## Algebra 2 (1st Edition)

$$\frac{8y}{x^3}$$
Using the rules of exponents, we find: $$\frac{48x^{-1}y^4}{6x^2y^3} \\ \frac{8y^4}{x^3y^3} \\ \frac{8y^{4-3}}{x^3}\\ \frac{8y}{x^3}$$