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?

  • $\begingroup$ It is not or, it is AND! If we have $|x|<1$ then $x<1$ and $x>-1$, not or. $\endgroup$ – Aqua Nov 9 '17 at 18:32
  • $\begingroup$ Agreed! $\left|x-4 \right| < \frac{4+2x}{-3}$ is definitely an AND situation - but tell me where the algebra is lacking? $\endgroup$ – Zediiiii Nov 9 '17 at 18:34
  • $\begingroup$ There is no lack. Since $x<8/5$ and $x>16$ there is no solution. $\endgroup$ – Aqua Nov 9 '17 at 18:36
  • 1
    $\begingroup$ When you got rid of absolute value bars, you forgot to add that in the first case $x \ge 4$ and in the second case $x < 4$. $\endgroup$ – Vasya Nov 9 '17 at 18:38
  • $\begingroup$ @Vasya: Not in this caes: the O.P. simply used that $|x-4|=\max(x-4, 4-x)$. $\endgroup$ – Bernard Nov 9 '17 at 18:52

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 $$-4-2x>-3(x-4)$$

  • $\begingroup$ Great response, and this gets at the heart of what I was missing. The definition of absolute value requires separate solutions for $x \geq 4$ and $x < 4$. Continuing shows that for $x \geq 4$ we have $x< \frac{8}{5}$ as mentioned, but since that is not within the restriction $x \geq 4$, it is not part of the solution. Likewise for $x<4$. $\endgroup$ – Zediiiii Nov 9 '17 at 18:39
  • $\begingroup$ Funny that of the 4 textbooks I use in my classroom, none of them mentions this rather important fact. Does anyone have a good resource recommendation that explains this well? $\endgroup$ – Zediiiii Nov 10 '17 at 15:47
  • $\begingroup$ have i not explained this well? $\endgroup$ – Dr. Sonnhard Graubner Nov 10 '17 at 15:49
  • $\begingroup$ see here analyzemath.com/Equations/Absolute_Value_Tutorial.html i hope this will help you $\endgroup$ – Dr. Sonnhard Graubner Nov 10 '17 at 15:51
  • $\begingroup$ Your answer was definitely sufficient and was accepted. Thanks! I get a frustrated when my bachelor's in mathematics has holes in it, (though they are inevitable in any major) because available resources gloss over details in the interest of better "understanding." ie, educators often skip depth in the interest of breadth - and one does not learn to swim well in the shallows. $\endgroup$ – Zediiiii Nov 11 '17 at 23:48

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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