Recalling that the absolute value of a real number means on the number line (real axis) the distance of the point from the origin, we have the following three rules which remove the absolute value signs from an inequality (we use the logical symbol ``$\lor$ '' for alternativeness `or'). Note that the symbols ``$\ge$ '' and ``$\le$ '' may also be without the equality bar.

1. $|a| \;\ge\; b\; \quad \Leftrightarrow \quad a \,\le\, -b \;\;\lor\;\;a \,\ge\, b$
2. $|a| \;\le\; b\; \quad \Leftrightarrow \quad -b \;\le\; a \;\le\; b$
3. $|a| \;\ge\; |b| \quad \Leftrightarrow \quad a^2 \;\ge\; b^2$

These rules are valid for all real values of $a$ and $b$ . For example, if one has a case $$|x| \;<\; -5$$ corresponding the rule 2, this inequality seems to be impossible since no absolute value is negative; but now also the result $-(-5) < x < -5$ given by the rule 2 is impossible -- no real number is simultaneously greater than $+5$ and less than $-5$ .

Examples. We solve some inequalities with absolute values.

a) $|2x\!+\!1| > 5x$
$2x\!+\!1 < -5x$ or $2x\!+\!1 > 5x$ (rule 1)
$7x < -1$ or $-3x > -1$
$x < -1/7$ or $x < 1/3$
$x < 1/3$ (combined)

b) $8|x|+|x\!-\!2| > 6$
$|8x| > 6-|x\!-\!2|$
$8x < -6+|x\!-\!2|$ or $8x > 6-|x\!-\!2|$ (rule 1)
$|x\!-\!2| > 8x\!+\!6$ or $|x\!-\!2| > 6\!-\!8x$
$x\!-\!2 < -8x\!-\!6$ or $x\!-\!2 > 8x\!+\!6$ or $x\!-\!2 < -6\!+\!8x$ or $x\!-\!2 > 6\!-\!8x$ (rule 1 twice)
$9x < -4$ or $-7x > 8$ or $-7x < -4$ or $9x > 8$
$x < -4/9$ or $x < -8/7$ or $x > 4/7$ or $x > 8/9$
$x < -4/9$ or $x > 4/7$ (from the number line)

c) $|1\!-\!5x| \le 3$
$-3 \le 1\!-\!5x \le 3$ (rule 2)
$-4 \le -5x \le 2$ (subtracted 1 from all parts)
$4/5 \ge x \ge -2/5$ (divided by $-5)$
$-2/5 \le x \le 4/5$ (rewritten from end to begin)