No of solutions for an equation with variable $X$

Question

IF $x-5=0$ then $x=5$ (One solution $x=5$) When multiplying $x$ to both sides we get (In a balance if you did same thing on both sides equation will not change) $x^2 -5x=0\cdot x$ (anything multiplied by $0$ is $0$) $$x^2-5x=0$$ Here $x$ has two solutions ($x=0$ and $x=5$) HOW WOULD THIS HAPPEN?

Answer 1

Because if you multiply anything by zero, the answer is zero, so you can't reverse the process and divide by zero. You have either $x=0$, which doesn't depend on what the original equation was, or the original equation holds.

You can't "do anything" to an equation and keep the same solutions. You have to do something reversible. Multiplying by zero is not reversible.

Answer 2

You're multiplying two things and getting $0:$ $$ x(x-5) = 0. $$ That means either $x=0$ or $x-5=0.$