Understanding Absolute Value Inequalities

Strategy for Solving Absolute Value Inequalities

When you see an absolute value inequality, think of it as a distance problem on the number line. The absolute value tells you how far a value is from zero.

Here's a simple 3-step approach to solve these problems:

1. Draw a picture on the number line to represent the distance relationship
2. Use your drawing to rewrite the problem without absolute values (usually resulting in two inequalities connected by AND or OR)
3. Solve the resulting inequalities and simplify your answer

💡 Remember this pattern: "sandwich" inequalities using < or ≤ typically produce AND statements, while "outward" inequalities using > or ≥ typically produce OR statements.

Don't try to memorize every possible case! Understanding the distance concept will help you visualize and solve any absolute value inequality correctly.

Examples with Greater Than (>)

Let's solve: |3x-1| > 3

First, draw a picture on the number line. Since we want values where the distance is greater than 3, we're looking for points that are more than 3 units away from zero on the number line.

This gives us two possibilities:

• Either $3x-1$ < -3
• Or $3x-1$ > 3

Rewrite this as: $3x-1$ < -3 OR $3x-1$ > 3

Now solve both parts:

• 3x < -2, so x < -2/3
• 3x > 4, so x > 4/3

Our answer is x < -2/3 OR x > 4/3, written as $-∞, -2/3$ ∪ $4/3, ∞$

The solution represents two separate regions on the number line where the original inequality is true.

Examples with Less Than or Equal (≤)

Let's solve: |2-x| ≤ 4

Draw a picture on the number line. Since we want values where the distance is less than or equal to 4, we're looking for all points within 4 units of zero.

This gives us:

• $2-x$ ≥ -4
• AND $2-x$ ≤ 4

Solving both parts:

• -x ≥ -6, so x ≤ 6
• -x ≤ 2, so x ≥ -2

Combining these: -2 ≤ x ≤ 6, written as $-2, 6$

💡 Notice how the "less than or equal to" inequality gave us an AND statement, resulting in a single continuous interval rather than separate regions.

Your solution represents all the values between -2 and 6, inclusive, where the original inequality is satisfied.

More Complex Example

Let's solve: 4 < |x-11/3| + 7/3

First, rearrange to isolate the absolute value: |x-11/3| > 5/3

Now draw a picture! We're looking for points where the distance from 0 is greater than 5/3.

This gives us:

• $x-11/3$ < -5/3 OR $x-11/3$ > 5/3

Solving both parts:

• x < 6/3 (which simplifies to x < 2)
• x > 16/3

Our answer is x < 2 OR x > 16/3, written as $-∞, 2$ ∪ $16/3, ∞$

💡 Be careful when working with fractions in absolute value inequalities. It's usually safer to keep working with fractions rather than converting to decimals.

This solution represents two regions: all numbers less than 2, and all numbers greater than 16/3.

Special Cases: Negative or Zero Right Side

Absolute value inequalities get interesting when the right side is negative or zero!

Example: |4+2x| < -2

Since absolute value always gives us a positive number (or zero), it's impossible for it to be less than a negative number. The answer is ∅ (empty set) - no solutions exist.

Example: |7-2x| ≤ 0

Since absolute value can't be less than 0, this is only true when |7-2x| = 0, which happens when 7-2x = 0.

Solving: x = 7/2

This gives a single point as our answer: {7/2}. No interval exists because we're finding exactly where the expression equals zero.

When dealing with special cases, always think about what's actually possible with absolute values!

More Special Cases

Example: |4+2x| ≥ -3

Since absolute value is always positive or zero, and -3 is negative, this inequality is always true! The absolute value will always be greater than or equal to -3.

The answer is (-∞,∞) - all real numbers satisfy this inequality.

Example: |x-3| > 0

This inequality is asking: "when is the distance from x to 3 positive?" Since distance is only zero when x = 3 (and positive everywhere else), our answer is all real numbers except 3.

The solution is $-∞,3$ ∪ (3,∞).

💡 Always draw a picture! Visualizing these problems on the number line helps you avoid common mistakes and understand what's really happening.

These examples show that absolute value inequalities can sometimes have unexpected answers when dealing with special cases.