# How do you simplify (2x^4y^-4z^-3)/(3x^2y^-3z^4) and write it using only positive exponents?

$\frac{2 {x}^{2}}{3 y {z}^{7}}$

#### Explanation:

$\frac{2 {x}^{4} {y}^{-} 4 {z}^{-} 3}{3 {x}^{2} {y}^{-} 3 {z}^{4}}$

I'm first going to rewrite this using the rule ${x}^{-} 1 = \frac{1}{x}$:

$\frac{2}{3} {x}^{4} {y}^{-} 4 {z}^{-} 3 {x}^{-} 2 {y}^{3} {z}^{-} 4$

and rearrange terms and group like terms:

$\frac{2}{3} \left({x}^{4} {x}^{-} 2\right) \left({y}^{-} 4 {y}^{3}\right) \left({z}^{-} 3 {z}^{-} 4\right)$

We can now use the rule ${x}^{a} \times {x}^{b} = {x}^{a + b}$

$\frac{2}{3} \left({x}^{4 - 2}\right) \left({y}^{- 4 + 3}\right) \left({z}^{- 3 - 4}\right)$

$\frac{2}{3} {x}^{2} {y}^{-} 1 {z}^{-} 7$

and now let's write this using only positive exponents:

$\frac{2 {x}^{2}}{3 y {z}^{7}}$