Question

7y+21/7y * 3/y^2-9

Answer 1

First you'll factor a 7 out of the first part making it:
7(y+3)/7y * 3/y^2-9
Then the 7s cancel and you get:
y+3/y*3/y^2-9
Factor the y^2-9 to get
y+3/y * 3/(y+3)(y-3)
Next just multiplication
3(y+3)/y(y+3)(y-3)
y+3 in Numerator and Denominator can be canceled.
3/y(y-3) or you can distribute and get 3/y^2-3

Answer 2

oops i meant 3/y^2-3y

Answer 3

Thanks!

Answer 4

No problem.