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

##### 2 Answers
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:
$2 {a}^{4 - 2} {b}^{3 - 1} = 2 {a}^{2} {b}^{2}$.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}