Question

$$\frac{4xy^{3}}{8x^{2}y^{4}}$$

Answer

1/(2*x*y)

Solution


Take out the constants.
\[\frac{4}{8}\times \frac{x{y}^{3}}{{x}^{2}{y}^{4}}\]
Simplify  \(\frac{4}{8}\)  to  \(\frac{1}{2}\).
\[\frac{1}{2}\times \frac{x{y}^{3}}{{x}^{2}{y}^{4}}\]
Simplify.
\[\frac{x{y}^{3}}{2{x}^{2}{y}^{4}}\]
Use Quotient Rule: \(\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}\).
\[\frac{{x}^{1-2}{y}^{3-4}}{2}\]
Simplify  \(1-2\)  to  \(-1\).
\[\frac{{x}^{-1}{y}^{3-4}}{2}\]
Simplify  \(3-4\)  to  \(-1\).
\[\frac{{x}^{-1}{y}^{-1}}{2}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\frac{\frac{1}{x}{y}^{-1}}{2}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\frac{\frac{1}{x}\times \frac{1}{y}}{2}\]
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{\frac{1\times 1}{xy}}{2}\]
Simplify  \(1\times 1\)  to  \(1\).
\[\frac{\frac{1}{xy}}{2}\]
Simplify.
\[\frac{1}{2xy}\]