Question

How do you simplify `(2y-6)/(3y^2-y^3)`?

Answer 1

`color(green)(-2/y^2)`

Explanation:

Note that `3y^2-y^3=-y^3+3y^2`

`color(white)("XXX")=(-y^2)(y-3)`

`color(white)("XXX")=(-y^2)(((2y-6))/2)`

`color(white)("XXX")=(-(y^2)/2)(2y-6)`

Therefore
`color(white)("XXX")(2y-6)/(-y^3+3y^2)=(cancel(2y-6))/(-y^2/2(cancel(2y-6)))=-2/y^2`

Answer 2

`=-2/y^2`

Explanation:

`(2y-6)/(3y^2-y^3)" "` factorise as far as possible

`=(2(y-3))/(y^2(3-y))`

Notice that the brackets are the same except for the signs.
Do a switch round:

`color(red)((3-y) = -(-3+y) = -(y-3))`

`=(2(y-3))/(y^2color(red)((3-y))`

`=(2(y-3))/(-y^2color(red)((y-3))`

`=(2cancel((y-3)))/(-y^2cancel((y-3))`

`=-2/y^2`