Absolute Value Inequality Domain Restrictions?
I'm almost embarrassed to post this elementary question, but I beg your indulgence.
For the inequality $$-3 \left| x-4 \right| +2 > 6 + 2x$$
a graphical examination immediately shows no solutions. However, solving this algebraically shows a solution that is the union of two intervals.
For the sake of completeness, I'll walk through it, though I will limit comments.
$$-3 \left| x-4 \right| > 4 + 2x$$
$$ \left| x-4 \right| < \frac{4+2x}{-3}$$
Which yields two inequalities
$$ +\left( x-4 \right) < \frac{4+2x}{-3}$$
and
$$ -\left( x-4 \right) < \frac{4+2x}{-3}$$
Solving the first
$$ x-4 < \frac{4+2x}{-3}$$
$$ -3x+12 > 4+2x$$
$$ 8 > 5x$$
$$ x<\frac{8}{5}$$
And the second
$$ -x+4 < \frac{4+2x}{-3}$$
$$ 3x-12 > 4+2x$$
$$ x > 16$$
so our algebraic solution is $x<\frac{8}{5}$ OR $ x > 16$. But this is obviously false based on the graph. Is there an absolute value domain restriction somewhere I'm missing?
2 Answers
$\begingroup$HINT: your inequality is equivalent to $$-4-2x>3|x-4|$$ and you have two cases: $$x\geq 4$$ and we get $$-4-2x>3(x-4)$$ or $$x<4$$ and we get $$-4-2x>-3(x-4)$$
$\endgroup$ 7 $\begingroup$The definition of the absolute value of a real number (or expression) $z$ is
$$|z| = \left\{ \begin{array}{ll} z & \mbox{if } z \geq 0 \\ -z & \mbox{if } z < 0 \end{array} \right. $$
Solving the equation in the OP for the absolute value portion yields $$ \left| x-4 \right| < \frac{4+2x}{-3}$$ Based on the definition of absolute value, $z$ is the "stuff" inside the absolute value bars: $z=x-4$. Applying the definition of aboslute value we have
$$|x-4| = \left\{ \begin{array}{ll} x-4 & \mbox{if } x-4 \geq 0 \\ -(x-4) & \mbox{if } x-4 < 0 \end{array} \right. $$
From this we see that for $x-4$, $x \geq 4$ and for $-(x-4)$, $x<4$.
Adding these two constraints to the algebraically derived solutions in the OP explains why there are actually no solutions. Specifically, $x<\frac{8}{5}$ AND $x \geq 4$ is impossible - likewise for $x>16$ AND $x<4$.
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