Problem 73 - Beginning and Intermediate Algebra

Question

The following exercises contain absolute value equations, linear inequalities, and both types of absolute value inequalities. Solve each. Write the solution set for equations in set notation and use interval notation for inequalities. $$|5 h-8|>7$$

Step-by-Step Solution

Answer

The solution set for the given absolute value inequality \(|5h - 8| > 7\) is: \((-\infty, \frac{1}{5}) \cup (3, \infty)\).

Step 1: Break down the absolute value inequality

Recall the definition of absolute value: if the expression inside the absolute value is positive or zero, the absolute value doesn't change it, but if it's negative, the absolute value reverses the sign. So, when considering inequalities with an absolute value, we're essentially concerned with two cases: when the expression inside the absolute value is positive or zero (and hence, equal to the absolute value), and when the expression is negative, and hence, equal to the negation of the absolute value. In our case, we have \(|5h - 8| > 7\). We can break this down into two cases: 1. The expression inside the absolute value is positive or zero: \(5h - 8 > 7\). 2. The expression inside the absolute value is negative and equal to the negation of its absolute value, which means: \(-(5h - 8) > 7\).

Step 2: Solve each inequality individually

Now, we'll solve each of the inequalities obtained in step 1 separately. 1. Solve \(5h - 8 > 7\): Add 8 to both sides of the inequality: \(5h > 15\) Divide both sides by 5: \(h > 3\) 2. Solve \(-(5h - 8) > 7\): Distribute the negative sign: \(-5h + 8 > 7\) Subtract 8 from both sides: \(-5h > -1\) Divide by -5 (and reverse the inequality sign since we're dividing by a negative number): \(h < \frac{1}{5}\)

Step 3: Combine the solutions and express them in interval notation

From step 2, we found that \(h > 3\) and \(h < \frac{1}{5}\). Since these inequalities are disjoint (i.e., there's no overlap between the intervals), we can write the solution in interval notation as: \((-\infty, \frac{1}{5}) \cup (3, \infty)\) This represents the solution set for the given absolute value inequality.

Key Concepts

Understanding Set Notation

Set notation is a way to describe a collection of elements. It's particularly useful when we want to express solutions to equations. In set notation, we use curly braces to list elements of the set. For example, the set of all solutions to an equation could be written as \( \{ x \mid x > 3 \} \), meaning "the set of all \( x \) such that \( x \) is greater than 3." This method is precise and works well for equations where solutions are discrete or can be specifically listed.
Set notation is often used together with absolute value equations to specify exact solution sets.

• Each element is explicitly listed or described.
• Useful for equations that yield specific results.
• Ensures clarity by highlighting the conditions each solution must satisfy.

While set notation is precise, it can become cumbersome with inequalities, which is where interval notation comes in handy.

Explaining Interval Notation

Interval notation is a streamlined way to express ranges of values, particularly useful for inequalities. It uses brackets and parentheses to denote open and closed intervals. For example, \((3, \infty)\) represents all numbers greater than 3, whereas \([3, \infty)\) would include 3 and all numbers greater than it.
This type of notation is especially applicable when dealing with inequalities, like the one in our original exercise.

• Parentheses \(()\) indicate values that are not included (open interval).
• Brackets \([]\) indicate values that are included (closed interval).
• \(\infty\) and \(-\infty\) always use parentheses because infinity is not a number that can be 'reached'.

With the exercise, the solution \((-\infty, \frac{1}{5}) \cup (3, \infty)\) indicates two separate ranges where the inequality holds true. These intervals are disjoint, showing two distinct solution sets.

Simplifying Linear Inequalities

Linear inequalities are similar to linear equations but instead of an equal sign, they use inequality signs like \(>\), \(<\), \(\geq\), and \(\leq\). When solving them, we aim to isolate the variable just as we do with regular equations. However, we must pay attention to reversing the inequality sign when multiplying or dividing by negative numbers.
In our exercise, the steps were clear and methodical. We broke down the absolute value into two linear inequalities:

• For \(5h - 8 > 7\), we solved step-by-step to isolate \(h\), resulting in \(h > 3\).
• In the case of \(-(5h - 8) > 7\), we distributed the negative, solved, and remembered to reverse the inequality when dividing by a negative, giving us \(h < \frac{1}{5}\).

These individual solutions are then combined systematically using interval notation, respecting the nature of the inequalities to express the full range of possible solutions.