Solving modulus inequality $|x - 1| + |x - 6|\le11$ geometrically

As you said, we are looking for points for which:

$$(\text{distance to $1$})+(\text{distance to $6$})\leq 11.$$ The first thing which now comes to my mind is an ellipse.

We first consider all $x\in \mathbb C$ for which $\vert x-1\vert +\vert x-6\vert \leq 11$. All points of this "filled" ellipse which lie on the real axis are the ones we want.

It is not very hard to imagine what these point will be. If we are on the real line then our furthest left point, $x_\ell$, will lie left of $1$. So the distance to $6$ will be at least five. So we have $$\vert x_\ell-1\vert+\vert x_\ell -6\vert=2\vert x_\ell-1\vert+5=11. $$ And since $x_\ell $ is to the left of $1$ this means that $x_\ell=-2$. Analogously, we find that $x_r=9$.

So we have found that $-2\leq x\leq 9$, through geometrical methods.

$$|1-x| + |x-6| \le 11$$

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}

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