$||x-1|-2|=|x-3|$.
Find the value of $x$.
In my attempt I got the critical values of the expression as $1$ and $3$.
But I’m not sure is we can just not consider the $2$ in the LHS.
My steps are Case 1: when $x>3$, $x-1-2=x-3$, $0=0$
Case 2: $1<x<3$, $X-1-2=-x+3 \implies 2x=6 \implies x= 3$
Case 3: $x<1$, $-x+1-2=-x+3$. Not possible.
Just sketch the graphs $y=||x-1|-2|$ and $y=|x-3|$.
To draw $y=|x-3|$, first draw $y=x-3$ then reflect the negative $y$-region up over the $x$-axis. (In other words, fold the graph up over the $x$-axis.) Leave the positive $y$-region as it is.
To draw $y=||x-1|-2|$, first draw $y=x-1$, then fold it up over the $x$-axis, slide the whole thing down by 2, then fold it up over the $x$-axis.
It should be quite obvious that $x \ge 1$ solves $||x-1|-2|=|x-3|$.
You must do case work. If $$x\geq 1$$ then we get $$\vert{x-3\vert}=|x-3|$$ which is true. If $$x<1$$ then we have $$|-x+1-2|=|x-3|$$, can you solve this equation?
Guide:
You have found the critical value to be $1$ and $3$, that's great, don't worry too much and work it out and see what needs to be done.
Consider $x<1$:
$$||x-1|-2|<|x-3|$$
$$|(1-x)-2|<3-x$$
$$|-1-x|<3-x$$
$$|x+1|<3-x$$
Now, as you see, you might need a new condition, just impose it and handle things as it comes by, now consider $x<-1$ and $x>-1$.
Move on to consider $1<x<3$, $x>3$, $x=1$, $x=3$ as well.
