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Solve Absolute Value Inequalities

Solve absolute value equations

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 \(|5|=5.\)

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.

\[\begin{array}{llllll}\text{If} & & & & & |x|=5 \\ \text{then} & & & & & \ x=-5\ \text{or}\ x=5\end{array}\]

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”.

Example

Try it.

Solve: ⓐ \(|x|=8\) ⓑ \(|y|=-6\) ⓒ \(|z|=0\)

Solution

ⓐ

	\(\|x\|=8\)
Write the equivalent equations.	\(x=-8\ \text{or}\ x=8\)
	\(x=\pm 8\)

ⓑ

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

ⓒ

	\(\|z\|=0\)
Write the equivalent equations.	\(z=-0\ \text{or}\ z=0\)
Since −0 = 0,	\(z=0\)
Both equations tell us that z = 0 and so there is only one solution.

How to Solve Absolute Value Equations

Try it.

Solve \(|5x-4|-3=8.\)

Solution

Example

Try it.

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.	\(\ x-7=\text{-}2\) or \(x-7=2\)
Solve each equation.	\(\ x=5\) or \(\ x=9\)
Check:

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

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.

\[\begin{array}{l}|x|=5 \\ x=-5\ \text{or}\ x=5\end{array}\]

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.

In a more general way, we can see that if \(|u|\le a,\) then \(\text{-}a\le u\le a.\)

This result is summarized here.

Absolute Value Inequalities with

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

\[\begin{array}{llllllllll}\text{if} & & & & |u|

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

Try it.

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.

Example

Try it.

Solve \(|5x-6|\le 4.\) Graph the solution and write the solution in interval notation.

Solution

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

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

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.

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.

In a more general way, we can see that if \(|u|\ge a,\) then \(u\le \text{-}a\) or \(u\ge a.\)

This result is summarized here.

Example

Try it.

Solve \(|x|>4.\) Graph the solution and write the solution in interval notation.

Solution

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

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

Example

Try it.

Solve \(|2x-3|\ge 5.\) Graph the solution and write the solution in interval notation.

Solution

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

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

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.

\[|\text{actual-ideal}|\le \text{tolerance}\]

Example

Try it.

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.	\(\|\text{actual-ideal}\|\le \text{tolerance}\)
	\(\|x-60\|\le 0.075\)
Rewrite as a compound inequality.	\(\text{-}0.075\le x-60\le 0.075\)
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.

Key Concepts

Absolute Value
The absolute value of a number is its distance from 0 on the number line.
The absolute value of a number n is written as \(|n|\) and \(|n|\ge 0\) for all numbers.
Absolute values are always greater than or equal to zero.
Absolute Value Equations
For any algebraic expression, u, and any positive real number, a,
\(\begin{array}{lllll}\text{if} & & & & |u|=a \\ \text{then} & & & & \ u=\text{-}a\ \text{or}\ u=a\end{array}\)
Remember that an absolute value cannot be a negative number.
How to Solve Absolute Value Equations
1. Isolate the absolute value expression.
2. Write the equivalent equations.
3. Solve each equation.
4. Check each solution.
Equations with Two Absolute Values
For any algebraic expressions, u and v,
\(\begin{array}{lllll}\text{if} & & & & |u|=|v| \\ \text{then} & & & & \ u=\text{-}v\ \text{or}\ u=v\end{array}\)
Absolute Value Inequalities with \(<\) or \(\le\)
For any algebraic expression, u, and any positive real number, a,
\begin{array}{lllllllllll}\text{if} & & & & & |u|
How To Solve Absolute Value Inequalities with \(<\) or \(\le\)
1. Isolate the absolute value expression.
2. Write the equivalent compound inequality.
  \begin{array}{lllllllllll}|u|
3. Solve the compound inequality.
4. Graph the solution
5. Write the solution using interval notation
Absolute Value Inequalities with \(>\) or \(\ge\)
For any algebraic expression, u, and any positive real number, a,
\(\begin{array}{lllllllllll}\text{if} & & & & & |u|>a, & & & & & \text{then}\ u<\text{-}a\ \text{or}\ u>a \\ \text{if} & & & & & |u|\ge a, & & & & & \text{then}\ u\le \text{-}a\ \text{or}\ u\ge a\end{array}\)
How To Solve Absolute Value Inequalities with \(>\) or \(\ge\)
1. Isolate the absolute value expression.
2. Write the equivalent compound inequality.
  \(\begin{array}{lllllllll}|u|>a & & & & \text{is equivalent to} & & & & u<\text{-}a\ \text{or}\ u>a \\ |u|\ge a & & & & \text{is equivalent to} & & & & u\le \text{-}a\ \text{or}\ u\ge a\end{array}\)
3. Solve the compound inequality.
4. Graph the solution
5. Write the solution using interval notation

이제 너 계산기는 이것을 해결하지 않지만, 그 조각은 계산 가능합니다. 아래의 하나를 시도하거나 자신의 것을 입력하십시오.

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Evaluate: \(\text{-}|7|.\)

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\(-7\)
Fill in \(<,>,\) or \(=\) for each of the following pairs of numbers.
ⓐ \(|-8|\text{\underline{\hspace{1.5em}}}-|-8|\) ⓑ \(12\text{\underline{\hspace{1.5em}}}-|-12|\) ⓒ \(|-6|\text{\underline{\hspace{1.5em}}}-6\) ⓓ \(\text{-}(-15)\text{\underline{\hspace{1.5em}}}-|-15|\)

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a. >; b. >; c. >; d. >
Simplify: \(14-2|8-3(4-1)|.\)

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\(12\)
Solve: ⓐ \(|x|=8\) ⓑ \(|y|=-6\) ⓒ \(|z|=0\)

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ⓐ
\(|x|=8\)
Write the equivalent equations. \(x=-8\ \text{or}\ x=8\)
\(x=\pm 8\)
ⓑ
\(|y|=-6\)
No solution
Since an absolute value is always positive, there are no solutions to this equation.
ⓒ
\(|z|=0\)
Write the equivalent equations. \(z=-0\ \text{or}\ z=0\)
Since −0 = 0, \(z=0\)
Both equations tell us that z = 0 and so there is only one solution.
Solve \(|5x-4|-3=8.\)
Solve: \(|3x-5|-1=6.\)

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\(x=4,x=-\frac{2}{3}\)
Solve: \(|4x-3|-5=2.\)

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\(x=-1,x=\frac{5}{2}\)
Solve \(2|x-7|+5=9.\)

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\(2|x-7|+5=9\)
Isolate the absolute value expression. \(2|x-7|=4\)
\(|x-7|=2\)
Write the equivalent equations. \(\ x-7=\text{-}2\) or \(x-7=2\)
Solve each equation. \(\ x=5\) or \(\ x=9\)
Check:
Solve: \(3|x-4|-4=8.\)

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\(x=8,x=0\)
Solve: \(2|x-5|+3=9.\)

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\(x=8,x=2\)
Solve: \(|\frac{2}{3}x-4|+11=3.\)

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\(|\frac{2}{3}x-4|+11=3\)
Isolate the absolute value term. \(|\frac{2}{3}x-4|=-8\)
An absolute value cannot be negative. No solution
Solve: \(|\frac{3}{4}x-5|+9=4.\)

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No solution
Solve: \(|\frac{5}{6}x+3|+8=6.\)

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No solution

Solve: \(|5x-1|=|2x+3|.\)

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

Solve: \(|7x-3|=|3x+7|.\)

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\(x=-\frac{2}{5},\) \(x=\frac{5}{2}\)
Solve: \(|6x-5|=|3x+4|.\)

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\(x=3,\) \(x=\frac{1}{9}\)
Solve \(|x|<7.\) Graph the solution and write the solution in interval notation.

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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.
Graph the solution and write the solution in interval notation: \(|x|<9.\)
Graph the solution and write the solution in interval notation: \(|x|<1.\)

Solve \(|5x-6|\le 4.\) Graph the solution and write the solution in interval notation.

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

Solve \(|2x-1|\le 5.\) Graph the solution and write the solution in interval notation:
Solve \(|4x-5|\le 3.\) Graph the solution and write the solution in interval notation:
Solve \(|x|>4.\) Graph the solution and write the solution in interval notation.

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\(|x|>4\)
Write the equivalent inequality. \(x<-4\ \text{or}\ x>4\)
Graph the solution.
Write the solution using interval notation. \((\text{-}\infty ,-4)\cup (4,\infty )\)
Check:
To verify, check a value in each section of the number line showing the solution. Choose numbers such as \(-6,\) 0, and 7.
Solve \(|x|>2.\) Graph the solution and write the solution in interval notation.
Solve \(|x|>1.\) Graph the solution and write the solution in interval notation.

Solve \(|2x-3|\ge 5.\) Graph the solution and write the solution in interval notation.

답을 드러내세요

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

Solve \(|4x-3|\ge 5.\) Graph the solution and write the solution in interval notation.
Solve \(|3x-4|\ge 2.\) Graph the solution and write the solution in interval notation.

답을 드러내세요

	Let x = the actual measurement.
Use an absolute value inequality to express this situation.	\(\|\text{actual-ideal}\|\le \text{tolerance}\)
	\(\|x-60\|\le 0.075\)
Rewrite as a compound inequality.	\(\text{-}0.075\le x-60\le 0.075\)
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.

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?

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The diameter of the rod can be between 79.991 and 80.009 mm.
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?

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The diameter of the rod can be between 74.95 and 75.05 mm.
\(|2x-3|-4=1\)
\(|4x-1|-3=0\)

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\(x=1,x=-\frac{1}{2}\)
\(|3x-4|+5=7\)

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어떻게: Solve Absolute Value Inequalities

Solve absolute value equations
Solve absolute value inequalities with “less than”
Solve absolute value inequalities with “greater than”
Solve applications with absolute value
Isolate the absolute value expression.
Write the equivalent equations.
Solve each equation.
Check each solution.

사람들이 묻는 질문

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

이 페이지의 일부는 다음에서 변경되었습니다. OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). 여기서 압축하고 다시 설명; 오류는 우리의. 소스가 비상업적 사용을 위해 라이선스되어 있기 때문에, 이 레슨은 무료이며, 아무것도 판매하지 않으며, 동일한 라이선스로 공유됩니다.

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