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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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The system does not like long strings of comments. I will delete all of mine. There is no real trouble when $a$ is negative, as long as you reverse all the inequalities. –  André Nicolas Feb 8 '13 at 19:51
    
@AndréNicolas But for $a > 0$, my two solutions are always valid, right? –  Šime Vidas Feb 8 '13 at 19:53
    
Yes, they are fine. In my answer I have also dealt with$a$ negative. Then, we need to reverse your end-inequalities, replace $\lt $ by $\gt$. –  André Nicolas Feb 8 '13 at 20:04
    
If c < 0 is a special case to handle here. –  JB King Feb 8 '13 at 22:01

2 Answers 2

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.

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Great way to make sense out of this :-) –  Šime Vidas Feb 8 '13 at 21:56

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.

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Hm, should inequality (i) be $ax > c - b$? So, $>$ instead of $<$. –  Šime Vidas Feb 8 '13 at 20:17
    
And, one quick side question: Can stuff like $|3-x| > 1$ be transformed into $|x-3| > 1$, i.e. are those two inequalities equivalent? –  Šime Vidas Feb 8 '13 at 20:19
    
Sure, $|-(3-x)|=|3-x|$. In general, $|-u|=|u|$. –  André Nicolas Feb 8 '13 at 20:32

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