## Algebra 1: Common Core (15th Edition)

a) The product is $\frac{15}{y^4}$ where $y \ne 0$ b) The product is $\frac{x(x+1)}{(x+1)(x-3)}$ where $x \ne -1$ and $x \ne 3$.
$a) \frac{5}{y}.\frac{3}{y^3}$ $=\frac{5*3}{y.y^3}$ $=\frac{15}{y^4}$ The product is $\frac{15}{y^4}$ where $y \ne 0$ $b) \frac{x}{x-2}.\frac{x+1}{x-3}$ $=\frac{x(x+1)}{(x+1)(x-3)}$ The product is $\frac{x(x+1)}{(x+1)(x-3)}$ where $x \ne -1$ and $x \ne 3$.