Absolute Value Inequalities Practice Problems & Solutions

Master solving absolute value inequalities with step-by-step practice problems. Learn geometric and algebraic methods with detailed solutions and examples.

πŸ“šMaster Absolute Value Inequalities with Interactive Practice
  • Solve absolute value inequalities using geometric method with number lines
  • Apply algebraic method by splitting into positive and negative cases
  • Determine solution ranges for |x-a| < b and |x-a| > b inequalities
  • Find common domains and mark valid ranges on number lines
  • Practice compound inequalities with absolute value expressions
  • Convert between geometric visualization and algebraic notation

Understanding Inequalities with Absolute Values

Complete explanation with examples

Inequality

When you come across signs like < < or ,> > , and even ≀≀ or β‰₯β‰₯ ,you will know it is an inequality.
Inequalities define ranges of possible values rather than single solutions, whether one value is less than, greater than, or equal to another, helping to describe situations where quantities can vary within certain limits. Therefore, The result of the inequality will be a certain range of values that you will have to find.

An important rule to keep in mind: when you double or divide both sides of the operation, the sign of the inequality is reversed!

Absolute Value Inequality

Absolute value inequality is an inequality that involves the absolute value of a variable or expression. The absolute value represents the distance of a number from zero on the number line, always as a non-negative quantity. In inequalities, this means solving for a range of values that satisfy the given condition, either greater than or less than a certain value, regardless of sign. Absolute value inequalities often split into two cases, one considering the positive side and the other the negative, providing two sets of possible solutions.

Number line visualizations of absolute value equations and inequalities, including |x| = 2, |x| β‰₯ 2, |x| ≀ 2, |x| > 2, and |x| < 2, with open and closed circles to represent boundary inclusion.

We can solve absolute value inequalities in 2 2 ways:

The geometric method
  • We will draw a number line, and mark the point from which we are interested in the distance to X X .
  • Then we find the points whose distance from the relevant point is exactly the distance mentioned in the condition.
  • Now, let's pay attention to the condition: > > or < < And we will find the results.
The algebraic method
  • Step 1: Divide into two cases:
    • Case 1: Assume the expression inside the absolute value is positive.
    • Case 2: Assume the expression inside the absolute value is negative.
  • Step 2: Return to the original inequality:
    • Case 1: Remove the absolute value and solve.
    • Case 2: Remove the negative sign, solve the inequality.
  • Step 3: Find the common domain for both cases.
  • Step 4: Use a number line to mark valid ranges.
  • Step 5: Determine all values of X X that satisfy the inequality.
Detailed explanation

Practice Inequalities with Absolute Values

Test your knowledge with 10 quizzes

Given:

\( |x+5| < 2 \)

Which of the following statements is necessarily true?

Examples with solutions for Inequalities with Absolute Values

Step-by-step solutions included
Exercise #1

Given:

∣2xβˆ’1∣>βˆ’10 \left|2x-1\right|>-10

Which of the following statements is necessarily true?

Step-by-Step Solution

Let's solve the problem:
Step 1: Recognize that the inequality we are dealing with is ∣2xβˆ’1∣>βˆ’10 \left|2x-1\right| > -10 .
Step 2: Consider the nature of absolute values: for any real xx, ∣2xβˆ’1∣\left|2x-1\right| is always non-negative (i.e., β‰₯0\geq 0).
Step 3: Observe that the right side of the inequality, βˆ’10-10, is negative. Therefore, the inequality ∣2xβˆ’1∣>βˆ’10\left|2x-1\right| > -10 is always true because ∣2xβˆ’1∣\left|2x-1\right| as a non-negative quantity will always be greater than any negative number.
Step 4: Since the inequality condition always holds true, this means that the statement is valid for all xx.

Therefore, the correct answer is that the inequality holds for all xx.

The solution to the problem is For all x.

Answer:

For all x

Video Solution
Exercise #2

Given:

∣4x+12∣>16 \left|4x+12\right|>16

Which of the following statements is necessarily true?

Step-by-Step Solution

To solve this inequality, we use the principle that if ∣A∣>B |A| > B , then A>B A > B or A<βˆ’B A < -B . Here's a detailed step-by-step solution:

  • Set up the two inequalities from the absolute value expression:
    1. 4x+12>16 4x + 12 > 16
    2. 4x+12<βˆ’16 4x + 12 < -16
  • For the first inequality, 4x+12>16 4x + 12 > 16 :
    - Subtract 12 from both sides to isolate the term with x x :
    4x>16βˆ’12 4x > 16 - 12
    4x>4 4x > 4
    - Divide both sides by 4:
    x>1 x > 1
  • For the second inequality, 4x+12<βˆ’16 4x + 12 < -16 :
    - Subtract 12 from both sides:
    4x<βˆ’16βˆ’12 4x < -16 - 12
    4x<βˆ’28 4x < -28
    - Divide both sides by 4:
    x<βˆ’7 x < -7
  • The solution is the union of both results:
    x>1 x > 1 or x<βˆ’7 x < -7

Therefore, the correct answer among the given choices is x>1 x > 1 or x<βˆ’7 x < -7 .

Answer:

x>1 x>1 or x<βˆ’7 x<-7

Video Solution
Exercise #3

Given:

∣5xβˆ’10∣>15 \left|5x-10\right|>15

Which of the following statements is necessarily true?

Step-by-Step Solution

To solve the inequality ∣5xβˆ’10∣>15 \left|5x-10\right| > 15 , we rewrite it as two separate inequalities:

  • First inequality: 5xβˆ’10>15 5x - 10 > 15
  • Second inequality: 5xβˆ’10<βˆ’15 5x - 10 < -15

Let's solve each one:

For the first inequality 5xβˆ’10>15 5x - 10 > 15 :
Add 10 to both sides: 5x>255x > 25
Divide both sides by 5: x>5x > 5

For the second inequality 5xβˆ’10<βˆ’15 5x - 10 < -15 :
Add 10 to both sides: 5x<βˆ’55x < -5
Divide both sides by 5: x<βˆ’1x < -1

Combining these solutions, we have:
x>5x > 5 or x<βˆ’1x < -1

Therefore, the correct statement regarding the solution set is: x>5 x > 5 or x<βˆ’1 x < -1 .

Answer:

x>5 x>5 or x<βˆ’1 x<-1

Video Solution
Exercise #4

Given:

∣3x+9∣<18 \left|3x+9\right|<18

Which of the following statements is necessarily true?

Step-by-Step Solution

To solve the inequality ∣3x+9∣<18\left|3x + 9\right| < 18, follow these steps:

  • Step 1: Remove the absolute value by expressing it as a double inequality:
    βˆ’18<3x+9<18 -18 < 3x + 9 < 18 .

  • Step 2: Simplify the inequality:
    First, subtract 9 from all parts:
    βˆ’18βˆ’9<3x+9βˆ’9<18βˆ’9 -18 - 9 < 3x + 9 - 9 < 18 - 9 ,
    which simplifies to βˆ’27<3x<9 -27 < 3x < 9 .

  • Step 3: Solve for xx by dividing the entire inequality by 3:
    βˆ’273<3x3<93 -\frac{27}{3} < \frac{3x}{3} < \frac{9}{3} ,
    resulting in βˆ’9<x<3 -9 < x < 3 .

Upon solving, we determine that the solution to the inequality is the interval:

βˆ’9<x<3 -9 < x < 3 .

Answer:

βˆ’9<x<3 -9 < x < 3

Video Solution
Exercise #5

Given:

∣2xβˆ’4∣<8 \left|2x-4\right|<8

Which of the following statements is necessarily true?

Step-by-Step Solution

To solve the absolute value inequality ∣2xβˆ’4∣<8 |2x - 4| < 8 , we begin by removing the absolute value expression. This gives us a compound inequality:

βˆ’8<2xβˆ’4<8-8 < 2x - 4 < 8.

We will solve this compound inequality by handling each part separately:

  • Start with the left inequality: βˆ’8<2xβˆ’4-8 < 2x - 4.
    • Add 4 to both sides to isolate the term with x x : βˆ’8+4<2x-8 + 4 < 2x.
    • Simplify: βˆ’4<2x-4 < 2x.
    • Finally, divide both sides by 2: βˆ’2<x-2 < x.
  • Now, solve the right inequality: 2xβˆ’4<82x - 4 < 8.
    • Add 4 to both sides to isolate the term with x x : 2xβˆ’4+4<8+42x - 4 + 4 < 8 + 4.
    • Simplify: 2x<122x < 12.
    • Finally, divide both sides by 2: x<6x < 6.

Combining the two solutions from the parts, we find:

βˆ’2<x<6-2 < x < 6.

The solution indicates that x x must be greater than -2 and less than 6. This form matches answer choice 4. Therefore, the correct solution is:

βˆ’2<x<6-2 < x < 6.

Answer:

βˆ’2<x<6 -2 < x < 6

Video Solution

Frequently Asked Questions

What is the difference between absolute value equations and inequalities?

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Absolute value equations like |x| = 2 have specific solutions (x = 2 or x = -2), while absolute value inequalities like |x| < 2 represent ranges of values (-2 < x < 2). Inequalities describe intervals on the number line rather than discrete points.

How do you solve |x - 3| < 7 step by step?

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For |x - 3| < 7: 1) Set up two cases: x - 3 β‰₯ 0 and x - 3 < 0, 2) Case 1: x - 3 < 7, so x < 10, 3) Case 2: -(x - 3) < 7, so x > -4, 4) Find intersection: -4 < x < 10.

When do you flip the inequality sign in absolute value problems?

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You flip the inequality sign when multiplying or dividing both sides by a negative number. This happens in the negative case when solving algebraically, such as when -x < 4 becomes x > -4 after dividing by -1.

What's the geometric method for solving absolute value inequalities?

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The geometric method uses a number line to visualize distance. For |x - a| < b, mark point 'a', then find all points within distance 'b' from 'a'. For |x - a| > b, find all points outside this distance range.

How do you solve absolute value inequalities with greater than signs?

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For |x - a| > b, the solution is x < a - b OR x > a + b (two separate intervals). For |x - a| < b, the solution is a - b < x < a + b (one continuous interval between the boundaries).

What are common mistakes when solving absolute value inequalities?

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Common errors include: 1) Forgetting to consider both positive and negative cases, 2) Not flipping inequality signs when dividing by negatives, 3) Using AND instead of OR for greater-than inequalities, 4) Incorrectly finding the intersection of solution sets.

Why do absolute value inequalities have two cases?

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Absolute value represents distance from zero, which is always non-negative. The expression inside can be positive or negative, so we consider both scenarios: when the expression equals itself (positive case) and when it equals its opposite (negative case).

How do you check your solution to an absolute value inequality?

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Test values from your solution interval and outside it in the original inequality. Values in your solution should make the inequality true, while values outside should make it false. Also verify boundary points match the inequality type (< vs ≀).

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