Problem 13 - Elementary and Intermediate Algebra

Question

For each absolute value equation or inequality, write an equivalent compound equation or inequality. a. \(|x|=8 \quad\) b. \(|x| \geq 8\) \(\begin{array}{ll}{\text { c. }|x| \leq 8} & {\text { d. }|5 x-1|=|x+3|}\end{array}\)

Step-by-Step Solution

Answer

a. \(x = 8\) or \(x = -8\). b. \(x \geq 8\) or \(x \leq -8\). c. \(-8 \leq x \leq 8\). d. \(x = 1\) or \(x = -\frac{1}{3}\).

Step 1: Solve for Absolute Value Equation |x| = 8

For the equation \(|x| = 8\), this means that the value of \(x\) can be 8 units away from 0 on the number line. Therefore, \(x\) can either be 8 or -8. This results in the equivalent equations: \(x = 8\) and \(x = -8\).

Step 2: Convert Absolute Value Inequality |x| ≥ 8

For \(|x| \geq 8\), the value of \(x\) needs to be 8 units away from 0 either in the positive or negative direction or farther. Thus, \(x \geq 8\) or \(x \leq -8\). These are the two scenarios that satisfy the inequality.

Step 3: Convert Absolute Value Inequality |x| ≤ 8

The inequality \(|x| \leq 8\) means that \(x\) must be within 8 units of 0, including 8 itself. Therefore, this gives the compound inequality \(-8 \leq x \leq 8\).

Step 4: Solve Absolute Value Equation involving two expressions |5x-1| = |x+3|

To solve \(|5x-1|=|x+3|\), consider two cases: **Case 1:** When \(5x-1 = x+3\). Solving this gives:\[5x - 1 = x + 3 \4x = 4 \x = 1\]**Case 2:** When \(5x-1 = -(x+3)\). Solving this gives:\[5x - 1 = -x - 3 \6x = -2 \x = -\frac{1}{3}\]Thus, the solutions are \(x = 1\) and \(x = -\frac{1}{3}\).

Key Concepts

Understanding Compound Equations

Compound equations involve two or more equations that are combined into one statement with an 'and' or an 'or'. This is especially useful when dealing with absolute value equations and inequalities.

• The purpose is to express the solutions to absolute value situations, where distances in opposite directions are considered, in a straightforward way.
• In terms of absolute value, if we have an equation like \(|x| = 8\), it indicates that \(x\) can be either 8 or -8. This is because 8 units away from 0 on the number line can be in both the positive and negative directions.
• The equivalent compound equation would be \(x = 8\) or \(x = -8\).

Recognizing compound equations is essential when solving absolute value problems because it encapsulates the idea that a solution set might extend in two directions from a fixed point.

Visualizing Solutions on a Number Line

A number line can be a useful tool to visualize solutions for equations and inequalities, particularly those involving absolute values.

• Consider each point on the number line as representing possible values for \(x\).
• When discussing absolute value inequalities, like \(|x| \geq 8\), you can see this as two arrows shooting outward from \(-8\) and \(+8\) on the number line, extending towards the negative and positive infinities.
• This representation helps one intuitively understand the boundaries and scope of solution sets.

By imagining a number line, it becomes much simpler to grasp why \(x \geq 8\) or \(x \leq -8\) satisfies the inequality. This is because any number further from 0 than 8 is included in these solutions. Similarly, for \(|x| \leq 8\), envisioning a "segment" from \(-8\) to \(+8\) encapsulates the solution set clearly.

Deciphering Inequality Solutions

Inequality solutions are all about establishing a set or range of values that satisfy a particular condition, often expressed in terms of absolute values.

• For example, a compound inequality like \(-8 \leq x \leq 8\) for \(|x| \leq 8\) suggests that \(x\) can be any number from \(-8\) to \(+8\), inclusive.
• This expression highlights the range where \(x\) is allowed to "move" if it must stay within 8 units of 0.
• Whenever interpreting absolute value inequalities, translating them into compound inequalities clarifies the entire scope of potential solutions.

With this approach, solving inequalities becomes more intuitive, as it neatly pinpoints all numbers that satisfy the initial condition, covering all possible real-world interpretations of the problem.