Cases of reversible and irreversible operations in algebra I am in high school algebra, solving typical equations such as rational, irrational, quadratic, etc., and I have come across the idea of extraneous solutions. My textbook does not touch upon the idea of extraneous solutions and how they relate to reversible and irreversible operations, and I can't find much online. Could I get an explanation of what exactly reversible and irreversible operations mean? I know that squaring is one, and multiplying both sides of an equation by $0$ is another, but why is multiplying both sides by $x$ (assuming $x$ is the variable in the equation) an irreversible operation? Is it also irreversible if I start with $2x^2 = x$, and divide by $x$ to get $2x = 1$, in which case $0$ is no longer a solution? Why is it irreversible if I have $\displaystyle \frac{x(2x + 1)}{x} = 0$ and cancel out $x$ to get $2x + 1 = 0$? Finally, why is it irreversible if I have $2x + 1 = 0$ and multiply both sides to get $\displaystyle \frac{x(2x + 1)}{x} = 0$? These are the four cases I am most interested in. I want to understand them specifically, because I don't want my maths to be ambiguous.
 A: If you are given Equation B and the operation that resulted in Equation B (pretend this is all you know) and you can determine the exact equation in the previous line (call it Equation A), then this is a reversible step. For example, if I told you that I added 3 to both sides of Equation A to get x=5 (x=5 would be Equation B in my explanation), then you could reverse this operation by subtracting 3 from both sides to find that Equation A is x-3=2.
If you can't be sure what the line before is, then this is an irreversible step.  For example, if I told you that I squared both sides of Equation A to get x^2=4, there's no way of knowing exactly what Equation A is.  It could be either x=-2 or x=2.  Or if I told you that I multiplied both sides of an equation by zero to get 0=0, then you have no idea what the previous line is because it could be anything. These steps are irreversible.
A: What you call operations on equations are logical inferences. If you "operate" on $A=B$ to get $A'=B'$ then what you have actually done is assert that $A=B\implies A'=B'.$ Assertions must be justifiable by the basic laws of arithmetic and the laws of logic. Instead of "operating", write in the "$\implies$"or "$\iff$ and ask "How do I know this inference is valid?
It might  be  that $A'=B'$ does $not$ imply $A=B.$  Reversibility means you have $A=B\iff A'=B'.$
E.g. $x=2\implies x^2=2^2\iff x\in \{-2,2\}$ is logical, but $x^2=2^2$ does not imply $x=2.$
E.g. $\frac {x(2x+1)}{x}=0$ implies $x\ne 0$ because if $x=0$ then the number $0$ cannot be equal the non-existent  $\frac {0(2 \cdot 0+1)}{0}.$ Therefore $$\frac {x(2x+1)}{x}=0\implies [\,\frac {x(2x+1)}{x}=0\land x\ne 0 \,].$$ And you know from basic arithmetic that if $x\ne 0$ then $\frac {xy}{x}=y.$
For the other direction we have $2x+1=0\implies x\ne 0$ (Why?) so $$2x+1=0\implies [2x+1=0\land x\ne 0]\implies 0=\frac {x\cdot 0}{x}=\frac {x(2x+1)}{x}.$$
