# 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.

• This question overlaps with yours : math.stackexchange.com/questions/1515492/… – arthur Nov 16 '15 at 3:09
• The reason why $f(x)=\frac{x(2x+1)}{x}$ and $g(x)=2x+1$ are considered to be different is because $f(0)$ is not defined and technically has a hole there (you are not allowed to divide by zero) whereas $g(0)$ is perfectly well defined. You do have that $f(x)=g(x)$ almost everywhere but due to that slight difference, we still say $f\neq g$. – JMoravitz Nov 16 '15 at 3:14
• The "reversible/irreversible" language may be nonstandard (at least I've not heard it before). Hence the difficulty searching for it. Are these terms from your textbook? – Daniel R. Collins Nov 16 '15 at 4:27