Solving |ax + b| = c (for c ≥ 0)

Introduction

Key Concepts

The Definition of Absolute Value

The absolute value of a number represents its distance from zero on the number line, regardless of direction. Mathematically, the absolute value of a real number $x$ is denoted as $|x|$ and is defined as:

$$ |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} $$

This definition implies that absolute values are always non-negative. For example, $|5| = 5$ and $|-5| = 5$.

Understanding the Equation |ax + b| = c

The equation $|ax + b| = c$ involves determining the value(s) of $x$ that satisfy the absolute value condition. Here, $a$ and $c$ are constants, with $c \geq 0$ to ensure the equation has real solutions. The general approach to solving such equations involves considering two cases based on the definition of absolute value:

1. Case 1: $ax + b \geq 0$
2. Case 2: $ax + b < 0$

By solving each case separately, we can find all possible solutions for $x$.

Solving the First Case: ax + b ≥ 0

When $ax + b \geq 0$, the absolute value equation simplifies as follows:

$$ |ax + b| = ax + b = c $$

To find $x$, we solve the linear equation:

$$ ax + b = c \\ ax = c - b \\ x = \frac{c - b}{a} $$

This provides one potential solution, provided that it satisfies the initial condition $ax + b \geq 0$.

Solving the Second Case: ax + b < 0

When $ax + b < 0$, the absolute value equation changes to:

$$ |ax + b| = -(ax + b) = c $$

Solving for $x$ involves:

$$ -(ax + b) = c \\ -ax - b = c \\ -ax = c + b \\ x = \frac{-(c + b)}{a} = \frac{-c - b}{a} $$

This yields the second potential solution, contingent on the condition $ax + b < 0$ being true.

Combined Solution

After solving both cases, the complete solution set for $x$ is:

$$ x = \frac{c - b}{a} \quad \text{or} \quad x = \frac{-c - b}{a} $$

However, it's essential to verify that each solution satisfies its respective case condition. If a solution does not meet its case's initial assumption, it should be discarded.

Example 1: Solving |2x + 3| = 7

Let's solve the equation $|2x + 3| = 7$ step by step.

• Case 1: $2x + 3 \geq 0$

$$ 2x + 3 = 7 \\ 2x = 7 - 3 \\ 2x = 4 \\ x = 2 $$

• Verification: $2(2) + 3 = 7 \geq 0$ ✅

• Case 2: $2x + 3 < 0$

$$ -(2x + 3) = 7 \\ -2x - 3 = 7 \\ -2x = 7 + 3 \\ -2x = 10 \\ x = -5 $$

• Verification: $2(-5) + 3 = -7 < 0$ ✅

Solutions: $x = 2$ and $x = -5$

Example 2: Solving | -3x + 6 | = 9

Consider the equation $| -3x + 6 | = 9$. We'll apply the same method:

• Case 1: $-3x + 6 \geq 0$

$$ -3x + 6 = 9 \\ -3x = 9 - 6 \\ -3x = 3 \\ x = -1 $$

• Verification: $-3(-1) + 6 = 9 \geq 0$ ✅

• Case 2: $-3x + 6 < 0$

$$ -(-3x + 6) = 9 \\ 3x - 6 = 9 \\ 3x = 15 \\ x = 5 $$

• Verification: $-3(5) + 6 = -9 < 0$ ✅

Solutions: $x = -1$ and $x = 5$

General Steps to Solve |ax + b| = c

1. Ensure that $c \geq 0$. If $c < 0$, the equation has no real solutions.
2. Set up two separate equations based on the definition of absolute value:
   • $ax + b = c$
   • $ax + b = -c$
3. Solve each equation for $x$.
4. Check each solution in the original equation to ensure validity.

Graphical Interpretation

Graphically, the equation $|ax + b| = c$ represents the points where the graph of $y = |ax + b|$ intersects the horizontal line $y = c$. Since absolute value functions produce a V-shape, there can be two, one, or no points of intersection based on the value of $c$ and the position of the V.

- If $c > 0$, there are typically two solutions, corresponding to the two arms of the V.

- If $c = 0$, there is one solution, the vertex of the V.

- If $c < 0$, no real solutions exist, as the absolute value cannot be negative.

Applications of Absolute Value Equations

Absolute value equations are used in various fields to model real-world situations where magnitude without regard to direction is important. Examples include:

• Distance Problems: Calculating the distance between two points on a number line.
• Engineering: Determining tolerances and deviations in manufacturing processes.
• Economics: Measuring fluctuations in stock prices or economic indicators.

Common Mistakes to Avoid

• Forgetting to consider both cases when solving absolute value equations.
• Neglecting to check if the solutions satisfy the original equation's conditions.
• Incorrectly handling negative coefficients within the absolute value expression.
• Misapplying the properties of absolute values during equation manipulation.

Practice Problems

To reinforce the understanding of solving absolute value equations, consider the following exercises:

1. Solve $|4x - 8| = 12$.
2. Solve $| -5x + 10 | = 15$.
3. Solve $|3x + 7| = 2$.
4. Solve $|2x - 4| = -5$ and explain why there is no solution.

Solutions:

1. $|4x - 8| = 12$
   • Case 1: $4x - 8 = 12$ $4x = 20$ $x = 5$
   • Case 2: $4x - 8 = -12$ $4x = -4$ $x = -1$
   Solutions: $x = 5$, $x = -1$
2. $| -5x + 10 | = 15$
   • Case 1: $-5x + 10 = 15$ $-5x = 5$ $x = -1$
   • Case 2: $-5x + 10 = -15$ $-5x = -25$ $x = 5$
   Solutions: $x = -1$, $x = 5$
3. $|3x + 7| = 2$
   • Case 1: $3x + 7 = 2$ $3x = -5$ $x = -\frac{5}{3}$
   • Case 2: $3x + 7 = -2$ $3x = -9$ $x = -3$
   Solutions: $x = -\frac{5}{3}$, $x = -3$
4. $|2x - 4| = -5$ Since the absolute value cannot be negative, there are no real solutions.

Word Problem Example

Problem: A mobile phone company's contract costs $|50x + 100|$ dollars per month, where $x$ represents the number of additional services subscribed to. If a customer wants to spend exactly $200$ dollars per month, find the number of additional services they can subscribe to.

Solution: Set up the equation: $$ |50x + 100| = 200 $$

• Case 1: $50x + 100 = 200$ $50x = 100$ $x = 2$
• Case 2: $50x + 100 = -200$ $50x = -300$ $x = -6$

Since the number of additional services cannot be negative, the only feasible solution is $x = 2$.

Answer: The customer can subscribe to 2 additional services.

Advanced Concepts

Theoretical Foundations of Absolute Value Equations

Delving deeper into absolute value equations, it's essential to explore their theoretical underpinnings. Absolute value functions are piecewise linear functions defined by different expressions over varying domains. This piecewise nature makes absolute value equations an excellent tool for introducing students to piecewise functions and their properties.

Derivation from Piecewise Functions

The absolute value function can be expressed as:

$$ f(x) = |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} $$

Similarly, for the equation $|ax + b| = c$, we derive two separate linear equations based on the interval in which $ax + b$ falls. This piecewise approach ensures comprehensive coverage of all possible scenarios, providing a foundation for understanding more complex functions and equations.

Solving Systems Involving Absolute Value Equations

In more advanced settings, absolute value equations often appear within larger systems of equations. Solving such systems requires integrating absolute value techniques with other algebraic methods.

Example: Solving a System

Solve the following system of equations:

$$ \begin{cases} |2x + 1| = 5 \\ x - y = 3 \end{cases} $$

Solution:

• From the first equation: $|2x + 1| = 5$
  • Case 1: $2x + 1 = 5$ $2x = 4$ $x = 2$
  • Case 2: $2x + 1 = -5$ $2x = -6$ $x = -3$
• From the second equation: $y = x - 3$

Substituting the $x$ values into the second equation:

• If $x = 2$, then $y = 2 - 3 = -1$
• If $x = -3$, then $y = -3 - 3 = -6$

Solutions: $(2, -1)$ and $(-3, -6)$

Graphical Representation of Solutions

Visualizing the solutions to absolute value equations can enhance understanding. Consider the equation $|ax + b| = c$ graphed against the corresponding functions:

• $y = |ax + b|$
• $y = c$

The points of intersection between these two graphs represent the solutions to the equation. Utilizing graphing tools or software can aid in verifying solutions and exploring the behavior of absolute value functions under different parameters.

Non-Linear Absolute Value Equations

While the equation $|ax + b| = c$ is linear within each case, absolute value equations can also involve non-linear expressions. For instance, $|ax^2 + bx + c| = d$ introduces quadratic elements, adding complexity to the solution process. Solving such equations often requires factoring, completing the square, or using the quadratic formula within each case.

Interdisciplinary Connections

Understanding absolute value equations isn't confined to pure mathematics. These equations find applications across various disciplines:

• Physics: Modeling oscillatory motion where displacement is measured as an absolute value.
• Engineering: Calculating tolerances and ensuring components stay within specified limits.
• Economics: Analyzing profit and loss scenarios where deviations from a target value are critical.

Complex Problem-Solving Techniques

Advanced problem-solving involving absolute value equations may require combining multiple algebraic techniques or integrating functions. Techniques such as substitution, elimination in systems, and leveraging inequalities are often employed to tackle intricate equations.

Example: Combining Absolute Values

Solve the equation $|x - 2| + |x + 3| = 10$.

Solution:

• Identify critical points where expressions inside absolute values change signs: $x = 2$ and $x = -3$.
• Divide the real line into intervals based on these points:

1. Interval 1: $x < -3$
   • $|x - 2| = -(x - 2) = -x + 2$
   • $|x + 3| = -(x + 3) = -x - 3$
   • Equation: $-x + 2 - x - 3 = 10$ $-2x - 1 = 10$ $-2x = 11$ $x = -\frac{11}{2}$
   • Verification: $-\frac{11}{2} < -3$ ✅
2. Interval 2: $-3 \leq x < 2$
   • $|x - 2| = -(x - 2) = -x + 2$
   • $|x + 3| = x + 3$
   • Equation: $-x + 2 + x + 3 = 10$ $5 = 10$ No solution in this interval.
3. Interval 3: $x \geq 2$
   • $|x - 2| = x - 2$
   • $|x + 3| = x + 3$
   • Equation: $x - 2 + x + 3 = 10$ $2x + 1 = 10$ $2x = 9$ $x = \frac{9}{2}$
   • Verification: $\frac{9}{2} \geq 2$ ✅

Solutions: $x = -\frac{11}{2}$ and $x = \frac{9}{2}$

Exploring Inequalities with Absolute Values

While the focus here is on equations, absolute value inequalities like $|ax + b| < c$ or $|ax + b| > c$ represent another layer of complexity. These inequalities define ranges of values for $x$ rather than specific solutions, broadening the scope of analysis.

For example, solving $|2x + 5| < 7$ involves finding all $x$ such that:

$$ -7 < 2x + 5 < 7 $$

Solving the compound inequality: $$ -7 - 5 < 2x < 7 - 5 \\ -12 < 2x < 2 \\ -6 < x < 1 $$

Thus, the solution set is $-6 < x < 1$.

Absolute Value in Complex Numbers

Extending beyond real numbers, absolute values play a pivotal role in complex analysis. The absolute value of a complex number $z = a + bi$ is defined as:

$$ |z| = \sqrt{a^2 + b^2} $$

This represents the magnitude of the complex number in the complex plane, analogous to the distance from the origin in a two-dimensional space. While this topic extends beyond the IGCSE syllabus, it showcases the versatility and importance of absolute value concepts in higher mathematics.

Historical Perspective

The concept of absolute value has its roots in ancient mathematics, where the need to measure magnitudes without direction arose in various scientific disciplines. Its formalization into mathematical definitions provided a foundation for modern algebra and analysis, influencing the development of calculus, geometry, and beyond.

Technological Tools for Solving Absolute Value Equations

With advancements in technology, various tools can aid in solving absolute value equations:

• Graphing Calculators: Visualizing absolute value functions and identifying solutions graphically.
• Computer Algebra Systems (CAS): Automating the solving process for complex equations.
• Online Solvers: Providing step-by-step solutions for educational purposes.

Comparison Table

Aspect	|ax + b| = c	|ax + b| < c	|ax + b| > c
Type of Solution	Discrete solutions (x-values)	Range of solutions (interval)	Two separate intervals of solutions
Graphical Representation	Intersection points with y = c	Region between y = -c and y = c	Regions outside y = -c and y = c
Number of Solutions	0, 1, or 2	Infinite solutions within a range	Infinite solutions in two ranges
Example	|2x + 3| = 7 → x = 2 or x = -5	|2x + 3| < 7 → -5 < x < 2	|2x + 3| > 7 → x < -5 or x > 2
Applications	Determining exact break-even points	Setting acceptable tolerance ranges	Identifying values that exceed thresholds

Summary and Key Takeaways