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Study Guides > College Algebra

Solve an absolute value inequality

Absolute value equations may not always involve equalities. Instead, we may need to solve an equation within a range of values. We would use an absolute value inequality to solve such an equation. An absolute value inequality is an equation of the form

[latex]|{A}|<{ B },|{ A }|\le{ B },|{ A }|>{ B },\text{ or } |{ A }|\ge { B }\\[/latex],

where an expression [latex]A\\[/latex] (and possibly but not usually [latex]B\\[/latex] ) depends on a variable [latex]x\\[/latex]. Solving the inequality means finding the set of all [latex]x\\[/latex] that satisfy the inequality. Usually this set will be an interval or the union of two intervals.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two functions. The advantage of the algebraic approach is it yields solutions that may be difficult to read from the graph.

For example, we know that all numbers within 200 units of 0 may be expressed as

[latex]|x|<{ 200 }\text{ or }{ -200 }<{ x }<{ 200 }\text{ }\\[/latex]

Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of values [latex]x\\[/latex] such that the distance between [latex]x\\[/latex] and 600 is less than 200. We represent the distance between [latex]x\\[/latex] and 600 as [latex]|{ x } - {600 }|\\[/latex].

[latex]|{ x } -{ 600 }|<{ 200 }\\[/latex]
OR
[latex]{ -200 }<{ x } - { 600 }<{ 200 }\\[/latex]
[latex]{-200 }+{ 600 }<{ x } - {600 }+{ 600 }<{ 200 }+{ 600 }\\[/latex]
[latex]{ 400 }<{ x }<{ 800 }\\[/latex]

This means our returns would be between $400 and $800.

Sometimes an absolute value inequality problem will be presented to us in terms of a shifted and/or stretched or compressed absolute value function, where we must determine for which values of the input the function’s output will be negative or positive.

How To: Given an absolute value inequality of the form [latex]|x-A|\le B\\[/latex] for real numbers [latex]a\\[/latex] and [latex]b\\[/latex] where [latex]b\\[/latex] is positive, solve the absolute value inequality algebraically.

  1. Find boundary points by solving [latex]|x-A|=B\\[/latex].
  2. Test intervals created by the boundary points to determine where [latex]|x-A|\le B\\[/latex].
  3. Write the interval or union of intervals satisfying the inequality in interval, inequality, or set-builder notation.

Example 6: Solving an Absolute Value Inequality

Solve [latex]|x - 5|\le 4\\[/latex].

Solution

With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where [latex]|x - 5|=4[/latex]. We do this because the absolute value is a function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve [latex]|x - 5|=4\\[/latex].

[latex]\begin{cases}x - 5=4 \hfill & \text{or}\hfill & {x - 5 }={ -4 }\\ \hfill {x }= {9}&\text{or}\hfill & \hfill{ x }={ 1 }\hfill \end{cases}\\[/latex]

After determining that the absolute value is equal to 4 at [latex]x=1\\[/latex] and [latex]x=9\\[/latex], we know the graph can change only from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals:

[latex]{ x }<{ 1 },\text{ }{ 1 }<{ x }<{ 9 },\text{ and }{ x }>{ 9 }\\[/latex].

To determine when the function is less than 4, we could choose a value in each interval and see if the output is less than or greater than 4, as shown in the table below.

Interval test [latex]x\\[/latex] [latex]f\left(x\right)\\[/latex] [latex]<4\\[/latex] or [latex]>4?\\[/latex]
[latex]{ x }<{ 1 }\\[/latex] 0 [latex]|0 - 5|=5\\[/latex] Greater than
[latex]{ 1 }<{ x }<{ 9 }\\[/latex] 6 [latex]|6 - 5|=1\\[/latex] Less than
[latex]{ x }>{ 9 }\\[/latex] 11 [latex]|11 - 5|=6\\[/latex] Greater than

Because [latex]1\le x\le 9\\[/latex] is the only interval in which the output at the test value is less than 4, we can conclude that the solution to [latex]|x - 5|\le 4[/latex] is [latex]1\le x\le 9\\[/latex], or [latex]\left[1,9\right]\\[/latex].

To use a graph, we can sketch the function [latex]f\left(x\right)=|x - 5|\\[/latex]. To help us see where the outputs are 4, the line [latex]g\left(x\right)=4\\[/latex] could also be sketched.
Graph of an absolute function and a vertical line, demonstrating how to see what outputs are less than the vertical line.Figure 11. Graph to find the points satisfying an absolute value inequality.

We can see the following:

  • The output values of the absolute value are equal to 4 at [latex]x=1\\[/latex] and [latex]x=9\\[/latex].
  • The graph of [latex]f\\[/latex] is below the graph of [latex]g\\[/latex] on [latex]1<x<9\\[/latex]. This means the output values of [latex]f\left(x\right)\\[/latex] are less than the output values of [latex]g\left(x\right)\\[/latex].
  • The absolute value is less than or equal to 4 between these two points, when [latex]1\le x\le 9\\[/latex]. In interval notation, this would be the interval [latex]\left[1,9\right]\\[/latex].

Analysis of the Solution

For absolute value inequalities,

[latex]\begin{cases}{|x-A| }<{ C },\hfill & \hfill &{ |x-A| }>{ C },\hfill \\{ -C }<{ x-A }<{ C },\hfill & \hfill &{ x-A }<{ -C }\text{ or }{ x-A }>{ C }.\hfill \end{cases}\\[/latex]

The [latex]<\\[/latex] or [latex]>\\[/latex] symbol may be replaced by [latex]\le \text{ or }\ge \\[/latex].

So, for this example, we could use this alternative approach.

[latex]\begin{cases}|x - 5|\le 4\hfill & \hfill & \hfill & \hfill \\ -4\le x - 5\le 4\hfill & \hfill & \hfill & \text{Rewrite by removing the absolute value bars}.\hfill \\ -4+5\le x - 5+5\le 4+5\hfill & \hfill & \hfill & \text{Isolate the }x.\hfill \\ 1\le x\le 9\hfill & \hfill & \hfill & \hfill \end{cases}\\[/latex]

Try It 6

Solve [latex]|x+2|\le 6\\[/latex].

How To: Given an absolute value function, solve for the set of inputs where the output is positive (or negative).

  1. Set the function equal to zero, and solve for the boundary points of the solution set.
  2. Use test points or a graph to determine where the function’s output is positive or negative.

Example 7: Using a Graphical Approach to Solve Absolute Value Inequalities

Given the function [latex]f\left(x\right)=-\frac{1}{2}|4x - 5|+3\\[/latex], determine the [latex]x\text{-}\\[/latex] values for which the function values are negative.

Solution

We are trying to determine where [latex]f\left(x\right)<0\\[/latex], which is when [latex]-\frac{1}{2}\text{ }|4x - 5|+3<0\\[/latex]. We begin by isolating the absolute value.

[latex]\begin{cases}-\frac{1}{2}|4x - 5|<-3 \hfill & \text{Multiply both sides by -2, and reverse the inequality}.\hfill \\ |4x - 5|>6\hfill & \hfill \end{cases}\\[/latex]

Next we solve for the equality [latex]|4x - 5|=6\\[/latex].

[latex]\begin{cases}4x - 5=6\hfill & \hfill & 4x - 5=-6\hfill \\ 4x - 5=6\hfill & \text{ or }\hfill & 4x=-1\hfill \\ x=\frac{11}{4}\hfill & \hfill & x=-\frac{1}{4}\hfill \end{cases}\\[/latex]
Graph of an absolute function with x-intercepts at -0.25 and 2.75.Figure 12

Now, we can examine the graph of [latex]f\\[/latex] to observe where the output is negative. We will observe where the branches are below the x-axis. Notice that it is not even important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at [latex]x=-\frac{1}{4}\\[/latex] and [latex]x=\frac{11}{4}\\[/latex] and that the graph has been reflected vertically.

We observe that the graph of the function is below the x-axis left of [latex]x=-\frac{1}{4}\\[/latex] and right of [latex]x=\frac{11}{4}\\[/latex]. This means the function values are negative to the left of the first horizontal intercept at [latex]x=-\frac{1}{4}\\[/latex], and negative to the right of the second intercept at [latex]x=\frac{11}{4}\\[/latex]. This gives us the solution to the inequality.

[latex]x<-\frac{1}{4}\text{ }\text{or}\text{ }x>\frac{11}{4}\\[/latex]

In interval notation, this would be [latex]\left(-\infty ,-0.25\right)\cup \left(2.75,\infty \right)\\[/latex].

Try It 7

Solve [latex]-2|k - 4|\le -6\\[/latex].

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  • Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1-introduction-to-functions. License: CC BY: Attribution. License terms: Download For Free at : http://cnx.org/contents/[email protected]..