Question

How do you solve `(2y) / (2y+6) + (9y-12) / (3y+9) = (9y+11) / (y+3)`?

Answer 1

Some of these fractions can be simplified.

Explanation:

`(2y)/(2y+6):2/2=y/(y+3)`

`(9y-12)/(3y+9):3/3=(3y-4)/(y+3)`

So now we have:
`y/(y+3)+(3y-4)/(y+3)=(9y+11)/(y+3)`

Since the numerators are all equal we may leave them out, on the condition `y!=-3` for this would make the numerators `=0` and the whole equation would be senseless.

`y+3y-4=9y+11->`

All the `y`'s to one side, all the numbers to the other:

`-5y=15->y=-3`

And this is exactly the solution that was forbidden.

Answer : no sensible solution.