Solving Absolute Value Inequalities

From Magnitude Bounds to Inequalities

An absolute value inequality restricts how large or how small a number’s magnitude can be. The previous tutorial used the reverse triangle inequality to compare magnitudes. Here we turn magnitude comparisons into conditions on the number itself, then use ordinary order rules to solve inequalities involving expressions such as \(|2x-3|\).

The key distinction is whether the absolute value is bounded above or below. A bound above confines a number between two endpoints. A bound below places it outside those endpoints. This distinction determines whether the solution is one interval or two separated regions.

We will use the Absolute-Value Bound Criterion from Absolute Value: for \(r\geq0\), \(|x|\leq r\) is equivalent to \(-r\leq x\leq r\). The strict and lower-bound versions below extend that idea. The distance result from Absolute Value as Distance gives the same interpretation for expressions of the form \(|x-a|\).

Bounds Above: Staying Between Two Endpoints

Proof. The non-strict statement is the Absolute-Value Bound Criterion established earlier. For the strict statement, first suppose \(|x|<r\). The basic properties of absolute value give \(-|x|\leq x\leq|x|\), so

The strict outer comparisons imply \(-r<x<r\). Conversely, suppose \(-r<x<r\). If \(x\geq0\), then \(|x|=x<r\). If \(x<0\), then \(|x|=-x\), and \(-r<x\) implies \(-x<r\) by order reversal under negation. Thus \(|x|<r\) in either case. This proves the equivalence. \(\square\)

The strict criterion requires \(r>0\), while the non-strict criterion allows \(r=0\). At \(r=0\), the non-strict inequality \(|x|\leq0\) holds exactly when \(x=0\); the strict inequality \(|x|<0\) has no solutions because absolute values are nonnegative.

The same reasoning works when the expression is centered at a number \(a\). The distance theorem from Absolute Value as Distance states that, for \(r\geq0\), \(|x-a|\leq r\) is equivalent to \(a-r\leq x\leq a+r\). With \(r>0\), the strict version is \(a-r<x<a+r\). Thus a distance bound describes the numbers within the specified distance of \(a\).

Bounds Below: Being Outside Two Endpoints

Proof. First consider the strict comparison. If \(|x|>r\), then either \(x\geq0\) or \(x<0\). When \(x\geq0\), \(|x|=x\), so \(x>r\). When \(x<0\), \(|x|=-x>r\), and order reversal under negation gives \(x<-r\). Hence one of the two stated alternatives holds. Conversely, if \(x>r\), then \(x>0\) and \(|x|=x>r\). If \(x<-r\), then \(x<0\) and \(|x|=-x>r\). Therefore either alternative implies \(|x|>r\).

For the non-strict comparison, suppose \(|x|\geq r\). If \(x\geq0\), then \(|x|=x\geq r\), so \(x\geq r\). If \(x<0\), then \(|x|=-x\geq r\), so \(x\leq-r\). Conversely, if \(x\geq r\), then \(x\geq0\) and \(|x|=x\geq r\); if \(x\leq-r\), then \(x\leq0\) and \(|x|=-x\geq r\). This proves both directions. The alternatives also cover \(r=0\): every real number satisfies \(x\leq0\) or \(x\geq0\), and every nonzero real number satisfies \(x<0\) or \(x>0\). \(\square\)

The word “or” is essential: a lower bound permits values on either side, not just values to the right. The result therefore usually gives two separate solution regions. For a shifted expression \(|x-a|\), the corresponding regions lie at least \(r\) units to the left or right of \(a\).

Coefficients, Thresholds, and Common Pitfalls

When solving a two-sided inequality, every operation must preserve the order. Adding or subtracting the same number preserves both comparisons. Multiplying or dividing by a positive number preserves their direction; multiplying or dividing by a negative number reverses their direction. For an exterior inequality, solve each of the two cases separately and apply the same rules within each case.

For example, if the coefficient of \(x\) is negative, the endpoints can be easy to reverse accidentally. Starting from \(|-2x+1|<3\), the inside-bound criterion gives \(-3<-2x+1<3\). Subtracting \(1\) gives \(-4<-2x<2\). Dividing by \(-2\) reverses both comparisons, yielding \(2>x>-1\), or equivalently \(-1<x<2\). The final form lists the smaller endpoint first.

It is also important to check the threshold before applying a criterion. Absolute values are always nonnegative, so thresholds below zero have immediate consequences. The following cases summarize what happens for real \(x\):

These cases follow directly from \(|x|\geq0\). For instance, if \(r<0\), then \(|x|>r\) for every \(x\), since a nonnegative number is greater than a negative one. At \(r=0\), however, \(|x|>0\) excludes \(x=0\), whereas \(|x|\geq0\) includes every real number. Strict and non-strict signs cannot be interchanged at the boundary.

A frequent mistake is to treat \(|x|>r\) as though it meant \(x>r\) alone. That discards all sufficiently negative numbers. For example, \(x=-10\) satisfies \(|x|>3\), even though \(-10\) is not greater than \(3\). The two-case form \(x<-r\) or \(x>r\) retains both possibilities. Conversely, an upper bound is a between condition, not an outside condition.

After solving, it is useful to test an interior point, an exterior point, and any proposed endpoint in the original inequality. This does not replace the equivalence used in the solution, but it can reveal a reversed sign or an incorrectly included boundary. In particular, equality holds at the endpoints of a non-strict bound, while a strict bound excludes them.