# How do you simplify (7x^4y^3)/(5xy)*(2xy^7)/(21y^5)?

##### 1 Answer
Feb 12, 2017

See the entire simplification process below:

#### Explanation:

First, use these rules of exponents to multiply the numerators and denominators:

$a = {a}^{\textcolor{red}{1}}$ and ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$\frac{7 {x}^{4} {y}^{3}}{5 x y} \cdot \frac{2 x {y}^{7}}{21 {y}^{5}} \to \frac{7 {x}^{4} {y}^{3}}{5 {x}^{1} {y}^{1}} \cdot \frac{2 {x}^{1} {y}^{7}}{21 {y}^{5}} \to \frac{14 {x}^{4 + 1} {y}^{3 + 7}}{105 {x}^{1} {y}^{1 + 5}} =$

$\frac{14 {x}^{5} {y}^{10}}{105 {x}^{1} {y}^{6}}$

Next, let's factor the constants:

$\frac{14 {x}^{5} {y}^{10}}{105 {x}^{1} {y}^{6}} = \frac{\left(7 \times 2\right) {x}^{5} {y}^{10}}{\left(7 \times 15\right) {x}^{1} {y}^{6}} = \frac{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} \times 2\right) {x}^{5} {y}^{10}}{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} \times 15\right) {x}^{1} {y}^{6}} =$

$\frac{2 {x}^{5} {y}^{10}}{15 {x}^{1} {y}^{6}}$

Now, we can use this rule for exponents to complete the simplification:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

$\frac{2 {x}^{5} {y}^{10}}{15 {x}^{1} {y}^{6}} = \frac{2 {x}^{5 - 1} {y}^{10 - 6}}{15} = \frac{2 {x}^{4} {y}^{4}}{15}$ or $\frac{2}{15} {x}^{4} {y}^{4}$