I thought of the equation (x-2)^2/(x-2)=0
Normally to solve it, I would cancel out (x-2) and then end up with x-2=0. I then would solve it by saying x=2.
However, I realized that at the same time x cannot be 2 due to the reciprocal statement. The question is whether if it is valid to divide straight away by the gcd so that the answer is still valid as x=2, or if we have to preserve the number which x cannot be, and then divide by gcd to get x=2, but then re-state that since x cannot be 2, there is actually no solutions.
but $$\frac{(x-2)^2}{x-2}$$ is only defined for $$x\ne 2$$