Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$ I'm solving the equation,
$$|x-1| + |x-2| = 1$$

I'm making cases, 

$C-1, \, x \in [2, \infty) $ 

So, $ x-1 + x-2 = 1 \Rightarrow x= 2$

$C-2, \, x \in [1, 2) $

$x-1 - x + 2 = 1 \Rightarrow 1 =1 \Rightarrow x\in [1,2) $

$C-3, \, x \in (- \infty, 1)$

$ - x + 1 - x+2 = 1 \Rightarrow x= 1 \notin (-\infty, 1) \Rightarrow x = \phi$ (null set)

Taking common of all three solution set, I get $x= \phi$ because of the last case. But the answer is supposed to be $x \in [1,2]$

But when I write this equation in graphing calculator, it shows $2$ lines $x=1$ and $ x= 2$ rather than a region between $[1,2]$



Someone explain this too?
Case 1 tells you, that the only $x$ in $[2,\infty)$ satisfying the equation is $x=2$.

Case 2 tells you, that all $x$ in $[1,2)$ satisfy the equation.

Case 3 tells you that no $x$ in $(-\infty,1)$ satisfy the equation.

So the union of those, $[1,2]$ is the set of all $x$ satisfying the equation.

In other words: Let $x \in \mathbb{R}$ satisfy the equation. Then either C1) $x\in [2,\infty)$ or C2) $x\in [1,2)$ or C3) $x\in (-\infty,1)$. You tackle all three cases, but $x$ only has to fulfill one of the three cases.

Apparently, desmos graphing calculator shows a wrong plot. See two other examples. The green curve is plottet the right way, the red one is obviously false.
By direct way we need to distinguish three cases


$x&lt;1 \implies |x-1| + |x-2| = 1-x+2-x=1 \implies -2x=-2\implies x=1$
$1\le x&lt;2 \implies |x-1| + |x-2| = x-1+2-x=1 \implies 1=1$
$x\ge 2 \implies |x-1| + |x-2| = x-1+x-2=1 \implies 2x=4\implies x=2$


therefore $1\le x\le 2$.