4.3 Linear Absolute Value Inequalities

Terrance Berg

Chapter 4: Inequalities

4.3 Linear Absolute Value Inequalities

Absolute values are positive magnitudes, which means that they represent the positive value of any number.

For instance, | −5 | and | +5 | are the same, with both having the same value of 5, and | −99 | and | +99 | both share the same value of 99.

When used in inequalities, absolute values become a boundary limit to a number.

Example 4.3.1

Consider $| x | < 4.$

This means that the unknown $x$ value is less than 4, so $| x | < 4$ becomes $x < 4.$ However, there is more to this with regards to negative values for $x.$

| −1 | is a value that is a solution, since 1 < 4.

However, | −5 | < 4 is not a solution, since 5 > 4.

The boundary of $| x | < 4$ works out to be between −4 and +4.

This means that $| x | < 4$ ends up being bounded as $-4 < x < 4.$

If the inequality is written as $| x | \le 4$ , then little changes, except that $x$ can then equal −4 and +4, rather than having to be larger or smaller.

This means that $| x | \le 4$ ends up being bounded as $-4 \le x \le 4.$

Example 4.3.2

Consider $|x| > 4.$

This means that the unknown $x$ value is greater than 4, so $|x| > 4$ becomes $x > 4.$ However, the negative values for $x$ must still be considered.

The boundary of $|x| > 4$ works out to be smaller than −4 and larger than +4.

This means that $|x| > 4$ ends up being bounded as $x < -4 \text{ or } 4 < x.$

If the inequality is written as $| x | \ge 4,$ then little changes, except that $x$ can then equal −4 and +4, rather than having to be larger or smaller.

This means that $|x| \ge 4$ ends up being bounded as $x \le -4 \text{ or } 4 \le x.$

When drawing the boundaries for inequalities on a number line graph, use the following conventions:

For ≤ or ≥, use [brackets] as boundary limits.

For < or >, use (parentheses) as boundary limits.

Equation	Number Line
$\| x \| <4$
$\| x \| \le 4$
$\| x \| > 4$
$\| x \| \ge 4$

When an inequality has an absolute value, isolate the absolute value first in order to graph a solution and/or write it in interval notation. The following examples will illustrate isolating and solving an inequality with an absolute value.

Example 4.3.3

Solve, graph, and give interval notation for the inequality $-4 - 3 | x | \ge -16.$

First, isolate the inequality:

$\begin{array}{rrrrrl} -4&-&3|x|& \ge & -16 &\\ +4&&&&+4& \text{add 4 to both sides}\\ \midrule &&\dfrac{-3|x|}{-3}& \ge & \dfrac{-12}{-3}&\text{divide by }-3 \text{ and flip the sense} \\ \\ &&|x|&\le & 4 && \end{array}$

At this point, it is known that the inequality is bounded by 4. Specifically, it is between −4 and 4.

This means that $-4 \le | x | \le 4.$

This solution on a number line looks like:

−4 ≤ | x | ≤ 4. Left square bracket at −4; right bracket at 4.

To write the solution in interval notation, use the symbols and numbers on the number line: $[-4, 4].$

Other examples of absolute value inequalities result in an algebraic expression that is bounded by an inequality.

Example 4.3.4

Solve, graph, and give interval notation for the inequality $| 2x - 4 | \le 6.$

This means that the inequality to solve is $-6\le 2x - 4\le 6$ :

$\begin{array}{rrrcrrr} -6&\le & 2x&-&4&\le & 6 \\ +4&&&+&4&&+4 \\ \midrule \dfrac{-2}{2}&\le &&\dfrac{2x}{2}&&\le & \dfrac{10}{2} \\ \\ -1 &\le &&x&&\le & 5 \end{array}$

−1 ≤ x ≤ 5. Left square bracket on −1; right bracket on 5.

To write the solution in interval notation, use the symbols and numbers on the number line: $[-1,5].$

Example 4.3.5

Solve, graph, and give interval notation for the inequality $9 - 2 | 4x + 1 | > 3.$

First, isolate the inequality by subtracting 9 from both sides:

$\begin{array}{rrrrrrr} 9&-&2|4x&+&1|&>&3 \\ -9&&&&&&-9 \\ \midrule &&-2|4x&+&1|&>&-6 \\ \end{array}$

Divide both sides by −2 and flip the sense:

$\begin{array}{rrr} \dfrac{-2|4x+1|}{-2}&>&\dfrac{-6}{-2} \\ \\ |4x+1|&<& 3 \end{array}$

At this point, it is known that the inequality expression is between −3 and 3, so $-3 < 4x + 1 < 3.$

All that is left is to isolate $x$ . First, subtract 1 from all three parts:

$\begin{array}{rrrrrrr} -3&<&4x&+&1&<&3 \\ -1&&&-&1&&-1 \\ \midrule -4&<&&4x&&<&2 \\ \end{array}$

Then, divide all three parts by 4:

$\begin{array}{rrrrr} \dfrac{-4}{4}&<&\dfrac{4x}{4}&<&\dfrac{2}{4} \\ \\ -1&<&x&<&\dfrac{1}{2} \\ \end{array}$

−1 < x < ½. Left parenthesis on −1; right parenthesis on ½.

In interval notation, this is written as $\left(-1,\dfrac{1}{2}\right).$

It is important to remember when solving these equations that the absolute value is always positive. If given an absolute value that is less than a negative number, there will be no solution because absolute value will always be positive, i.e., greater than a negative. Similarly, if absolute value is greater than a negative, the answer will be all real numbers.

This means that:

$\begin{array}{c} | 2x - 4 | < -6 \text{ has no possible solution } (x \ne \mathbb{R}) \\ \\ \text{and} \\ \\ | 2x - 4 | > -6 \text{ has every number as a solution and is written as } (-\infty, \infty) \end{array}$