# How do you simplify (12xy^4)/(8x^2y^3)?

Jul 6, 2015

$\frac{3 y}{2 x}$ with $\left(x \ne 0 , y \ne 0\right)$

#### Explanation:

$\frac{12 x {y}^{4}}{8 {x}^{2} {y}^{3}}$ Note : $\left(x \ne 0 , y \ne 0\right)$

Decompose the expression :

$\frac{12 x {y}^{4}}{8 {x}^{2} {y}^{3}} = \frac{2 \cdot 2 \cdot 3 \cdot x \cdot y \cdot y \cdot y \cdot y}{2 \cdot 2 \cdot 2 \cdot x \cdot x \cdot y \cdot y \cdot y}$

And cancels pairs :

$\frac{\textcolor{g r e e n}{\cancel{2} \cdot \cancel{2} \cdot 3} \cdot \textcolor{b l u e}{\cancel{x}} \cdot \textcolor{red}{\cancel{y} \cdot \cancel{y} \cdot \cancel{y} \cdot y}}{\textcolor{g r e e n}{\cancel{2} \cdot \cancel{2} \cdot 2} \cdot \textcolor{b l u e}{\cancel{x} \cdot x} \cdot \textcolor{red}{\cancel{y} \cdot \cancel{y} \cdot \cancel{y}}} = \frac{3 y}{2 x}$

Then : $\frac{12 x {y}^{4}}{8 {x}^{2} {y}^{3}} = \frac{3 y}{2 x}$ with $\left(x \ne 0 , y \ne 0\right)$