# Question #935d4

Apr 9, 2017

$\frac{2 {x}^{3} {y}^{3}}{6 {x}^{-} 4 {y}^{4}} = {x}^{7} / \left(3 y\right)$

#### Explanation:

Given: $\frac{2 {x}^{3} {y}^{3}}{6 {x}^{-} 4 {y}^{4}}$

We can see the integer multipliers: $\frac{2}{6} = \frac{1}{3}$

The $x$ component is ${x}^{3} / {x}^{-} 4 = {x}^{3} / \left(\frac{1}{x} ^ 4\right) = {x}^{7}$ $\to$after invert and multiply

The $y$ component is ${y}^{3} / {y}^{4} = \frac{1}{y}$ $\to$dividing exponents = subtract

Then we can multiply all the components:

$\frac{2 {x}^{3} {y}^{3}}{6 {x}^{-} 4 {y}^{4}} = \left(\frac{1}{3}\right) {x}^{7} \left(\frac{1}{y}\right) = {x}^{7} / \left(3 y\right)$