# How do you simplify (x^2 y^3 + x y^2) / (xy)?

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Mar 9, 2018

color(blue)(=> y(xy +1)

#### Explanation:

$\frac{{x}^{2} {y}^{3} + x {y}^{2}}{x y}$

$\implies \frac{\left(x y\right) \left(x {y}^{2} + y\right)}{x y}$ taking common term (xy) outside.

$\implies \frac{\cancel{x y} \cdot \left(x {y}^{2} + y\right)}{\cancel{x y}}$

$\implies y \left(x y + 1\right)$ taking “y” outside as common.

Then teach the underlying concepts
Don't copy without citing sources
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#### Explanation

Explain in detail...

#### Explanation:

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1
Mar 9, 2018

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\frac{{x}^{2} {y}^{3}}{x y} + \frac{x {y}^{2}}{x y}$

Now, cancel common terms in the numerators and denominators:

$\frac{{x}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} 1} {y}^{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{3}}} 2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{y}}}} + \frac{\textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{x}}} {y}^{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{2}}} 1}}{\textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{x}}} \textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{y}}}} \implies$

${x}^{1} {y}^{2} + y \implies$

$x {y}^{2} + y$ Where $x \ne 0$ and $y \ne 0$

Or

$y \left(x y + 1\right)$ Where $x \ne 0$ and $y \ne 0$

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