Consider the real numbers $a$ and $b$ such that the equation in $x$ $$| x - 3 | + | x - 6 | = a x + b \tag{1}$$ has a solution. Set the left side of (1) as $y = | x - 3 | + | x - 6 |$. This can be represented without using the absolute value signs in the following way. $$\begin{array} { l l }
\text { If } x < \mathbf { A } , & \text { then } y = - \mathbf { B } x + \mathbf { C } ; \\
\text { if } \mathbf { A } \leqq x < \mathbf { D } , & \text { then } y = \mathbf { E } ; \\
\text { if } \mathbf { D } \leqq x , & \text { then } y = \mathbf { F } x - \mathbf { G } .
\end{array}$$ Next, let us consider the common point(s) of the graph of this function and the straight line $y = a x + b$ on the $x y$-plane. Then we see the following: (i) If $a = 1$, then the range of the values of $b$ such that (1) has one or more solutions is $$b \geqq \mathbf { H I } .$$ (ii) If $b = 6$, then the range of the values of $a$ such that (1) has two different solutions is $$\mathbf { J K } < a < \mathbf { L }.$$
Consider the real numbers $a$ and $b$ such that the equation in $x$
$$| x - 3 | + | x - 6 | = a x + b \tag{1}$$
has a solution.
Set the left side of (1) as $y = | x - 3 | + | x - 6 |$. This can be represented without using the absolute value signs in the following way.
$$\begin{array} { l l }
\text { If } x < \mathbf { A } , & \text { then } y = - \mathbf { B } x + \mathbf { C } ; \\
\text { if } \mathbf { A } \leqq x < \mathbf { D } , & \text { then } y = \mathbf { E } ; \\
\text { if } \mathbf { D } \leqq x , & \text { then } y = \mathbf { F } x - \mathbf { G } .
\end{array}$$
Next, let us consider the common point(s) of the graph of this function and the straight line $y = a x + b$ on the $x y$-plane. Then we see the following:
(i) If $a = 1$, then the range of the values of $b$ such that (1) has one or more solutions is
$$b \geqq \mathbf { H I } .$$
(ii) If $b = 6$, then the range of the values of $a$ such that (1) has two different solutions is
$$\mathbf { J K } < a < \mathbf { L }.$$