# How do you simplify (2a^4b^3 )/(a^2b)?

May 30, 2018

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$2 \left({a}^{4} / {a}^{2}\right) \left({b}^{3} / b\right)$

Next, use this rule of exponents to rewrite the denominator of the $b$ term:

$x = {x}^{\textcolor{b l u e}{1}}$

$2 \left({a}^{4} / {a}^{2}\right) \left({b}^{3} / {b}^{\textcolor{b l u e}{1}}\right)$

Now, use this rule of exponents to simplify the $a$ and $b$ terms:

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

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

$2 {a}^{\textcolor{red}{4} - \textcolor{b l u e}{2}} {b}^{\textcolor{red}{3} - \textcolor{b l u e}{1}} \implies$

$2 {a}^{2} {b}^{2}$

May 30, 2018

$2 {a}^{2} {b}^{2}$

#### Explanation:

Remember that exponents underneath the fraction are the same as negative exponents above the fraction. So this expression is the same as $2 {a}^{4} {b}^{3} {a}^{- 2} {b}^{- 1}$, which one then adds as normal:
$2 {a}^{4 - 2} {b}^{3 - 1} = 2 {a}^{2} {b}^{2}$.