2.7 Solve Absolute Value Inequalities

Learning objectives

Learning objectives

• Solve absolute value equations

• Solve absolute value inequalities with “less than”

• Solve absolute value inequalities with “greater than”

• Solve applications with absolute value

Solve Absolute Value Equations

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

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.

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.

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

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

Example 2.68

Solution

ⓐ

$\ · x\ · =8$ Write the equivalent equations. · $x=-8\,\mathrm{or}\,x=8$

$x=\pm{}8$

ⓑ

$\ · y\ · =-6$

No solution

Since an absolute value is always positive, there are no solutions to this equation.

ⓒ

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

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

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

Example 2.70

Solution

Remember, an absolute value is always positive!

Example 2.71

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

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

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

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

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

Example 2.72

Solution

$\ · 5x-1\ · =\ · 2x+3\ ·$ Write the equivalent equations.Solve each equation. · $\begin{array}{cccrlllrll} & & & \,5x-1 & = & -(2x+3) & \,\mathrm{or}\, & 5x-1 & = & 2x+3 \\ & & & \,5x-1 & = & -2x-3 & \,\mathrm{or}\, & 3x-1 & = & 3 \\ & & & \,7x-1 & = & -3 & & 3x & = & 4 \\ & & & \,7x & = & -2 & & x & = & \frac{4}{3} \\ & & & \,x & = & -\frac{2}{7} & \,\mathrm{or}\, & x & = & \frac{4}{3} \\ \end{array}$

Check.

We leave the check to you.

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.

What about the inequality $|x|\le{}5?$ 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.

In a more general way, we can see that if $|u|\le{}a,$ then $-a\le{}u\le{}a.$ See Figure 2.8.

This result is summarized here.

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

Solution

Check:

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

Example 2.74

Solution

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 $|x|\le{}5.$ 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.

Now we want to look at the inequality $|x|\ge{}5.$ 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.

In a more general way, we can see that if $|u|\ge{}a,$ then $u\le-a$ or $u\ge{}a.$ See Figure 2.11.

This result is summarized here.

Example 2.75

Solution

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

Example 2.76

Solution

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.

Example 2.77

Solution

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

Section 2.7 Exercises

Section 2.7 Exercises

Solve Absolute Value Equations

In the following exercises, solve.

Solve Absolute Value Inequalities with “less than”

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

Solve Absolute Value Inequalities with “greater than”

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

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

Solve Applications with Absolute Value

In the following exercises, solve.

Writing Exercises

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

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

1. Solve: ⓐ $|x|=2$ ⓑ $|y|=-4$ ⓒ $|z|=0$

2. Solve: ⓐ $|x|=11$ ⓑ $|y|=-5$ ⓒ $|z|=0$

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

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

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

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

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

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

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

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

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

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

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

2. Solve $|4x-5|\le{}3.$ Graph the solution and write the solution in interval notation:

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

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

1. Solve $|4x-3|\ge{}5.$ Graph the solution and write the solution in interval notation.

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

3. 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?

4. 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?

1. ⓐ $|x|=4$ ⓑ $|y|=-5$ ⓒ $|z|=0$

2. ⓐ $|x|=3$ ⓑ $|y|=-1$ ⓒ $|z|=0$

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

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

1. $3|x-4|+2=11$

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

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

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

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

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

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

4. $|x|<1$

1. $|x|\le{}3$

2. $|2x-5|\le{}3$

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

4. $|6x-5|<7$

1. $|5x+1|\le-2$

2. $|x|>6$

3. $|x|\ge{}5$

4. $|x-5|>-2$

1. $|2x-1|>5$

2. $|x-7|\ge{}1$

3. $5|x|+6\ge{}1$

4. $|6x-5|=|2x+3|$

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

2. $|3x+1|-3=7$

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

4. $|x-7|>-3$

1. 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?

2. 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?

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