Solve Absolute Value Inequalities - Intermediate Algebra

2.7 Solve Absolute Value Inequalities

Learning Objectives

By the end of this section, you will be able to:

Solve absolute value equations
Solve absolute value inequalities with “less than”
Solve absolute value inequalities with “greater than”
Solve applications with absolute value

Be Prepared 2.17

Before you get started, take this readiness quiz.

Evaluate: $− | 7 | .$
If you missed this problem, review Example 1.12.

Be Prepared 2.18

Fill in $<, >,$ or $=$ for each of the following pairs of numbers.
ⓐ $| −8 |___- | −8 |$ ⓑ $12___- | −12 |$ ⓒ $| −6 |___- 6$ ⓓ $− (−15)___- | −15 |$
If you missed this problem, review Example 1.12.

Be Prepared 2.19

Simplify: $14 - 2 | 8 - 3 (4 - 1) | .$
If you missed this problem, review Example 1.13.

Solve Absolute Value Equations

As we prepare to solve absolute value equations, we review our definition of absolute value.

Absolute Value

The absolute value of a number is its distance from zero on the number line.

The absolute value of a number n is written as $| n |$ and $| n | \geq 0$ for all numbers.

Absolute values are always greater than or equal to zero.

We learned that both a number and its opposite are the same distance from zero on the number line. Since they have the same distance from zero, they have the same absolute value. For example:

$−5$ is 5 units away from 0, so $| −5 | = 5 .$

$5$ is 5 units away from 0, so $| 5 | = 5 .$

Figure 2.6 illustrates this idea.

The figure is a number line with tick marks at negative 5, 0, and 5. The distance between negative 5 and 0 is given as 5 units, so the absolute value of negative 5 is 5. The distance between 5 and 0 is 5 units, so the absolute value of 5 is 5.

Figure 2.6 The numbers 5 and

−5

are both five units away from zero.

For the equation $| x | = 5,$ we are looking for all numbers that make this a true statement. We are looking for the numbers whose distance from zero is 5. We just saw that both 5 and $−5$ are five units from zero on the number line. They are the solutions to the equation.

\begin{matrix} If & | x | = 5 \\ then & x = −5 or x = 5 \end{matrix}

The solution can be simplified to a single statement by writing $x = ± 5 .$ This is read, “x is equal to positive or negative 5”.

We can generalize this to the following property for absolute value equations.

Absolute Value Equations

For any algebraic expression, u, and any positive real number, a,

\begin{matrix} if & | u | = a \\ then & u = − a or u = a \end{matrix}

Remember that an absolute value cannot be a negative number.

Example 2.68

Solve: ⓐ $| x | = 8$ ⓑ $| y | = −6$ ⓒ $| z | = 0$

Solution

ⓐ

	$\| x \| = 8$
Write the equivalent equations.	$x = −8 or x = 8$
	$x = ± 8$

Illustrates how to write equivalent equations for the absolute value equation |x|=8, showing its two solutions.

ⓑ

	$\| y \| = −6$
	No solution
Since an absolute value is always positive, there are no solutions to this equation.

This table illustrates the absolute value equation |y| = -6, demonstrating it has no solution as absolute values are always non-negative.

ⓒ

	$\| z \| = 0$
Write the equivalent equations.	$z = −0 or z = 0$
Since −0 = 0,	$z = 0$
Both equations tell us that z = 0 and so there is only one solution.

Steps for solving the absolute value equation |z|=0, illustrating the transformation to equivalent equations and the unique solution z=0.

Try It 2.135

Solve: ⓐ $| x | = 2$ ⓑ $| y | = −4$ ⓒ $| z | = 0$

Try It 2.136

Solve: ⓐ $| x | = 11$ ⓑ $| y | = −5$ ⓒ $| z | = 0$

To solve an absolute value equation, we first isolate the absolute value expression using the same procedures we used to solve linear equations. Once we isolate the absolute value expression we rewrite it as the two equivalent equations.

Example 2.69

How to Solve Absolute Value Equations

Solve $| 5 x - 4 | - 3 = 8 .$

Solution

Step 2 is to write the equivalent equations, 5 x minus 4 is equal to negative 11 and 5 x minus 4 is equal to 11.

Step 3 is to solve each equation. Add 4 to each side. 5 x is equal to negative 7 or 5 x is equal to 15. Divide each side by 5. The result is x is equal to negative seven-fifths or x is equal to 3.

Step 4 is to check each solution. Substitute 3 and negative seven-fifths into the original equation, the difference between the absolute value of the quantity 5 x minus 4 and 3 is equal to 8. Substitute 3 for x. Is the difference between the absolute value of the quantity 5 times 3 minus 4 and 3 equal to 8? Is the difference between the absolute value of the quantity 15 minus 4 and 3 equal to 8? Is the difference between the absolute value of the 11 and 3 equal to 8? Is 11 minus 3 equal to 8? 8 is equal to 8, so the solution x is equal to 3 checks. Substitute negative seven-fifths for x. Is the difference between the absolute value of the quantity 5 times negative seven-fifths minus 4 and 3 equal to 8? Is the difference between the absolute value of the quantity negative 7 minus 4 and 3 equal to 8? Is the difference between the absolute value of the negative 11 and 3 equal to 8? Is 11 minus 3 equal to 8? 8 is equal to 8, so the solution x is equal to negative seven-fifths checks.

Try It 2.137

Solve: $| 3 x - 5 | - 1 = 6 .$

Try It 2.138

Solve: $| 4 x - 3 | - 5 = 2 .$

The steps for solving an absolute value equation are summarized here.

How To

Solve absolute value equations.

Step 1. Isolate the absolute value expression.
Step 2. Write the equivalent equations.
Step 3. Solve each equation.
Step 4. Check each solution.

Example 2.70

Solve $2 | x - 7 | + 5 = 9 .$

Solution

	$2 \| x - 7 \| + 5 = 9$
Isolate the absolute value expression.	$2 \| x - 7 \| = 4$
	$\| x - 7 \| = 2$
Write the equivalent equations.	$x - 7 = − 2$ or $x - 7 = 2$
Solve each equation.	$x = 5$ or $x = 9$
Check:

Try It 2.139

Solve: $3 | x - 4 | - 4 = 8 .$

Try It 2.140

Solve: $2 | x - 5 | + 3 = 9 .$

Remember, an absolute value is always positive!

Example 2.71

Solve: $| \frac{2}{3} x - 4 | + 11 = 3 .$

Solution

	$\| \frac{2}{3} x - 4 \| + 11 = 3$
Isolate the absolute value term.	$\| \frac{2}{3} x - 4 \| = −8$
An absolute value cannot be negative.	No solution

Steps for solving an absolute value equation resulting in no solution.

Try It 2.141

Solve: $| \frac{3}{4} x - 5 | + 9 = 4 .$

Try It 2.142

Solve: $| \frac{5}{6} x + 3 | + 8 = 6 .$

Some of our absolute value equations could be of the form $| u | = | v |$ where u and v are algebraic expressions. For example, $| x - 3 | = | 2 x + 1 | .$

How would we solve them? If two algebraic expressions are equal in absolute value, then they are either equal to each other or negatives of each other. The property for absolute value equations says that for any algebraic expression, u, and a positive real number, a, if $| u | = a,$ then $u = − a$ or $u = a .$

This tells us that

\begin{matrix} if & | u | = | v | \\ then & u = − v & or u = v \end{matrix}

This leads us to the following property for equations with two absolute values.

Equations with Two Absolute Values

For any algebraic expressions, u and v,

\begin{matrix} if & | u | = | v | \\ then & u = − v or u = v \end{matrix}

When we take the opposite of a quantity, we must be careful with the signs and to add parentheses where needed.

Example 2.72

Solve: $| 5 x - 1 | = | 2 x + 3 | .$

Solution

	$\| 5 x - 1 \| = \| 2 x + 3 \|$
Write the equivalent equations. Solve each equation.	$\begin{matrix} 5 x - 1 & = & − (2 x + 3) & or & 5 x - 1 & = & 2 x + 3 \\ 5 x - 1 & = & −2 x - 3 & or & 3 x - 1 & = & 3 \\ 7 x - 1 & = & −3 & 3 x & = & 4 \\ 7 x & = & −2 & x & = & \frac{4}{3} \\ x & = & - \frac{2}{7} & or & x & = & \frac{4}{3} \end{matrix}$
Check.
We leave the check to you.

This table illustrates the step-by-step process of solving the absolute value equation |5x - 1| = |2x + 3|, from setting up equivalent equations to finding the solutions.

Try It 2.143

Solve: $| 7 x - 3 | = | 3 x + 7 | .$

Try It 2.144

Solve: $| 6 x - 5 | = | 3 x + 4 | .$

Solve Absolute Value Inequalities with “Less Than”

Let’s look now at what happens when we have an absolute value inequality. Everything we’ve learned about solving inequalities still holds, but we must consider how the absolute value impacts our work.

Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line. For the equation $| x | = 5,$ we saw that both 5 and $−5$ are five units from zero on the number line. They are the solutions to the equation.

\begin{matrix} | x | = 5 \\ x = −5 or x = 5 \end{matrix}

What about the inequality $| x | \leq 5 ?$ Where are the numbers whose distance is less than or equal to 5? We know $−5$ and 5 are both five units from zero. All the numbers between $−5$ and 5 are less than five units from zero. See Figure 2.7.

The figure is a number line with negative 5, 0, and 5 displayed. There is a left bracket at negative 5 and a right bracket at 5. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x which is less than or equal to 5.

Figure 2.7

In a more general way, we can see that if $| u | \leq a,$ then $− a \leq u \leq a .$ See Figure 2.8.

The figure is a number line with negative a 0, and a displayed. There is a left bracket at negative a and a right bracket at a. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is less than or equal to a, then negative a is less than or equal to u which is less than or equal to a.

Figure 2.8

This result is summarized here.

Absolute Value Inequalities with $<$ or $\leq$

For any algebraic expression, u, and any positive real number, a,

\begin{matrix} if & | u | < a, & then − a < u < a \\ if & | u | \leq a, & then − a \leq u \leq a \end{matrix}

After solving an inequality, it is often helpful to check some points to see if the solution makes sense. The graph of the solution divides the number line into three sections. Choose a value in each section and substitute it in the original inequality to see if it makes the inequality true or not. While this is not a complete check, it often helps verify the solution.

Example 2.73

Solve $| x | < 7 .$ Graph the solution and write the solution in interval notation.

Solution


Write the equivalent inequality.
Graph the solution.
Write the solution using interval notation.

Check:

To verify, check a value in each section of the number line showing the solution. Choose numbers such as $−8,$ 1, and 9.

The figure is a number line with a left parenthesis at negative 7, a right parenthesis at 7 and shading between the parentheses. The values negative 8, 1, and 9 are marked with points. The absolute value of negative 8 is less than 7 is false. It does not satisfy the absolute value of x is less than 7. The absolute value of 1 is less than 7 is true. It does satisfy the absolute value of x is less than 7. The absolute value of 9 is less than 7 is false. It does not satisfy the absolute value of x is less than 7.

Try It 2.145

Graph the solution and write the solution in interval notation: $| x | < 9 .$

Try It 2.146

Graph the solution and write the solution in interval notation: $| x | < 1 .$

Example 2.74

Solve $| 5 x - 6 | \leq 4 .$ Graph the solution and write the solution in interval notation.

Solution

Step 1. Isolate the absolute value expression. It is isolated.	$\| 5 x - 6 \| \leq 4$
Step 2. Write the equivalent compound inequality.	$−4 \leq 5 x - 6 \leq 4$
Step 3. Solve the compound inequality.	$2 \leq 5 x \leq 10$ $\frac{2}{5} \leq x \leq 2$
Step 4. Graph the solution.
Step 5. Write the solution using interval notation.	$[\frac{2}{5}, 2]$
Check: The check is left to you.

Try It 2.147

Solve $| 2 x - 1 | \leq 5 .$ Graph the solution and write the solution in interval notation:

Try It 2.148

Solve $| 4 x - 5 | \leq 3 .$ Graph the solution and write the solution in interval notation:

How To

Solve absolute value inequalities with < or ≤.

Step 1. Isolate the absolute value expression.
Step 2.
Write the equivalent compound inequality.

$\begin{matrix} | u | < a & is equivalent to & − a < u < a \\ | u | \leq a & is equivalent to & − a \leq u \leq a \end{matrix}$
Step 3. Solve the compound inequality.
Step 4. Graph the solution
Step 5. Write the solution using interval notation.

Solve Absolute Value Inequalities with “Greater Than”

What happens for absolute value inequalities that have “greater than”? Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line.

We started with the inequality $| x | \leq 5 .$ We saw that the numbers whose distance is less than or equal to five from zero on the number line were $−5$ and 5 and all the numbers between $−5$ and 5. See Figure 2.9.

The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its right and a right bracket at 5 with shading to its left. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x is less than or equal to 5.

Figure 2.9

Now we want to look at the inequality $| x | \geq 5 .$ Where are the numbers whose distance from zero is greater than or equal to five?

Again both $−5$ and 5 are five units from zero and so are included in the solution. Numbers whose distance from zero is greater than five units would be less than $−5$ and greater than 5 on the number line. See Figure 2.10.

The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its left and a left bracket at 5 with shading to its right. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is greater than or equal to 5, then x is less than or equal to negative 5 or x is greater than or equal to 5.

Figure 2.10

In a more general way, we can see that if $| u | \geq a,$ then $u \leq − a$ or $u \geq a .$ See Figure 2.11.

The figure is a number line with negative a, 0, and a displayed. There is a right bracket at negative a that has shading to its left and a left bracket at a with shading to its right. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is greater than or equal to a, then u is less than or equal to negative a or u is greater than or equal to a.

Figure 2.11

This result is summarized here.

Absolute Value Inequalities with > or ≥

For any algebraic expression, u, and any positive real number, a,

\begin{matrix} if & | u | > a, & then u < - a or u > a \\ if & | u | \geq a, & then u \leq − a or u \geq a \end{matrix}

Example 2.75

Solve $| x | > 4 .$ Graph the solution and write the solution in interval notation.

Solution

	$\| x \| > 4$
Write the equivalent inequality.	$x < −4 or x > 4$
Graph the solution.
Write the solution using interval notation.	$(− \infty, −4) \cup (4, \infty)$
Check:

To verify, check a value in each section of the number line showing the solution. Choose numbers such as $−6,$ 0, and 7.

The figure is a number line with a right parenthesis at negative 4 with shading to its left and a left parenthesis at 4 shading to its right. The values negative 6, 0, and 7 are marked with points. The absolute value of negative 6 is greater than negative 4 is true. It does not satisfy the absolute value of x is greater than 4. The absolute value of 0 is greater than 4 is false. It does not satisfy the absolute value of x is greater than 4. The absolute value of 7 is less than 4 is true. It does satisfy the absolute value of x is greater than 4.

Try It 2.149

Solve $| x | > 2 .$ Graph the solution and write the solution in interval notation.

Try It 2.150

Solve $| x | > 1 .$ Graph the solution and write the solution in interval notation.

Example 2.76

Solve $| 2 x - 3 | \geq 5 .$ Graph the solution and write the solution in interval notation.

Solution

	$\| 2 x - 3 \| \geq 5$
Step 1. Isolate the absolute value expression. It is isolated.
Step 2. Write the equivalent compound inequality.	$2 x - 3 \leq −5 or 2 x - 3 \geq 5$
Step 3. Solve the compound inequality.	$2 x \leq − 2 or 2 x \geq 8$ $x \leq − 1 or x \geq 4$
Step 4. Graph the solution.
Step 5. Write the solution using interval notation.	$(− \infty, −1] \cup [4, \infty)$
Check: The check is left to you.

Try It 2.151

Solve $| 4 x - 3 | \geq 5 .$ Graph the solution and write the solution in interval notation.

Try It 2.152

Solve $| 3 x - 4 | \geq 2 .$ Graph the solution and write the solution in interval notation.

How To

Solve absolute value inequalities with > or ≥.

Step 1. Isolate the absolute value expression.
Step 2.
Write the equivalent compound inequality.

$\begin{matrix} | u | > a is equivalent to u < - a or u > a \\ | u | \geq a is equivalent to u \leq − a or u \geq a \end{matrix}$
Step 3. Solve the compound inequality.
Step 4. Graph the solution
Step 5. Write the solution using interval notation.

Solve Applications with Absolute Value

Absolute value inequalities are often used in the manufacturing process. An item must be made with near perfect specifications. Usually there is a certain tolerance of the difference from the specifications that is allowed. If the difference from the specifications exceeds the tolerance, the item is rejected.

| actual-ideal | \leq tolerance

Example 2.77

The ideal diameter of a rod needed for a machine is 60 mm. The actual diameter can vary from the ideal diameter by $0.075$ mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

Solution

	Let x = the actual measurement.
Use an absolute value inequality to express this situation.	$\| actual-ideal \| \leq tolerance$
	$\| x - 60 \| \leq 0.075$
Rewrite as a compound inequality.	$− 0.075 \leq x - 60 \leq 0.075$
Solve the inequality.	$59.925 \leq x \leq 60.075$
Answer the question.	The diameter of the rod can be between 59.925 mm and 60.075 mm.

Demonstrates solving an absolute value inequality problem, from setup to finding the final measurement range.

Try It 2.153

The ideal diameter of a rod needed for a machine is 80 mm. The actual diameter can vary from the ideal diameter by 0.009 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

Try It 2.154

The ideal diameter of a rod needed for a machine is 75 mm. The actual diameter can vary from the ideal diameter by 0.05 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

Media

Access this online resource for additional instruction and practice with solving linear absolute value equations and inequalities.

Solving Linear Absolute Value Equations and Inequalities

Section 2.7 Exercises

Practice Makes Perfect

Solve Absolute Value Equations

In the following exercises, solve.

434.

ⓐ $| x | = 6$ ⓑ $| y | = −3$ ⓒ $| z | = 0$

435.

ⓐ $| x | = 4$ ⓑ $| y | = −5$ ⓒ $| z | = 0$

436.

ⓐ $| x | = 7$ ⓑ $| y | = −11$ ⓒ $| z | = 0$

437.

ⓐ $| x | = 3$ ⓑ $| y | = −1$ ⓒ $| z | = 0$

438.

$| 2 x - 3 | - 4 = 1$

439.

$| 4 x - 1 | - 3 = 0$

440.

$| 3 x - 4 | + 5 = 7$

441.

$| 4 x + 7 | + 2 = 5$

442.

$4 | x - 1 | + 2 = 10$

443.

$3 | x - 4 | + 2 = 11$

444.

$3 | 4 x - 5 | - 4 = 11$

445.

$3 | x + 2 | - 5 = 4$

446.

$−2 | x - 3 | + 8 = −4$

447.

$−3 | x - 4 | + 4 = −5$

448.

$| \frac{3}{4} x - 3 | + 7 = 2$

449.

$| \frac{3}{5} x - 2 | + 5 = 2$

450.

$| \frac{1}{2} x + 5 | + 4 = 1$

451.

$| \frac{1}{4} x + 3 | + 3 = 1$

452.

$| 3 x - 2 | = | 2 x - 3 |$

453.

$| 4 x + 3 | = | 2 x + 1 |$

454.

$| 6 x - 5 | = | 2 x + 3 |$

455.

$| 6 - x | = | 3 - 2 x |$

Solve Absolute Value Inequalities with “less than”

In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation.

456.

$| x | < 5$

457.

$| x | < 1$

458.

$| x | \leq 8$

459.

$| x | \leq 3$

460.

$| 3 x - 3 | \leq 6$

461.

$| 2 x - 5 | \leq 3$

462.

$| 2 x + 3 | + 5 < 4$

463.

$| 3 x - 7 | + 3 < 1$

464.

$| 4 x - 3 | < 1$

465.

$| 6 x - 5 | < 7$

466.

$| x - 4 | \leq −1$

467.

$| 5 x + 1 | \leq −2$

Solve Absolute Value Inequalities with “greater than”

In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation.

468.

$| x | > 3$

469.

$| x | > 6$

470.

$| x | \geq 2$

471.

$| x | \geq 5$

472.

$| 3 x - 8 | > − 1$

473.

$| x - 5 | > − 2$

474.

$| 3 x - 2 | > 4$

475.

$| 2 x - 1 | > 5$

476.

$| x + 3 | \geq 5$

477.

$| x - 7 | \geq 1$

478.

$3 | x | + 4 \geq 1$

479.

$5 | x | + 6 \geq 1$

In the following exercises, solve. For each inequality, also graph the solution and write the solution in interval notation.

480.

$2 | x + 6 | + 4 = 8$

481.

$| 6 x - 5 | = | 2 x + 3 |$

482.

$| 3 x - 4 | \geq 2$

483.

$| 2 x - 5 | + 2 = 3$

484.

$| 4 x - 3 | < 5$

485.

$| 3 x + 1 | - 3 = 7$

486.

$| 7 x + 2 | + 8 < 4$

487.

$5 | 2 x - 1 | - 3 = 7$

488.

$| 8 - x | = | 4 - 3 x |$

489.

$| x - 7 | > − 3$

Solve Applications with Absolute Value

In the following exercises, solve.

490.

A chicken farm ideally produces 200,000 eggs per day. But this total can vary by as much as 25,000 eggs. What is the maximum and minimum expected production at the farm?

491.

An organic juice bottler ideally produces 215,000 bottle per day. But this total can vary by as much as 7,500 bottles. What is the maximum and minimum expected production at the bottling company?

492.

In order to insure compliance with the law, Miguel routinely overshoots the weight of his tortillas by 0.5 gram. He just received a report that told him that he could be losing as much as $100,000 per year using this practice. He now plans to buy new equipment that guarantees the thickness of the tortilla within 0.005 inches. If the ideal thickness of the tortilla is 0.04 inches, what thickness of tortillas will be guaranteed?

493.

At Lilly’s Bakery, the ideal weight of a loaf of bread is 24 ounces. By law, the actual weight can vary from the ideal by 1.5 ounces. What range of weight will be acceptable to the inspector without causing the bakery being fined?

Writing Exercises

494.

Write a graphical description of the absolute value of a number.

495.

In your own words, explain how to solve the absolute value inequality, $| 3 x - 2 | \geq 4 .$

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve absolute value equations. In row 3, the I can was solve absolute value inequalities with “less than.” In row 4, the I can was solve absolute value inequalities with “greater than.” In row 5, the I can was solve applications with absolute value.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

2.7 Solve Absolute Value Inequalities

Learning Objectives

Be Prepared 2.17

Be Prepared 2.18

Be Prepared 2.19

Solve Absolute Value Equations

Absolute Value

Absolute Value Equations

Example 2.68

Solution

Try It 2.135

Try It 2.136

Example 2.69

How to Solve Absolute Value Equations

Solution

Try It 2.137

Try It 2.138

How To

Solve absolute value equations.

Example 2.70

Solution

Try It 2.139

Try It 2.140

Example 2.71

Solution

Try It 2.141

Try It 2.142

Equations with Two Absolute Values

Example 2.72

Solution

Try It 2.143

Try It 2.144

Solve Absolute Value Inequalities with “Less Than”

Absolute Value Inequalities with < < or ≤ ≤

Example 2.73

Solution

Try It 2.145

Try It 2.146

Example 2.74

Solution

Try It 2.147

Try It 2.148

How To

Solve absolute value inequalities with < or ≤.

Solve Absolute Value Inequalities with “Greater Than”

Absolute Value Inequalities with > or ≥

Example 2.75

Solution

Try It 2.149

Try It 2.150

Example 2.76

Solution

Try It 2.151

Try It 2.152

How To

Solve absolute value inequalities with > or ≥.

Solve Applications with Absolute Value

Example 2.77

Solution

Try It 2.153

Try It 2.154

Media

Section 2.7 Exercises

Practice Makes Perfect

Writing Exercises

Self Check

Absolute Value Inequalities with $<$ or $\leq$