Topics
1. Equations & Inequalities in One Variable
topic badge
United States of AmericaTennessee
Integrated Math 1

1.08 Absolute value inequalities

Lesson
Practice
Lesson

Concept summary

The absolute value of a number is a measure of the size of a number, and is equal to its distance from zero (0), which is always a non-negative value. Absolute value is sometimes called "magnitude".

An absolute value inequality is an inequality containing the absolute value of one more variable expressions.

Inequality

A mathematical relation that compares two non-equivalent expressions

Solutions to absolute value inequalities usually involve multiple inequalities joined by one of the keywords "and" or "or". Solutions with two overlapping regions joined by "and" can be rewritten as a single compound inequality:

\left|x\right| \geq 1 has solutions of x \leq -1 \text{ or } x \geq 1 or in interval notation \left(-\infty, -1\right] \cup \left[-1,\infty\right).

\left|x\right| < 3 has solutions of x > -3 \text{ and } x < 3 which is equivalent to -3 < x < 3 or in interval notation \left(-3,3\right).

In general, for an algebraic expression p(x) and k>0, we have:

  • \left|p(x)\right|<k can be written as -k<p(x)<k
  • \left|p(x)\right|\leq k can be written as -k\leq p(x) \leq k
  • \left|p(x)\right|> k can be written as p(x)<-k or p(x)> k
  • \left|p(x)\right|\geq k can be written as p(x)\leq-k or p(x)\geq k

Solutions to absolute value inequalities can also be represented graphically using number lines:

-4-3-2-101234
\left|x\right| \geq 1
-4-3-2-101234
\left|x\right| < 3

Worked examples

Example 1

Consider the inequality \left|x\right| > 2:

a

Represent the inequality \left|x\right| > 2 on a number line.

Approach

This inequality represents values of x which are "more than 2 units away from 0". To plot this, we will need to use two regions. Also note that this inequality doesn't include the endpoints, so we will use unfilled points to show this.

Solution

-5-4-3-2-1012345

Reflection

We can give a quick check of the answer by thinking about whether this is an "and"-type inequality or an "or"-type inequality.

In this case, the inequality \left|x\right| > 2 has a solution of x < -2 \text{ or } x > 2. Our solution on the numberline has two distinct parts, which matches what we expect for an "or"-type inequality.

b

Rewrite the solution to \left|x\right| > 2 in interval notation.

Approach

We can use either the number line or x < -2 \text{ or } x > 2 to help write this in interval notation.

Solution

The x < -2 part can be written as \left(-\infty, -2\right).

The x >2 part can be written as \left(-2,-\infty\right).

Since this is an "or", not "and", we will find the union of these two sets to give: \left(-\infty, -2\right) \cup \left(-2,-\infty\right)

Example 2

Consider the inequality \left|\dfrac{4}{3}x - 5\right| \leq 3.

a

Solve the inequality for x. Express your solution using interval notation.

Approach

In order to solve this inequality, it will be easier to first remove the absolute value by rewriting the inequality as a compound inequality. We can then solve as normal.

Solution

\displaystyle \left|\dfrac{4}{3}x - 5\right|\displaystyle \leq\displaystyle 3
\displaystyle -3 \leq \dfrac{4}{3}x - 5\displaystyle \leq\displaystyle 3Rewrite as a compound inequality
\displaystyle 2 \leq \dfrac{4}{3}x\displaystyle \leq\displaystyle 8Addition property of inequality
\displaystyle 6 \leq 4x\displaystyle \leq\displaystyle 24Multiplication property of inequality
\displaystyle \frac{3}{2} \leq x\displaystyle \leq\displaystyle 6Division property of inequality

So the solutions are "all values of x between \dfrac{3}{2} and 6 inclusive". We can express this using interval notation as the interval \left[\frac{3}{2},\, 6\right]

b

Represent the solution set on a number line.

Approach

The solution set for this inequality consists of a single interval, so it will only need one region on the numberline. The endpoints are also included this time, which we represent using filled points.

Solution

-1012345678

Reflection

We can give a quick check of the answer by thinking about whether this is an "and"-type inequality or an "or"-type inequality.

In this case, we found the solution to the inequality \left|\dfrac{4}{3}x - 5\right| \leq 3 to be the compound inequality \dfrac{3}{2} \leq x \leq 6. A compound inequality is a way of writing an "and"-type inequality, which matches the fact that the numberline has one region between two endpoints.

c

Determine whether x=2.5 is a valid solution to the absolute value inequality.

Approach

Plot the solution set togther with the indicated point to check if the point lies inside or outside the solution set. Or test algebraically by substituting the value into either the original or rearranged inequality and check if the resulting statement is true.

Solution

Graphically:

Consider the plot of the solution set together with the point at x=2.5:

-1012345678

The point x=2.5 lies on a section of the line that in the solution set, indicated by the green interval, and therefore is a valid solution.

Algebraically:

Substituting x=2.5 into the original inequality:

\displaystyle \left|\dfrac{4}{3}x - 5\right|\displaystyle \leq\displaystyle 3
\displaystyle \left|\dfrac{4}{3}\left(2.5\right) - 5\right|\displaystyle \leq\displaystyle 3Substitute x=2.5 into the inequality
\displaystyle \left|\dfrac{10}{3}-5\right|\displaystyle \leq\displaystyle 3Evaluate the multiplication
\displaystyle \left|-\dfrac{5}{3}\right|\displaystyle \leq\displaystyle 3Evaluate the subtraction
\displaystyle \dfrac{5}{3}\displaystyle \leq\displaystyle 3Evaluate the absolute value

As this statement is true, x=2.5 is a valid solution.

Example 3

Write an absolute value inequality to represent the set of "all real numbers x which are at least 5 units away from 12".

Approach

We want to break down the wording of the set into its key parts. These key parts are:

  • at least
  • away from 12
  • 5 units away

Each of these key parts tells us information about the form of the inequality.

Solution

Let's interpret these key parts:

  • "At least" is another way of saying "greater than or equal to", which tells us the inequality symbol that we want to use.
  • "Away from 12" means that we want to compare the distance between x and 12. We can represent this as \left|x - 12\right|.
  • "5 units away from" means that the absolute value will be compared to 5.

Putting this all together, we get "the distance between x and 12 is greater than or equal to 5", which we can represent as the absolute value inequality \left|x - 12\right| \geq 5

Outcomes

M1.N.Q.A.1

Use units as a way to understand real-world problems.*

M1.N.Q.A.1.C

Define and justify appropriate quantities within a context for the purpose of modeling.*

M1.A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems in a real-world context.*

M1.A.CED.A.3

Create individual and systems of equations and/or inequalities to represent constraints in a contextual situation, and interpret solutions as viable or non-viable.*

M1.A.REI.B.2

Solve linear and absolute value equations and inequalities in one variable.

M1.A.REI.B.2.B

Solve absolute value equations and inequalities in one variable. Represent solutions algebraically and graphically.

M1.MP1

Make sense of problems and persevere in solving them.

M1.MP2

Reason abstractly and quantitatively.

M1.MP3

Construct viable arguments and critique the reasoning of others.

M1.MP4

Model with mathematics.

M1.MP6

Attend to precision.

M1.MP7

Look for and make use of structure.

M1.MP8

Look for and express regularity in repeated reasoning.

What is Mathspace

About Mathspace