# inequality with absolute values

Recalling that the absolute value     (http://planetmath.org/AbsoluteValue) 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 “$\geq$” and “$\leq$” may also be without the equality bar.

1. 1.

$|a|\;\geq\;b\;\quad\Leftrightarrow\quad a\,\leq\,-b\;\;\lor\;\;a\,\geq\,b$

2. 2.

$|a|\;\leq\;b\;\quad\Leftrightarrow\quad-b\;\leq\;a\;\leq\;b$

3. 3.

$|a|\;\geq\;|b|\quad\Leftrightarrow\quad a^{2}\;\geq\;b^{2}$

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

Title inequality with absolute values InequalityWithAbsoluteValues 2013-03-22 16:57:20 2013-03-22 16:57:20 pahio (2872) pahio (2872) 7 pahio (2872) Topic msc 97D40 AbsoluteValue AbsoluteValueInequalities OrderOfSixMeans