Answer
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 $x \ne 3$.
Work Step by Step
$a) \frac{5}{y}.\frac{3}{y^3}$
$=\frac{5*3}{y.y^3}$
$=\frac{15}{y^4}$
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 $x \ne 3$.
