# How do you simplify  (16q^0 r^-6) /( 4q^-3 r^-7)?

Jan 24, 2016

$\frac{16 {q}^{0} {r}^{- 6}}{4 {q}^{- 3} {r}^{- 7}} = 4 {q}^{3} r$

with exclusions $q \ne 0$ and $r \ne 0$

#### Explanation:

If $a \ne 0$ and $b$ and $c$ are integers, then ${a}^{b} / {a}^{c} = {a}^{b - c}$

So:

$\frac{16 {q}^{0} {r}^{- 6}}{4 {q}^{- 3} {r}^{- 7}}$

$= \left(\frac{16}{4}\right) \left({q}^{0} / {q}^{- 3}\right) \left({r}^{- 6} / {r}^{- 7}\right)$

$= 4 {q}^{0 - \left(- 3\right)} {r}^{\left(- 6\right) - \left(- 7\right)}$

$= 4 {q}^{3} {r}^{1}$

$= 4 {q}^{3} r$

with exclusions $q \ne 0$ and $r \ne 0$

The exclusions are required since if $q = 0$ or $r = 0$ then at least one of ${q}^{- 3}$, ${r}^{- 6}$, ${r}^{- 7}$ is undefined.