<s>[INST] Does |x| = |y| requires checking conditions while solving? I am trying to solve this equation $\lvert 2x \rvert = \lvert x - 2 + y \rvert$ (specifically, find set of all points $(x, y)$ satisfying equation). 
$\lvert 2x \rvert = \lvert x - 2 + y \rvert$ is equivalent to $2x = x - 2 + y \lor 2x = -(x - 2 + y)$. 
If $2x = x - 2 + y$, then I am assuming that $2x \ge 0$ and $x - 2 + y \ge 0$, but do I have to check if solutions meet those conditions? 
In general, while solving any equation with absolute values you have to check conditions, but sometimes you don't have to. 
For example,
$$\lvert\lvert x-1\rvert +5\rvert = 7$$ 
I can write it as $\lvert x - 1\rvert + 5 = 7$ or $\lvert x - 1\rvert + 5 = -7$. In this case, I don't have to check conditons.
 [/INST] The absolute value $\lvert x \rvert$ is defined as
$$
\lvert x \rvert =
\begin{cases}
x & \text{if } x \ge 0 \\
-x & \text{if } x < 0
\end{cases}
$$
So one has to treat two cases. Your example involves two such expressions, so up to $2\cdot 2 = 4$ cases can occur.


*

*$-,-$: $x < 0$, $x-2+y < 0 \iff y < 2 - x$. Equality condition is $-2x = -(x-2+y) \iff y = 2 + x$. Combining them gives the condition $2 + x < 2 - x \iff x < 0$. So this case gives the line segment $y = 2 + x$ for $x < 0$.

*$0+,0+$: $x\ge 0$, $x-2+y \ge 0  \iff y \ge 2 - x$. Equality condition is $y = 2 + x$ again. Combination: $2+x \ge 2-x \iff x \ge 0$. So the whole line $y = 2 + x$ is part of the solution.

*$-,0+$: $x<0$, $y \ge 2-x$. Equality: $-2x = x-2+y \iff y = 2-3x$. Combination: $2-3x \ge 2-x \iff 0 \ge 2x \iff x \le 0$. This case gives the line segment $y=2-3x$ for $x < 0$.

*$0+,-$: $x\ge 0$, $y<2-x$. Equality: $y=2-3x$. Combination: $2-3x<2-x\iff 0<2x \iff x > 0$. This gives the line segment $y=2-3x$ for $x > 0$.
Thus the solution are both lines $y = 2 + x$ and $y = 2-3x$.
Do you have to consider all cases like above? The alternatives shown here are to employ properties of $\lvert . \rvert$ to operate on a slightly more abstract level (mathlove's answer) or to manipulate the logical conditions in a different order (Graham's answer). Of the above I prefer that I know I have handled all possible cases.
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