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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.

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

À vous Aucune calculatrice ne tranche celle-ci, mais ses éléments sont calculables. Essayez-en un ci-dessous, ou tapez le vôtre.

Exercices (40)

Essayez chacun d'abord sur papier. Révélez la réponse pour vérifier; les vérifiés peuvent être ouverts dans le solveur pour chaque étape.

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

    Révèle la réponse

    \(-7\)

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

    Révèle la réponse

    a. >; b. >; c. >; d. >

  3. Simplify: \(14-2|8-3(4-1)|.\)

    Révèle la réponse

    \(12\)

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

    Révèle la réponse

    ⓐ

    \(|x|=8\)
    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.

  5. Solve: ⓐ \(|x|=2\) ⓑ \(|y|=-4\) ⓒ \(|z|=0\)

    Révèle la réponse

    ⓐ \(\pm 2\) ⓑ no solution ⓒ 0

  6. Solve: ⓐ \(|x|=11\) ⓑ \(|y|=-5\) ⓒ \(|z|=0\)

    Révèle la réponse

    ⓐ \(\pm 11\) ⓑ no solution ⓒ 0

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

  8. Solve: \(|3x-5|-1=6.\)

    Révèle la réponse

    \(x=4,x=-\frac{2}{3}\)

  9. Solve: \(|4x-3|-5=2.\)

    Révèle la réponse

    \(x=-1,x=\frac{5}{2}\)

  10. Solve \(2|x-7|+5=9.\)

    Révèle la réponse

    \(2|x-7|+5=9\)
    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:

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

    Révèle la réponse

    \(x=8,x=0\)

  12. Solve: \(2|x-5|+3=9.\)

    Révèle la réponse

    \(x=8,x=2\)

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

    Révèle la réponse

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

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

    Révèle la réponse

    No solution

  15. Solve: \(|\frac{5}{6}x+3|+8=6.\)

    Révèle la réponse

    No solution

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

    Révèle la réponse

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

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

    Révèle la réponse

    \(x=-\frac{2}{5},\) \(x=\frac{5}{2}\)

  18. Solve: \(|6x-5|=|3x+4|.\)

    Révèle la réponse

    \(x=3,\) \(x=\frac{1}{9}\)

  19. Solve \(|x|<7.\) Graph the solution and write the solution in interval notation.

    Révèle la réponse

    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.

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

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

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

    Révèle la réponse

    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.

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

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

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

    Révèle la réponse

    \(|x|>4\)
    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.

  26. Solve \(|x|>2.\) Graph the solution and write the solution in interval notation.

  27. Solve \(|x|>1.\) Graph the solution and write the solution in interval notation.

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

    Révèle la réponse

    \(|2x-3|\ge 5\)
    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.

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

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

  31. 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?

    Révèle la réponse

    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.

  32. 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?

    Révèle la réponse

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

  33. 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?

    Révèle la réponse

    The diameter of the rod can be between 74.95 and 75.05 mm.

  34. ⓐ \(|x|=6\) ⓑ \(|y|=-3\) ⓒ \(|z|=0\)

  35. ⓐ \(|x|=4\) ⓑ \(|y|=-5\) ⓒ \(|z|=0\)

    Révèle la réponse

    ⓐ \(x=4,x=-4\) ⓑ no solution ⓒ \(z=0\)

  36. ⓐ \(|x|=7\) ⓑ \(|y|=-11\) ⓒ \(|z|=0\)

  37. ⓐ \(|x|=3\) ⓑ \(|y|=-1\) ⓒ \(|z|=0\)

    Révèle la réponse

    ⓐ \(x=3,x=-3\) ⓑ no solution ⓒ \(z=0\)

  38. \(|2x-3|-4=1\)

  39. \(|4x-1|-3=0\)

    Révèle la réponse

    \(x=1,x=-\frac{1}{2}\)

  40. \(|3x-4|+5=7\)
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Comment faire: Solve Absolute Value Inequalities

  1. Solve absolute value equations
  2. Solve absolute value inequalities with “less than”
  3. Solve absolute value inequalities with “greater than”
  4. Solve applications with absolute value
  5. Isolate the absolute value expression.
  6. Write the equivalent equations.
  7. Solve each equation.
  8. Check each solution.

Questions posées par les gens

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.

Certaines parties de cette page sont adaptées de OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensés et réexpliqués ici ; les erreurs sont les nôtres. Comme sa source est autorisée à des fins non commerciales, cette leçon est gratuite, ne vend rien et est partagée avec la même licence.

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