Problem 17 - Thomas Calculus

Question

Solve the equations in Exercises \(13-18\) $$ |8-3 s|=\frac{9}{2} $$

Step-by-Step Solution

Answer

Solutions: \(s = \frac{7}{6}\) and \(s = \frac{25}{6}\).

Step 1: Understand the Absolute Value Equation

We need to solve the equation \(|8 - 3s| = \frac{9}{2}\). The definition of absolute value \(|x|\) states that it represents the distance of a number \(x\) from zero on the number line. Therefore, we consider both the positive and negative scenarios: 1. \(8 - 3s = \frac{9}{2}\) 2. \(8 - 3s = -\frac{9}{2}\)

Step 2: Solving the Positive Case

First, let's solve the equation \(8 - 3s = \frac{9}{2}\). - Subtract 8 from both sides: \( -3s = \frac{9}{2} - 8 \) - Since 8 can be expressed as \(\frac{16}{2}\), we have: \( -3s = \frac{9}{2} - \frac{16}{2}\) \( -3s = -\frac{7}{2}\) - Divide both sides by -3 to solve for \(s\): \( s = \frac{-\frac{7}{2}}{-3}\) \( s = \frac{7}{6}\)

Step 3: Solving the Negative Case

Next, solve the equation \(8 - 3s = -\frac{9}{2}\). - Subtract 8 from both sides: \( -3s = -\frac{9}{2} - 8 \) - Simplify 8 as \(\frac{16}{2}\): \( -3s = -\frac{9}{2} - \frac{16}{2}\) \( -3s = -\frac{25}{2}\) - Divide both sides by -3: \( s = \frac{-\frac{25}{2}}{-3}\) \( s = \frac{25}{6}\)

Step 4: Solution Summary

The solutions to the equation \(|8 - 3s| = \frac{9}{2}\) are \(s = \frac{7}{6}\) and \(s = \frac{25}{6}\). Both solutions maintain the equality when substituted back, confirming their validity.

Key Concepts

Solving Equations

When solving equations, the main goal is to find the value or values of the variable that make the equation true. In the exercise, we encounter an absolute value equation which means we're focusing on finding the values of 's' that satisfy \( |8 - 3s| = \frac{9}{2} \).To solve such equations, we need to consider two scenarios because of the nature of absolute values:

• The expression inside the absolute value equals the positive case, \(8 - 3s = \frac{9}{2} \).
• The expression inside the absolute value equals the negative case, \(8 - 3s = -\frac{9}{2} \).

By setting up these separate equations, we can find the potential solutions for 's'. Remember that solving each case involves standard techniques such as isolating the variable and using inverse operations like subtraction and division. It's crucial to perform operations on both sides of the equation to maintain equality. Once solved, always check the solutions by substituting them back into the original equation to ensure they satisfy it.

Absolute Value

The concept of absolute value is fundamental in understanding and solving equations like the one given. Absolute value, denoted as \(|x|\), measures the distance a number \(x\) is from zero on the number line, without considering direction. This means that \(|x|\) is always non-negative regardless of whether \(x\) is positive or negative.For example, both \(|3|\) and \(|-3|\) would equal 3, highlighting that absolute value only captures magnitude, not a specific side of zero. In absolute value equations, it is critical to analyze both positive and negative scenarios. This ensures that both possible values for the expression inside the absolute value are considered. For our exercise, this results in two equations to solve, one for each possible state of the expression's sign inside \(|8 - 3s|\). Understanding this concept helps bridge the solution process, making sense of why we approach these equations by considering two cases.

Distance on Number Line

Understanding the absolute value as a measure of distance on the number line is another useful way to comprehend the solution process.When we refer to the distance on a number line, we focus on how far a point (or number) is from zero. Therefore, the absolute value indicates this distance without considering whether it's to the left or right of zero. For instance, in our problem, \(|8 - 3s| = \frac{9}{2}\) sets the condition that the distance of \(8 - 3s\) from zero is \(\frac{9}{2}\).This perspective can help visualize why breaking down the equation into both a positive and a negative scenario leads to valid solutions. By understanding the roles of distance and direction on a number line, it becomes clearer that for \(|x| = d\), both \(x = d\) and \(x = -d\) satisfy this distance condition, reflecting the fundamental property of absolute value.