<s>[INST] Solving modulus inequality $|x - 1| + |x - 6|\le11$ geometrically Find all possible values of $x$ for which $x$ for which the inequality $$|x - 1| + |x - 6|\le11$$ is true.
I know this can be easily solved by taking $3$ cases for$x$ and then taking the intersection of those $3$ cases. The solution will be $-2\ge x\le9$.
But suppose if I interpret this in this way:

What number $x$ satisfy the condition that the distance between $x$ and $6$ plus the distance between $x$ and $1$ is less than or equal to $11$?

I would be better to get an idea to solve these types of problems by geometrically by intuition  using number line.
 [/INST] $$|1-x| + |x-6| \le 11$$
Let $A, X, B \in \mathbb R^n$ (Euclidean $n$-space). If $X\in \overline{AB}$ ( $X$ is on the line segment $\overline{AB}$ ), we say that $X$ is between $A$ and $B$, and we write this as $A-X-B$.
$$A-X-B \; \text{ if and only if } \; \|A-X\| + \|X-B\| = \|A-B\|$$
On $\mathbb R^1$ (the real number line), every point is on the line through points $1$ and $6$. So there are three possibilities for $1, x$, and $6$ : $x-1-6$, $1-x-6$, or $1-6-x$.
CASE: $x-1-6$
Then $ x \le 1 \le 6$
\begin{align}
  |1-x| + |x-6| &\le 11 \\
  (1-x) + (6-x) &\le 11\\
  -2x &\le 4\\
  x &\ge -2\\
  x &\in [-2,1]
\end{align}
CASE: $1-x-6$
Then $1 \le x \le 6$ and $|1-x| + |x-6| = |1-6| = 5$
\begin{align}
  |1-x| + |x-6| &\le 11 \\
  5 &\le 11 \\
  x &\in [1,6]
\end{align}
CASE: $1-6-x$
Then $1 \le 6 \le x$
\begin{align}
  |1-x| + |x-6| &\le 11 \\
  (x-1) + (x-6) &\le 11 \\
  2x &\le 18\\
  x &\le 9\\
  x &\in [6,9]
\end{align}

So $x \in [-2,9]$
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