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2.07 Absolute value inequalities

Introduction

The concepts of absolute value equations and inequalities are combined here. We know that both concepts are useful for modeling real world phenomena, and we can use and build on those ideas with absolute value inequalities.

Absolute value inequalities

Recall that the absolute value of a number is a measure of the size of a number, and is equal to its distance from 0, which is always a non-negative value. Absolute value is sometimes called the magnitude.

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

Exploration

Plot the range of values that satisfy each of the following inequalities on number lines.

Form \left|x\right|>a:\left|x\right|>1,\, \left|x\right|>2,\, \left|x\right|>5,\, \left|x\right|>100

Form \left|x\right|<a:\left|x\right|<1,\, \left|x\right|<2,\, \left|x\right|<5,\, \left|x\right|<100

  1. What do you notice about absolute value inequalities in the form \left|x\right|>a?

  2. What do you notice about absolute value inequalities in the form \left|x\right|<a?

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.

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\left|x\right| \geq 4 is read as the absolute value of some value, x, is 4 or more spaces from zero. The inequality has solutions of x \leq -4 \text{ or } x \geq 4 or in interval notation \left(-\infty, -4\right] \cup \left[4,\infty\right).

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\left|x\right| < 3 is read as the absolute value of some value, x, is within 3 spaces of zero. The inequality 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 order to solve an absolute value inequality that results in a compound inequality solution, we begin by isolating the absolute value expression. Once the expression is isolated, we write and solve two separate inequalities without the absolute value bars. This is similar to solving an absolute value equation, but the second inequality's symbol will be reversed.

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

  • \left|p\left(x\right)\right|<k is within k spaces of 0 and can be written as -k<p\left(x\right)<k
  • \left|p\left(x\right)\right|> k is more than k spaces of 0 and can be written as p\left(x\right)<-k or p\left(x\right)> k
  • For the absolute value inequality a\left|p\left(x\right)\right|+b < c,the expression \left|p\left(x\right)\right| must be isolated using inverse operations before rewriting the inequality as a compound inequality

Examples

Example 1

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

a

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

Worked Solution
Create a strategy

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.

Apply the idea
-5-4-3-2-1012345
Reflect and check

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 number line 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.

Worked Solution
Create a strategy

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

Apply the idea

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" inequality 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.

Worked Solution
Create a strategy

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.

Apply the idea
\displaystyle \left|\dfrac{4}{3}x - 5\right|\displaystyle \leq\displaystyle 3Original inequality
\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.

Worked Solution
Create a strategy

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

Apply the idea
-1012345678
Reflect and check

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 number line has one region between two endpoints.

c

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

Worked Solution
Create a strategy

Plot the solution set together 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 checking if the resulting statement is true.

Apply the idea

Graphically:

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

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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:

Substitute x=2.5 into the original inequality.

\displaystyle \left|\dfrac{4}{3}x - 5\right|\displaystyle \leq\displaystyle 3Original inequality
\displaystyle \left|\dfrac{4}{3}\left(2.5\right) - 5\right|\displaystyle \leq\displaystyle 3Substitute x=2.5
\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

Considering \dfrac{5}{3} \approx 1.67 this statement is true and x=2.5 is a valid solution.

Example 3

Consider the inequality 4-\dfrac{1}{2}\left|x+2\right|>-3.

a

Solve the inequality.

Worked Solution
Create a strategy

In order to solve this inequality, we need to isolate the absolute value expression and then rewrite the inequality as a compound inequality. We can then solve by applying the properties of inequality.

Apply the idea
\displaystyle 4-\dfrac{1}{2}|x+2|\displaystyle >\displaystyle -3Original inequality
\displaystyle -\dfrac{1}{2}|x+2|\displaystyle >\displaystyle -7Subtraction property of inequality
\displaystyle |x+2|\displaystyle <\displaystyle 14Multiplication property of inequality

Now, we can write |x+2|<14 as a compound inequality and solve.

\displaystyle -14\displaystyle <\displaystyle x+2 < 14Rewrite as a compound inequality
\displaystyle -16\displaystyle <\displaystyle x < 12Subtraction property of equality
b

Represent the solution on a number line.

Worked Solution
Create a strategy

The solution set for this inequality consists of a single interval, so it will only need one region on the number line. Because neither inequality has an "equal to" the endpoints are not included in the solution, which we represent using unfilled points.

Apply the idea
-20-15-10-505101520

Example 4

In a survey, 76 \% of people asked stated they would likely vote yes for new neighborhood park plans. The margin of error is 4 percentage points.

a

Write and solve an absolute value inequality that represents the scenario.

Worked Solution
Create a strategy

A margin of error indicates that the percentages for the poll could be off by 4 percentage points in either direction of 76 \%. It may be helpful to first graph possible solutions to the inequality on a number line and then use the number line to help us write an inequality.

Apply the idea
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Similar to writing and solving absolute value equations, understanding the distance of the possible solutions from the middle or average value will help us write our statement.

Let v= \text{the percentage of people who said they will vote yes}. If the solutions are up to 4 percentage points from 76 \%, an inequality with the possible solutions of 72 \% to 80 \% is |v-76| \leq 4.

Reflect and check

By solving the inequality we wrote, we can confirm that our inequality makes sense in context and that its solution matches the number line solution set.

\displaystyle |v-76|\displaystyle \leq\displaystyle 4Original inequality
\displaystyle -4 \leq v-76\displaystyle \leq\displaystyle 4Rewrite as a compound inequaltiy
\displaystyle -4+76 \leq v-76+76\displaystyle \leq\displaystyle 4+76Addition property of inequality
\displaystyle 72 \leq v\displaystyle \leq\displaystyle 80Evaluate the addition
b

Determine what the viable solutions to the inequality mean in context.

Worked Solution
Apply the idea

The margin of error would mean that anywhere from 72 \% to 80 \% of people would likely vote yes on the survey.

Idea summary

For absolute value inequalities with an algebraic expression p\left(x\right) and k>0.

  • \left|p\left(x\right)\right|<k is within k spaces of 0 and can be written as -k<p\left(x\right)<k
  • \left|p\left(x\right)\right|> k is more than k spaces of 0 and can be written as p\left(x\right)<-k or p\left(x\right)> k

For the absolute value inequality a\left|p\left(x\right)\right|+b < c,the expression \left|p\left(x\right)\right| must be isolated using inverse operations before rewriting the inequality as a compound inequality.

Outcomes

A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems.

A.REI.B.3

Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

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