division of fraction simplification

Question

The expression is this:

${{y^2 - y} \over 1 {}} \div {{y^2 - 1} \over 3}$

The first step is to swap the second expression round to:

${{y^2 - y} \over 1 {}} \div {3 \over {y^2 - 1}}$

The answer in the text book is ${3y \over {y + 1}}$

I don't get how they got this answer I don't see how ${y^2 -y}$ can be cancelled out by ${y^2 - 1}$ as they are made up of different expressions of ${- y}$ and ${- 1}$.

Answer 1

Stolen from Michael Galuza's comment:

When you "swap around" the second term you should also make it a multiplication, not a division. You then get a common factor $y-1$ $$\frac{y^2-y}{1}\times\frac{3}{y^2-1}=\frac{3y(y-1)}{(y-1)(y+1)}=\frac{3y}{y+1}‌​$$

Answer 2

${{y^2 - y} \over 1 {}} \div {{y^2 - 1} \over 3}$
Now whenever you divide by something, you are essentially multiplying the reciprocal of the something.
So ${{y^2 - y} \over 1 {}} * {{3} \over y^2-1}$
$=\frac{y(y-1)}{1}*\frac{3}{(y-1)(y+1)}$
Cancel $y-1$ to get your required answer.