(4 - a\right)}{3 \left(4 + a\right) \left(4 - a\right)} =$
$= \frac{9 - 20 + 5 a}{3 \left(4 + a\right) \left(4 - a\right)} = \frac{5 a - 11}{3 \left(4 + a\right) \left(4 - a\right)}$