Solving \( k|ax + b| > c \) and \( k|ax + b| \leq c \) for \( c > 0 \)

Introduction

Key Concepts

Understanding Absolute Value Inequalities

Absolute value inequalities involve expressions where the absolute value of a linear expression is compared to a constant. The general forms are:

• \( k|ax + b| > c \)
• \( k|ax + b| \leq c \)

Here, \( k \), \( a \), and \( b \) are constants, and \( c \) is a positive constant (\( c > 0 \)). Solving these inequalities requires understanding the properties of absolute values and how they interact with inequalities.

Basic Properties of Absolute Values

The absolute value of a real number \( y \), denoted by \( |y| \), is defined as:

• If \( y \geq 0 \), then \( |y| = y \).
• If \( y

This definition implies that \( |y| \) is always non-negative. When solving inequalities involving absolute values, it's essential to consider both the positive and negative scenarios.

Solving \( k|ax + b| > c \)

To solve the inequality \( k|ax + b| > c \), follow these steps:

1. Isolate the Absolute Value: If \( k \neq 0 \), divide both sides by \( k \) to simplify. $$ |ax + b| > \frac{c}{k} $$
2. Consider Two Cases:
   • Case 1: \( ax + b > \frac{c}{k} \)
   • Case 2: \( ax + b
3. Solve Each Case Separately:
   • For Case 1: $$ ax + b > \frac{c}{k} $$ $$ ax > \frac{c}{k} - b $$ $$ x > \frac{\frac{c}{k} - b}{a} $$
   • For Case 2: $$ ax + b
4. Combine the Solutions: The solution set is the union of the two cases. $$ x > \frac{\frac{c}{k} - b}{a} \quad \text{or} \quad x

Solving \( k|ax + b| \leq c \)

Solving the inequality \( k|ax + b| \leq c \) involves a similar approach with slight differences in the inequalities:

1. Isolate the Absolute Value: Divide both sides by \( k \) if \( k \neq 0 \). $$ |ax + b| \leq \frac{c}{k} $$
2. Set Up Compound Inequality: $$ -\frac{c}{k} \leq ax + b \leq \frac{c}{k} $$
3. Solve the Compound Inequality:
   • $$ -\frac{c}{k} \leq ax + b \leq \frac{c}{k} $$ Subtract \( b \) from all parts: $$ -\frac{c}{k} - b \leq ax \leq \frac{c}{k} - b $$ Divide by \( a \) (assuming \( a > 0 \); if \( a

Thus, the solution for \( k|ax + b| \leq c \) is:

Examples

Example 1: Solve \( 2|3x + 4| > 10 \)

1. Isolate the absolute value: $$ |3x + 4| > \frac{10}{2} $$ $$ |3x + 4| > 5 $$
2. Set up two cases:
   • Case 1: \( 3x + 4 > 5 \)
   • Case 2: \( 3x + 4
3. Solve Case 1: $$ 3x + 4 > 5 $$ $$ 3x > 1 $$ $$ x > \frac{1}{3} $$
4. Solve Case 2: $$ 3x + 4
5. Combine solutions: $$ x > \frac{1}{3} \quad \text{or} \quad x

Example 2: Solve \( -1|2x - 5| \leq 7 \)

1. Isolate the absolute value (note that \( k = -1 \)): $$ |2x - 5| \geq \frac{7}{-1} $$ Since dividing by a negative reverses the inequality: $$ |2x - 5| \geq -7 $$ However, since absolute values are always non-negative and \( -7 \) is negative, the inequality \( |2x - 5| \geq -7 \) is always true for all real numbers \( x \).

Graphical Interpretation

Graphing absolute value inequalities provides a visual understanding of the solution sets. For \( k|ax + b| > c \), the graph of \( y = k|ax + b| \) is compared to \( y = c \). The solution consists of the values of \( x \) where the graph of \( y = k|ax + b| \) lies above (for \( > \)) or below (for \( \leq \)) the horizontal line \( y = c \).

For instance, in \( 2|3x + 4| > 10 \), the graph of \( y = 2|3x + 4| \) will intersect \( y = 10 \) at \( x = \frac{1}{3} \) and \( x = -3 \). The solution \( x > \frac{1}{3} \) or \( x

Advanced Concepts

The Role of the Constant \( k \) in Inequalities

The constant \( k \) plays a pivotal role in determining the behavior of the absolute value expression in an inequality. Its value affects both the direction and the steepness of the graph.

• Positive \( k \): If \( k \) is positive, the graph of \( y = k|ax + b| \) opens upwards, maintaining the V-shape characteristic of absolute value functions.
• Negative \( k \): If \( k \) is negative, the graph reflects over the x-axis, turning the V-shape into an inverted V.

When \( k \) is negative in inequalities, it necessitates reversing the inequality sign when dividing or multiplying by \( k \). This reversal is crucial for maintaining the inequality's truth value.

Multi-Step Problem Solving

Solving absolute value inequalities can sometimes involve multiple steps, especially when nested absolute values or additional variables are present. Consider the inequality:

\( 3|2|x - 1| + 4| > 10 \)

To solve this, follow these steps:

1. Isolate the absolute value expression: $$ 2|x - 1| + 4 > \frac{10}{3} $$ $$ 2|x - 1| > \frac{10}{3} - 4 $$ $$ 2|x - 1| > \frac{-2}{3} $$ Since the right side is negative and absolute values are non-negative: $$ 2|x - 1| > \frac{-2}{3} $$ This inequality is always true for all real numbers \( x \).

Interdisciplinary Connections

Absolute value inequalities are not confined to pure mathematics; they have applications across various disciplines:

• Physics: Absolute values are used to describe quantities that are inherently non-negative, such as distance and speed. Inequalities involving absolute values can represent constraints in physical systems.
• Economics: In cost analysis, absolute value inequalities can model scenarios where costs must remain within certain bounds.
• Engineering: Tolerances in manufacturing processes often utilize absolute value inequalities to ensure components stay within specified limits.

Mathematical Derivations and Proofs

Delving deeper into the theory, consider deriving the solution set for \( k|ax + b| > c \).

Starting with:

Assuming \( k > 0 \):

By definition of absolute value:

Handling Complex Absolute Value Expressions

In certain problems, absolute value expressions may be nested or combined with other operations. For example:

\( |2|x + 3| - 4| \leq 5 \)

Solving such inequalities involves breaking down each absolute value layer:

1. Let \( y = |x + \frac{3}{2}| \), then the inequality becomes: $$ |2y - 4| \leq 5 $$
2. Solve the inner absolute value: $$ -5 \leq 2y - 4 \leq 5 $$ $$ -1 \leq 2y \leq 9 $$ $$ -\frac{1}{2} \leq y \leq \frac{9}{2} $$
3. Substitute back \( y = |x + \frac{3}{2}| \): $$ |x + \frac{3}{2}| \leq \frac{9}{2} $$ $$ -\frac{9}{2} \leq x + \frac{3}{2} \leq \frac{9}{2} $$ $$ -6 \leq x \leq 3 $$

This layered approach simplifies the process of solving complex absolute value inequalities.

Real-World Applications

Understanding absolute value inequalities is essential in modeling real-life situations where constraints and limitations are present. Examples include:

• Budgeting: Ensuring expenditures do not exceed a certain limit.
• Tolerances in Construction: Maintaining measurements within specified deviations.
• Quality Control: Keeping product defects below acceptable thresholds.

Comparison Table

Summary and Key Takeaways