Help to solve equation with absolute value. Solve the equation $\mid \mid x \mid - 1 \mid + \mid x-2 \mid = 2$, please can I get help with this? Should I divide it into cases ?
 A: Hint: Yes, you should. Observe that $$|x-2|=\begin{cases}x-2, & \text{ if } x\ge 2\\ 2-x, & \text{ if } x<2 \end{cases}, \qquad |x|=\begin{cases}x, & \text{ if } x\ge 0\\ -x, & \text{ if } x<0 \end{cases}$$ and finally $$||x|-1|=\begin{cases}|x-1|, & \text{ if } x\ge 0\\ |-x-1|, & \text{ if } x<0 \end{cases}=\begin{cases}x-1, & \text{ if } x\ge 1\\ -x+1, & \text{ if } 0\le x<1 \\ x+1,& \text{ if } -1\le x<0 \\ -x-1, & \text{ if } x<-1 \end{cases}$$
Now, see if you can take together some cases and proceed for each case separately. In the end you have to check whether the solution belongs to the interval you started with. Otherwise it is not acceptable. 
A: Yes, you should divide it into cases. However, if you think of the problem geometrically on the number line, thinking of $|a-b|$ as the distance between $a$ and $b$ on the number line, you need only two cases.
If $x\ge 0$ then $|x|=x$, and the equation becomes
$$|x-1|+|x-2|=2$$
which means the sum of the distances from $x$ to $1$ and to $2$ equals $2$. If you think of this geometrically, there are only two answers, both of them obvious. Namely, $x=\frac 12$ and $x=\frac 52$, where the distances are $\frac 12$ and $\frac 32$.


If $x<0$ then $|x|=-x$, $||x|-1|=|-x-1|=|x+1|=|x-(-1)|$ and the equation becomes
$$|x-(-1)|+|x-2|=2$$
which means the sum of the distances from $x$ to $-1$ and to $2$ equals $2$. If you think of this geometrically, it is clear that this sum must be at least $3$, so there is no solution in this case.

Therefore the solution set is
$$x\in \left\{\frac 12, \frac 52 \right\}$$
A: i will try to explain the solution geometrically first before using algebra.
(a) the graph of $y = |x-2|$ is $v$-shaped with the vertex at $(2,0)$ with rays going out with the slopes $\pm 1.$
(b) the graph $y = |\, |\, x\, | - 1|$ is $w$-shaped with vertices at $(-1,0), (0,1)$ and $(1,0)$
(c) both functions are nonnegative and at least one of them is $\ge 2$ for $x \le 0$ and $x \ge 3.$ this implies the solutions are in the interval $(0, 3).$ 
(d) the graph $y = |x-2| + |\, |\, x\, | - 1|$ made up of line segments with possible corners at $x = -1, 0, 1$ and $2$
(e) we will make a table of values 
$$\begin{array}{|c|c|c|c|} \hline
x&0&1&2&3 \\ \hline
y&3&1&1&3\\\hline
\end{array} $$
you can see that at the midpoint $x = \frac 12, y = 2$ on the lines connecting $(0,3)$ and $(1,1)$ and again at the midpoint $x = \frac 52, y = 2$ on the lines connecting $(2,1)$ and $(3,3)$
there are two solutions $x = \frac 12, \frac 52$ for $|x-2| + |\, |\, x\, | - 1| = 2.$
