Can Extraneous Roots be Introduced by Elimination? Suppose you have two equtaions:
$$2xy + y^2 = 0$$
$$x^2 + 2xy + 1 = 0$$
Subtracting the second from the first yields $y^2 - x^2 - 1 = 0$. Isolating y, we discover that $y = \pm\sqrt{x^2 + 1}$. However, by inspection we can wee that the $2xy$ term in both equations must be negative, which means that a single value of x cannot have both a positive and negative corresponding y-value. (i.e. if x is positive, then y must be negative). It seems that the equation $y = \pm\sqrt{x^2 + 1}$ contains an "extraneous root", but I'm struggling to wrap my mind around how that could be. After all, I solved for y w.r.t. x without squaring both sides. Could anyone help me understand what is going on?
 A: The problem is that your reasoning isn't reversible. Your two equations together imply $y^2-x^2-1=0$, but the converse is not true: that equation does not imply your original system.
Compare your question to the following argument. The system $$x+y=1$$ $$2x+y=1$$
yields, by subtraction, that $x=0$. But this second equation allows $y$ to be arbitrary! So we get a whole lot of "extraneous solutions": $(0,y)$, for any $y$. The problem, of course, is that the equation $x=0$ does not, in turn, imply the original system. I take it you do not find this situation puzzling.
You generally lose information when you replace a system of equations with a linear combination of them. The combination will include the solutions of your original equation, but it likely will include non-solutions, as well. 
The form of the reasoning is: "any solution of the original equations will be a solution of this equation, too." Yes, but then you've only found a superset containing the solution set. To characterize the solution set exactly, you must worry about the converse.
A: When you solve a system of equations (whether it be by elimination or other means), you cannot discard the individual equations you started with. The reason is that each one of these equations carry more information (restrictions for example) than the one equation you end up with.
This goes for other example as well. The function $f(x) = \sqrt{x} - \sqrt{x}$ is not equal to $0$. The reason being any $x < 0 $ is not in the domain of $f$. By eliminating $\sqrt{x}$ , you have lost a key piece of information about this function (its domain).
A: I think the issue here is at least partially that you're assuming the rest of the solution will go something like:

We know that $y=\pm\sqrt{x^2+1}$. We'll use this to find the possible values for $x$, then substitute in for $y$.

and this is missing something: When we go to solve for $x$, we're going to need to know whether $y$ is positive or negative. That is, we need to split into cases $y=\sqrt{x^2+1}$ or $y=-\sqrt{x^2+1}$ and then solve for $x$ by substituting in $y$. This is to say that when you write:

After all, I solved for y w.r.t. x without squaring both sides.

you're missing something. You didn't solve for $y$. You reduced one statement to saying that $y$ was among two possibilities. You have to proceed to the end of the argument with each possibility separate, since you can't substitute "either this or that" in for $y$.
To be sure, what elimination does is we start with a system like
$$2xy+y^2=0$$
$$x^2+2xy+1=0$$
And then you've arranged into an equivalent statement:
$$y=\pm\sqrt{x^2+1}$$
$$x^2+2xy+1=0$$
- or whatever you take the second equation to be. This system is equivalent to the first since all of your steps are reversible (noting that you've correctly used $y=\pm\sqrt{z}$ as the equivalent to $y^2=z$). So, you aren't introducing extraneous roots - it's just that your formula for $x$ depends on the $\pm$, so you don't get to choose $x$, then choose the sign of $y$. Another way to say this is that your error is in this statement:
A more simple example might be a system like:
$$y^2=1$$
$$x+y=0$$
You get that $y=\pm 1$, but you can't really substitute that into the second equation itself since it depends on a variable $\pm$.
A: You have taken a square root. Whenever you take a square root you can introduce an extraneous root. Even for the simple system $x^2=y^2$, $x=y$ if you took a square root of the first equation you'd get an extraneous root. You must always consider all the equations as they all contain information about the solution. If you had gone about solving it via elimination or substitution (rather than merging the two equations together randomly like you did) then you would have not encountered this.
If you solve your equation fully (and use complex numbers) you get two pairs of roots:
$$\left(\frac{1}{\sqrt3},-\frac{2}{\sqrt3}\right), \left(-\frac{1}{\sqrt3},\frac{2}{\sqrt3}\right),(i,0),(-i,0)$$
The second pair of roots does fit your reduced equation (with a small rearrangement), namely $x=\pm\sqrt{y^2-1}$.
