# How do you simplify (6r^2p^3)/(4rp^4) and find any non permissible values?

Jun 29, 2017

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\left(\frac{6}{4}\right) \left({r}^{2} / r\right) \left({p}^{3} / {p}^{4}\right) \implies$

$\left(\frac{3 \times 2}{2 \times 2}\right) \left({r}^{2} / r\right) \left({p}^{3} / {p}^{4}\right) \implies$

$\left(\frac{3 \times \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}{2 \times \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right) \left({r}^{2} / r\right) \left({p}^{3} / {p}^{4}\right) \implies$

$\frac{3}{2} \left({r}^{2} / r\right) \left({p}^{3} / {p}^{4}\right)$

Next, use these rules of exponents to simplify the $r$ terms:

$a = {a}^{\textcolor{b l u e}{1}}$ and ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$ and ${a}^{\textcolor{red}{1}} = a$

$\frac{3}{2} \left({r}^{\textcolor{red}{2}} / {r}^{\textcolor{b l u e}{1}}\right) \left({p}^{3} / {p}^{4}\right) \implies$

$\frac{3}{2} \left({r}^{\textcolor{red}{2} - \textcolor{b l u e}{1}}\right) \left({p}^{3} / {p}^{4}\right) \implies$

$\frac{3}{2} \left({r}^{1}\right) \left({p}^{3} / {p}^{4}\right) \implies$

$\frac{3}{2} \left(r\right) \left({p}^{3} / {p}^{4}\right) \implies$

$\frac{3 r}{2} \left({p}^{3} / {p}^{4}\right)$

Now, use these rules of exponents to simplify the $p$ terms:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$ and ${a}^{\textcolor{red}{1}} = a$

$\frac{3 r}{2} \left({p}^{\textcolor{red}{3}} / {p}^{\textcolor{b l u e}{4}}\right) \implies$

$\frac{3 r}{2} \left(\frac{1}{p} ^ \left(\textcolor{b l u e}{4} - \textcolor{red}{3}\right)\right) \implies$

$\frac{3 r}{2} \left(\frac{1}{p} ^ 1\right) \implies$

$\frac{3 r}{2} \left(\frac{1}{p}\right) \implies$

$\frac{3 r}{2 p}$

From the original expression, because we cannot divide by $0$:

$4 r {p}^{4} \ne 0$

Or

$r \ne 0$ and $p \ne 0$