Question

One more help please. Thanks!

Answer 1

First we start out by factoring everything we can so our original problem changes to become : $\frac{y^2}{y-3}$ x $\frac{(y-3)(y+2)}{y(y+1)}$
Next we cancel terms so since to have a (y-3) on top and the bottom they cancel out leaving you with : $\frac{y^2(y+2)}{y(y+1)}$
Again with a lone y on top and bottom we cancel terms leaving your answer as : $\frac{y(y+2)}{y+1}$

Answer 2

After factoring this gives

 y ( y + 2)              y^2 + 2
------------  =           ---------   y not equal  3 , 0 or -1
   y + 1                   y + 1


Its B