10.2 Absolute Value Inequalities

Recall that \(|x|= \begin {cases} x & \text {if}\quad x\geq 0\\\\ -x & \text {if}\quad x<0\\ \end {cases} \)

Consider solving the inequality \(|x|<k\) where \(k\) is positive real number.
There are two cases to consider either \(x\) is positive or \(x\) is negative.

If \(x>0\) then \(x<k\) is the solution set.
Since \(|x|=x\).
If \(x<0,\) then \(|x|=-x\) so that \(|x|<k \implies -x<k\implies x>-k\).

Hence the two solutions imply that \(-k<x\) and \(x<k\).

Thus, \(|x|<k\) if and only if \(-k<x<k\).

Both conditions must hold at once, so the solution is their intersection.

Take the intersection of the two solutions: \((-k,k)=\{x\in \mathbb {R}:-k<x<k\}\)

Theorem 10.3 (Absolute value inequalities). For \(k>0\), \[\left |x\right |<k \iff -k<x<k,\] \[\left |x\right |>k \iff x<-k \quad \text {or}\quad x>k .\] The same holds with \(\leq \) and \(\geq \) throughout.

Note 10.4. The two cases behave oppositely and this is the single most common error in the chapter. The “less than” case gives one interval, joined by and; the “greater than” case gives two, joined by or. Writing \(-k>x>k\) for the second is meaningless — no number is both less than \(-k\) and greater than \(k\).

The distance reading settles it without memorising. \(\left |x\right |\) is how far \(x\) is from \(0\). Points closer than \(k\) form the single stretch around the origin; points further than \(k\) form two pieces, one at each end. More generally \(\left |x-a\right |<k\) is the interval of radius \(k\) centred at \(a\).

Note also that \(k\) must be positive for the theorem to say anything. If \(k<0\) then \(\left |x\right |<k\) has no solution at all, since an absolute value is never negative, and \(\left |x\right |>k\) is satisfied by every real \(x\).

Example 10.5.

Find the solution set of \(|2x-3|\leq 5\).

Solution.

\(|2x-3|\leq 5\implies -5\leq 2x-3\leq 5\)

two sections

\(-5\leq 2x-3\cdots (1)\)

\(2x-3\leq 5\cdots (2)\)

Solve the two inequalities and take the intersection of their solution sets:

Now \begin {align*} -5 &\leq 2x-3\\ \implies -5+3 & \leq 2x\\ \implies -2 & \leq 2x\\ \implies -1 & \leq x \end {align*}