$$|x|=y...(i)$$
$$\implies \pm x=y...(ii)$$
Can I write this? If not, what line can follow (i) that doesn't include a $||$ but includes a $\pm$?
I ask this because in desmos, the graphs of $|x|=y$ and that of $x=y$ & $x=-y$ aren't the same. The graphs of $x=y$ & $x=-y$ contain extra values. So, I figured that (ii) can't follow (i). Am I correct?
On
No. It is not true.
$$|x|=y\require{cancel}\hspace{1em} \cancel{\hspace{-1em}{\implies}\hspace{-1em}}\hspace{1em} y=±x$$
Because, if $x<0$ (or $x>0$) then you get
$$y=±(-x)=±x, ~x>0~\text{or} \\ y=±x, ~x>0$$
But, this is false. Because, $y≥0$ must be.
The correct result can be considered as follows:
$$|x|=y,y≥0\iff x=±y$$
Also, we can write
$$|x|=y\implies y=-x~\text{if}~x<0$$
and
$$|x|=y\implies y=x~\text{if}~x≥0.$$
In this answer, the definition of $±$ means $x ~~\text{«and»}-x.$
Writing $y=\pm x$ is a shorthand for "$y=x$ or $y=-x$".
It is true that $|x|=y\implies y=\pm x$.
But the other direction is false. The condition $y=\pm x$ may be met when $y$ is a negative number. And then it would be impossible for $|x|=y$.
This is precisely why the two Desmos graphs don't match. The two relations are not equivalent. It's only the case that one relation implies the other, so one graph is a subset of the other graph.