Question

How do you solve `(y+2)/y=1/(y-5)`?

Answer 1

First, note that `0` and `-5` cannot be the answers.

Then, multiply by `y*(y-5)` :

#(y+2)/y = 1/(y-5) <=> (y-5) (y+2) = y#

`<=> y^2 -4y -10 =0`

And you get a simple quadratic equation.