# How do you simplify (7x^-4 y^5) (2x^-3 y^-4)?

Apr 4, 2017

See the entire simplification process below:

#### Explanation:

First, rewrite the expression as:

$\left(7 \times 2\right) \left({x}^{-} 4 \cdot {x}^{-} 3\right) \left({y}^{5} \cdot {y}^{-} 4\right) = 14 \left({x}^{-} 4 \cdot {x}^{-} 3\right) \left({y}^{5} \cdot {y}^{-} 4\right)$

Next, use this rule of exponents to multiply the variables:

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

$14 \left({x}^{\textcolor{red}{- 4}} \cdot {x}^{\textcolor{b l u e}{- 3}}\right) \left({y}^{\textcolor{red}{5}} \cdot {y}^{\textcolor{b l u e}{- 4}}\right) = 14 {x}^{\textcolor{red}{- 4} + \textcolor{b l u e}{- 3}} {y}^{\textcolor{red}{5} + \textcolor{b l u e}{- 4}} = 14 {x}^{-} 7 {y}^{1}$

Then, use this rule of exponents to complete the simplification for the $y$ variable:

${a}^{\textcolor{red}{1}} = a$

$14 {x}^{-} 7 {y}^{\textcolor{red}{1}} = 14 {x}^{-} 7 y$

Now, use this rule of exponents to complete the simplification:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$

$14 {x}^{\textcolor{red}{- 7}} y = \frac{14 y}{x} ^ \textcolor{red}{- - 7} = \frac{14 y}{x} ^ 7$