Does $| a x + b | > c$ always result in two solutions, $x \gt \dfrac{c - b}{a}$, and $x \lt\dfrac{-c - b}{a}$?
If I understand correctly, the first solution, $x > \dfrac{c - b}{a}$, is only true for $x \gt -\dfrac{b}{a}$, while the second solution, $x \lt \dfrac{-c - b}{a}$, is only true for $x \lt -\dfrac{b}{a}$. Is this correct?
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Assuming $a\neq0\neq c$, if you want to immediately (graphically) see that there must be two half lines of solutions you can see it in this way: draw the straight line $r:y=ax+b$. When r goes down the $x$ axis, reverse $r$, as if the $x$ axis was a mirror (that's what $|\cdot| $ does). Now draw the straight line $y=c$. You see that there are two intersections (that's where $|ax+b|=c$) and the inequality holds exactly on the two half lines on the left (right) of the first (second) intersection.
Assume that $a\ne 0$.
The absolute value on the left is always non-negative. So if $c\lt 0$, the inequality holds for all $x$.
Let us assume then from now on that $c\ge 0$.
Then $|ax+b|\gt c$ if and only if $ax+b \gt c$ or $ax+b\lt -c$.
The inequality (i) can be rewritten as $ax\lt c-b$.
If $a\gt 0$, this is equivalent to $x\lt \frac{c-b}{a}$.
If $a\lt 0$, this is equivalent to $x\gt \frac{c-b}{a}$.
The inequality (ii) is handled in a similar way.
Remark: The algebra will take care of things correctly even if $c\lt 0$. But then the two intervals we get from our inequalities overlap, and cover the whole real line.