Problem 78 - A Graphical Approach to Precalculus with Limits

Question

Explain how to solve an equation of the form \(|a x+b|=|c x+d|\) analytically.

Step-by-Step Solution

Answer

Solve using cases: \\(ax+b=cx+d\\) and \\(ax+b=-(cx+d)\\). Solutions are \\(x = \frac{d-b}{a-c}\\) and \\(x = \frac{-d-b}{a+c}\\).

Step 1: Understanding Absolute Value

Absolute value expressions, such as \(|ax+b|\) and \(|cx+d|\), represent the distance of a number from zero on a number line. This means that each expression can potentially equal a number or its negative counterpart. To solve \(|ax+b|=|cx+d|\), consider both scenarios where the contents inside the absolute value bars are equal and where they are negatives of each other.

Step 2: Set Up the System of Equations

To solve \(|ax+b|=|cx+d|\), consider the cases where \(ax+b = cx+d\) and the case where \((ax+b) = -(cx+d)\), which handles the symmetry of absolute values. Thus, create these two separate equations to solve: 1. \((1)\; ax + b = cx + d\)2. \(2)\; ax + b = -(cx + d)\)

Step 3: Solve the First Equation

For the equation \(1)\; ax + b = cx + d\), start by moving all terms involving \(x\) to one side and constant terms to the other side. This yields: \(ax - cx = d - b\)Now, factor out \(x\) and solve: \((a-c)x = d-b\)\(x = \frac{d-b}{a-c}\).

Step 4: Solve the Second Equation

For the equation \(2)\; ax + b = -(cx + d)\), first distribute the negative sign:\(ax + b = -cx - d\).Rearrange to combine like terms:\(ax + cx = -d - b\).Factor out \(x\): \((a+c)x = -d-b\).Solve for \(x\):\(x = \frac{-d-b}{a+c}\).

Step 5: Check the Solutions

Substitute the obtained solutions, \(x = \frac{d-b}{a-c}\) and \(x = \frac{-d-b}{a+c}\), back into the original equation \(|ax+b|=|cx+d|\) to verify they are valid solutions. Both should satisfy the original equation, unless a solution drives the absolute value expression to be undefined or contradicts the absolute value definitions.

Key Concepts

Solving Equations

When tackling equations that have absolute values, understanding the nature of these values is crucial. Absolute value represents the distance a number is from zero on the number line. Because of this, when you have an equation like \(|a x+b| = |c x+d|\), you're dealing with two main scenarios. These scenarios involve either equating the expressions inside the absolute value or considering them as negatives of each other.
To begin solving, you set up two separate equations based on these scenarios. For the first scenario, you'll create an equation where these expressions are directly equal:

• \(a x + b = c x + d\)

For the second scenario, consider the negative equality:

• \(a x + b = -(c x + d)\)

By creating these equations, we consider the full range of possible solutions that take into account the distances on the number line.

Analytical Methods

To solve an absolute value equation analytically, breaking down the problem into manageable parts is crucial. Analytical methods guide you through the arithmetic and algebraic manipulations. By acting systematically, you avoid errors and ensure all solutions are found.
For the equation \(a x + b = c x + d\), rearrange the terms to gather all variables on one side. Perform operations such as subtraction or addition to isolate terms with \(x\):

• Move terms involving \(x\) to one side: \(ax - cx\)
• Move constant terms to the other side: \(d - b\)

Next, factor \(x\) out of the terms to solve for \(x\):

• \((a-c)x = d-b\)
• Solution: \(x = \frac{d-b}{a-c}\)

For the second equation, \(a x + b = -(c x + d)\), proceed in a similar manner by distributing and rearranging terms:

• Distribute the negative sign: \(ax + b = -cx - d\)
• Combine like terms and solve for \(x\): \(ax + cx = -d - b\)
• Solution: \(x = \frac{-d-b}{a+c}\)

By using these methods, you ensure that each potential solution is extracted correctly.

Mathematical Proofs

Verifying solutions through mathematical proofs is an essential part of solving equations. Once potential solutions are found, substitute them back into the original equation to confirm. For an equation involving absolute values like \(|a x+b|=|c x+d|\), each solution must not only satisfy the numerical equality but also adhere to absolute value properties.
For example, take the solution \(x = \frac{d-b}{a-c}\) from the equation \(a x + b = c x + d\). Substituting back, you should ensure:

• Both sides of the equation result in the same absolute value
• No contradictions to absolute value are present, such as undefined expressions

Perform the same check for the solution \(x = \frac{-d-b}{a+c}\) derived from \(a x + b = -(c x + d)\). This step validates that all derived solutions are consistent and correct under the properties of absolute values.

Precalculus Concepts

Analyzing absolute value equations incorporates several precalculus concepts, equipping students with necessary problem-solving skills in mathematics. The main precalculus concept here is the deep understanding of absolute value properties as elucidated above.
Incorporating these concepts:

• Helps visualize solutions on a number line
• Encourages rearranging terms to isolate variables in an equation
• Fosters the skill of checking solutions for contradictions

Such an analysis is important in preparing for calculus, where deeper understanding of functions and limits require precision and clarity in foundational algebra and precalculus skills.