Inequalities involving absolute values are inequalities in which the unknown appears inside one or more absolute values. Solving them correctly requires more than simply memorising mechanical rules: one must understand what the absolute value actually means and rewrite each inequality in an equivalent form that is free of absolute value signs.
Recall first the definition of absolute value:
\[ |x|= \begin{cases} x, & \text{if } x\ge 0,\\ -x, & \text{if } x<0. \end{cases} \]
This definition states that the absolute value of any real number is always non-negative:
\[ |x|\ge 0 \qquad \text{for every } x\in\mathbb{R}. \]
Moreover:
\[ |x|=0 \iff x=0. \]
Table of Contents
- Geometric meaning of absolute value
- Inequalities of the form \(|A(x)|<k\)
- Inequalities of the form \(|A(x)|\le k\)
- Inequalities of the form \(|A(x)|>k\)
- Inequalities of the form \(|A(x)|\ge k\)
- What happens when the right-hand side is negative
- The definition method
- Inequalities with multiple absolute values
- Worked examples
Geometric meaning of absolute value
The absolute value has a fundamental geometric interpretation: it represents a distance on the real number line.
Specifically, \(|x|\) is the distance from the point \(x\) to the origin:
\[ |x|=d(x,0). \]
More generally, the expression:
\[ |x-a| \]
represents the distance between \(x\) and \(a\):
\[ |x-a|=d(x,a). \]
For instance, the inequality:
\[ |x-3|<2 \]
states that the distance between \(x\) and \(3\) must be less than \(2\). Therefore \(x\) must lie between \(3-2\) and \(3+2\), that is:
\[ 1<x<5. \]
This geometric reading is particularly useful because it makes the difference between "less than" and "greater than" inequalities immediately clear.
Inequalities of the form \(|A(x)|<k\)
Consider an inequality of the form:
\[ |A(x)|<k \]
Assume for now that:
\[ k>0. \]
Saying that \(|A(x)|<k\) means that \(A(x)\) must be within a distance \(k\) of \(0\). Hence \(A(x)\) must be strictly between \(-k\) and \(k\):
\[ |A(x)|<k \iff -k<A(x)<k. \]
An absolute value inequality of this type therefore reduces to a compound inequality.
Example. Solve:
\[ |2x-3|<5. \]
Since the right-hand side is positive, we write:
\[ -5<2x-3<5. \]
Adding \(3\) throughout:
\[ -2<2x<8. \]
Dividing by \(2\):
\[ -1<x<4. \]
The solution set is therefore:
\[ S=(-1,4). \]
Inequalities of the form \(|A(x)|\le k\)
When the symbol is \(\le\), the reasoning is identical. For \(k>0\), one has:
\[ |A(x)|\le k \iff -k\le A(x)\le k. \]
Here the endpoints are included, since the inequality also allows the absolute value to equal \(k\) exactly.
Example. Solve:
\[ |x-4|\le 2. \]
We write the equivalent inequality:
\[ -2\le x-4\le 2. \]
Adding \(4\):
\[ 2\le x\le 6. \]
Therefore:
\[ S=[2,6]. \]
Inequalities of the form \(|A(x)|>k\)
Now consider an inequality of the form:
\[ |A(x)|>k, \]
with:
\[ k>0. \]
Saying that \(|A(x)|>k\) means that \(A(x)\) must lie at a distance greater than \(k\) from \(0\). Thus \(A(x)\) must be either less than \(-k\) or greater than \(k\):
\[ |A(x)|>k \iff A(x)<-k \quad \text{or} \quad A(x)>k. \]
Unlike the previous case, we do not obtain a compound inequality but rather the union of two separate conditions.
Example. Solve:
\[ |3x+1|>7. \]
We write:
\[ 3x+1<-7 \quad \text{or} \quad 3x+1>7. \]
Solving the first inequality:
\[ 3x<-8 \iff x<-\frac{8}{3}. \]
Solving the second:
\[ 3x>6 \iff x>2. \]
Therefore:
\[ S=\left(-\infty,-\frac{8}{3}\right)\cup(2,+\infty). \]
Inequalities of the form \(|A(x)|\ge k\)
When the symbol is \(\ge\), for \(k>0\) one has:
\[ |A(x)|\ge k \iff A(x)\le -k \quad \text{or} \quad A(x)\ge k. \]
The endpoints are included because the absolute value is allowed to equal \(k\).
Example. Solve:
\[ |2x+5|\ge 1. \]
We write:
\[ 2x+5\le -1 \quad \text{or} \quad 2x+5\ge 1. \]
First inequality:
\[ 2x\le -6 \iff x\le -3. \]
Second inequality:
\[ 2x\ge -4 \iff x\ge -2. \]
Therefore:
\[ S=(-\infty,-3]\cup[-2,+\infty). \]
What happens when the right-hand side is negative
The absolute value is always greater than or equal to zero. When the right-hand side is negative, one must therefore reason carefully.
If \(k<0\), then:
\[ |A(x)|<k \]
has no solutions, since a non-negative number can never be less than a negative number.
Thus:
\[ |A(x)|<k, \quad k<0 \implies S=\varnothing. \]
Likewise:
\[ |A(x)|\le k, \quad k<0 \implies S=\varnothing. \]
On the other hand, if \(k<0\), the inequality:
\[ |A(x)|>k \]
is satisfied for every value in the domain of \(A(x)\), since the absolute value is always at least \(0\) and therefore certainly greater than any negative number.
Thus:
\[ |A(x)|>k, \quad k<0 \implies S=D, \]
where \(D\) denotes the domain of the expression \(A(x)\).
Similarly:
\[ |A(x)|\ge k, \quad k<0 \implies S=D. \]
Examples
The inequality:
\[ |x-1|<-3 \]
has no solutions:
\[ S=\varnothing. \]
By contrast:
\[ |2x+1|>-5 \]
holds for every real number:
\[ S=\mathbb{R}. \]
The definition method
When the inequality contains only a single linear absolute value, the rules above are usually the most efficient approach. However, in the presence of more complex expressions or multiple absolute values, it is often safer to work directly from the definition.
The general definition is:
\[ |A(x)|= \begin{cases} A(x), & \text{if } A(x)\ge 0,\\ -A(x), & \text{if } A(x)<0. \end{cases} \]
To remove an absolute value sign, one must therefore know where its argument is positive and where it is negative.
For example, consider:
\[ |x-2|. \]
The argument vanishes at:
\[ x-2=0 \iff x=2. \]
Therefore:
\[ |x-2|= \begin{cases} -(x-2), & \text{if } x<2,\\ x-2, & \text{if } x\ge 2, \end{cases} \]
that is:
\[ |x-2|= \begin{cases} -x+2, & \text{if } x<2,\\ x-2, & \text{if } x\ge 2. \end{cases} \]
Inequalities with multiple absolute values
When more than one absolute value appears, one must first identify all the points at which the arguments of the absolute values vanish. These points partition the real line into intervals. On each interval, every argument maintains a constant sign, so each absolute value can be removed correctly.
The general procedure is as follows:
- set each argument of an absolute value equal to zero;
- arrange the resulting points in order on the real line;
- study the inequality separately on each interval;
- remove each absolute value using the sign of its argument on that interval;
- solve the resulting inequality;
- intersect the solution with the interval under consideration;
- take the union of all the partial solutions.
This method involves more steps, but it is also the most general approach and minimises the risk of errors.
Worked examples
Example 1. Solve:
\[ |x+3|<4. \]
Since the right-hand side is positive, we write:
\[ -4<x+3<4. \]
Subtracting \(3\):
\[ -7<x<1. \]
Therefore:
\[ S=(-7,1). \]
Example 2. Solve:
\[ |2x-1|\ge 5. \]
Since the right-hand side is positive, we obtain:
\[ 2x-1\le -5 \quad \text{or} \quad 2x-1\ge 5. \]
Solving the first inequality:
\[ 2x\le -4 \iff x\le -2. \]
Solving the second:
\[ 2x\ge 6 \iff x\ge 3. \]
Therefore:
\[ S=(-\infty,-2]\cup[3,+\infty). \]
Example 3. Solve:
\[ |3x+2|\le 1. \]
We write:
\[ -1\le 3x+2\le 1. \]
Subtracting \(2\):
\[ -3\le 3x\le -1. \]
Dividing by \(3\):
\[ -1\le x\le -\frac{1}{3}. \]
Therefore:
\[ S=\left[-1,-\frac{1}{3}\right]. \]
Example 4. Solve:
\[ |x-5|>2. \]
We write:
\[ x-5<-2 \quad \text{or} \quad x-5>2. \]
Solving:
\[ x<3 \quad \text{or} \quad x>7. \]
Therefore:
\[ S=(-\infty,3)\cup(7,+\infty). \]
Example 5. Solve:
\[ |x+1|+|x-2|\le 4. \]
The arguments of the absolute values are:
\[ x+1, \qquad x-2. \]
We find the points at which they vanish:
\[ x+1=0 \iff x=-1, \qquad x-2=0 \iff x=2. \]
The critical points are therefore:
\[ -1, \qquad 2. \]
They partition the real line into three intervals:
\[ (-\infty,-1), \qquad [-1,2), \qquad [2,+\infty). \]
Case 1: \(x<-1\)
If \(x<-1\), then:
\[ x+1<0, \qquad x-2<0. \]
Hence:
\[ |x+1|=-(x+1)=-x-1, \qquad |x-2|=-(x-2)=-x+2. \]
The inequality becomes:
\[ -x-1-x+2\le 4. \]
Simplifying:
\[ -2x+1\le 4. \]
Subtracting \(1\):
\[ -2x\le 3. \]
Dividing by \(-2\) reverses the inequality sign:
\[ x\ge -\frac{3}{2}. \]
Intersecting with the condition \(x<-1\):
\[ -\frac{3}{2}\le x<-1. \]
Case 2: \(-1\le x<2\)
If \(-1\le x<2\), then:
\[ x+1\ge 0, \qquad x-2<0. \]
Therefore:
\[ |x+1|=x+1, \qquad |x-2|=-(x-2)=-x+2. \]
The inequality becomes:
\[ x+1-x+2\le 4, \]
that is:
\[ 3\le 4. \]
This is always true, so the entire interval is part of the solution:
\[ -1\le x<2. \]
Case 3: \(x\ge 2\)
If \(x\ge 2\), then:
\[ x+1>0, \qquad x-2\ge 0. \]
Hence:
\[ |x+1|=x+1, \qquad |x-2|=x-2. \]
The inequality becomes:
\[ x+1+x-2\le 4. \]
Simplifying:
\[ 2x-1\le 4. \]
Adding \(1\):
\[ 2x\le 5. \]
Dividing by \(2\):
\[ x\le \frac{5}{2}. \]
Intersecting with the condition \(x\ge 2\):
\[ 2\le x\le \frac{5}{2}. \]
Taking the union of the three cases:
\[ S= \left[-\frac{3}{2},-1\right) \cup [-1,2) \cup \left[2,\frac{5}{2}\right]. \]
Since these intervals are adjacent, we can write more simply:
\[ S=\left[-\frac{3}{2},\frac{5}{2}\right]. \]
Summary
For \(k>0\), the following equivalences hold:
\[ |A(x)|<k \iff -k<A(x)<k, \]
\[ |A(x)|\le k \iff -k\le A(x)\le k, \]
\[ |A(x)|>k \iff A(x)<-k \quad \text{or} \quad A(x)>k, \]
\[ |A(x)|\ge k \iff A(x)\le -k \quad \text{or} \quad A(x)\ge k. \]
If instead \(k<0\), one must bear in mind that:
\[ |A(x)|\ge 0. \]
As a consequence:
\[ |A(x)|<k \quad \text{and} \quad |A(x)|\le k \]
have no solutions, whereas:
\[ |A(x)|>k \quad \text{and} \quad |A(x)|\ge k \]
hold for every value in the domain of the expression.
Inequalities involving absolute values rest on one simple but fundamental idea: the absolute value measures a distance. For this reason, inequalities of the form \(|A(x)|<k\) or \(|A(x)|\le k\) express a condition of proximity to the origin, while inequalities of the form \(|A(x)|>k\) or \(|A(x)|\ge k\) express a condition of being far from the origin.
In straightforward cases one applies the fundamental equivalences directly. In more involved cases — especially when multiple absolute values appear — the safest approach is to use the definition of absolute value, partitioning the real line into the intervals determined by the zeros of each argument.