Range of $|x+3|+|x-3|=6$ I can solve this equation by som tedious algebra, I got $x_1=3$ and $x_2=-3$. But according to the book the solutions are given by $x\in[-3,3]$, which means that for example $x=1$ and $x=2$ are solutions as well. How can I algebraically show this? Or can I interpret the absolutes as distances along the x-axis and somehow proceed from there?

$$|x+3|=\begin{cases}
x+3, & \text{ if } x\geq -3\\
-x-3,& \text{ if } x<-3
\end{cases}$$
and 
$$|x-3|=\begin{cases}
x-3, & \text{ if } x> 3\\
-x+3,& \text{ if } x\leq3
\end{cases}$$
Thus,
$$|x+3|+|x-3|=\begin{cases}
2x, & \text{ if } x> 3\\
6, &\text{ if }-3 \leq x \leq3\\
-2x,& \text{ if } x<-3
\end{cases}$$
When $x>3$, then $2x>6.$ Similarly if $x<-3,$ then $-2x>6.$
Thus, the set of solutions is $[-3,3].$
