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?
 A: If you consider the interval $[-3,3]$ (whose total length is $6$) and pick a point $x$ in that interval, then $|x-3|+|x+3|$ is the sum of the distances of this point from the two end-points $\pm 3$. But this sum is simply the length of the interval (which is $6$ as mentioned above) hence every point in the interval is a solution.
If you pick $x$ outside the interval $[-3,3]$, then it's distance from one of the end points (depending on which side $x$ is) will already be greater than $6$, hence the equality cannot occur for points outside the interval.
A: More mechanically: for each of the three ranges $x<-3$, $x\in[ -3, 3]$, and $x>3$, simplify the expressions involving absolute value signs.  For instance, when $x<-3$, you have $|x+3|+|x-3| = (-x-3) + (-x +3) = -2x$.  On $[-3,3]$ you have $|x+3| = x+3$ but $|x-3| = 3-x$, so $|x+3|+|x-3| = 6$ there.  And similarly for $x>3$, resulting in $2x$.
A: If $x<-3$:
$$|x+3|+|x-3|=-(x+3)-(x-3)=-2x>6\implies\text{no solutions}$$
If $x>3$:
$$|x+3|+|x-3|=x+3+x-3=2x>6\implies \text{no solutions}$$
If $-3\leq x\leq 3$:
$$|x+3|+|x-3|=x+3-(x-3)=6\implies \text{true}$$
A: $$|x+3|=\begin{cases}
x+3, & \text{ if } x\geq -3\\
-x-3,& \text{ if } x<-3
\end{cases}$$
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 $[-3,3].$
A: If you want to solve this kind of equations more or less methodically, you have to consider in separate cases the possible signs of the expressions inside the absolute value.


*

*Let $x>3$.


Then $x-3>0$ and $x+3>0$, so $|x+3|+|x-3|=(x+3)+(x-3)=2x$, and the only solution to $|x+3|+|x-3|=6$ is $2x=6$, or $x=3$. But this is outside the range $x>3$, so it must be disregarded.


*Let $x<3$.


Now both expressions are negative and, as before, there is no solution.


*Let $x\in[-3,3]$.


Then $x+3\ge0$ and $x-3\le 0$, so $|x+3|+|x-3|=(x+3)-(x-3)=6$. In other words, for every point in this interval the equation holds, so the solution is the interval $[-3,3]$.
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A: By triangle inequality we have:
$$|x+3|+|x-3| = |x+3|+|3-x|\geq |x+3+3-x| = 6$$
and equality is achived when $x+3$ and $3-x$ are of equal sign. 
So $x+3\geq 0$ and $3-x\geq 0$ so $x\in [-3,3]$ 
or  $x+3< 0$ and $3-x< 0$ which is impossible. 
