Problem 27 - Algebra and Trigonometry

Question

\(23-48\) Solve the inequality. Express the answer using interval notation. $$ |x-5| \leq 3 $$

Step-by-Step Solution

Answer

The solution is \([2, 8]\).

Step 1: Understanding Absolute Value Inequality

The inequality \(|x-5| \leq 3\) involves an absolute value, meaning it can express two conditions: \((x-5) \leq 3\) and \( -(x-5) \leq 3\). These two conditions will help us define the range of \(x\) that satisfies the inequality.

Step 2: Solving First Inequality

The first inequality \((x-5) \leq 3\) implies adding 5 to both sides: \(x \leq 8\). This tells us that \(x\) can be no greater than 8.

Step 3: Solving Second Inequality

The second inequality \(-(x-5) \leq 3\) simplifies to \(-x + 5 \leq 3\). Subtract 5 from both sides to get \(-x \leq -2\). Subsequently, multiplying the entire inequality by -1 (which reverses the inequality sign), we reach \(x \geq 2\). This indicates \(x\) must be at least 2.

Step 4: Combining Results

Combine the results from the inequalities. Since \(x\) needs to satisfy both \(x \leq 8\) and \(x \geq 2\), we write this as the compound inequality \(2 \leq x \leq 8\).

Step 5: Expressing Interval Notation

Now we write the solution in interval notation based on the compound inequality \(2 \leq x \leq 8\). In interval notation, this solution is expressed as \([2, 8]\). It signifies that \(x\) includes all values from 2 to 8, inclusive.

Key Concepts

Interval Notation

Interval notation is a system of writing the set of solutions for an inequality or range of values. It employs brackets and parentheses to denote closed and open intervals.
This method is widely used because it is compact and clear.

• Closed Interval: When values are included in the solution set, we use square brackets, such as \[a, b\]. This means that all values between \(a\) and \(b\), including \(a\) and \(b\) themselves, are solutions.
• Open Interval: To express that endpoints are not included, round brackets are used: \((a, b)\). This implies that values between \(a\) and \(b\) are in the solution, but \(a\) and \(b\) themselves are not.

For the inequality \(|x-5| \leq 3\), after solving it, the interval notation \[2, 8\] denotes that \(x\) can take any value from 2 to 8 including the boundaries. This is a closed interval because both endpoints are part of the solution.

Compound Inequalities

A compound inequality involves multiple inequalities tied together. The aim is to find the values that solve all parts of the inequality at once.
For absolute value inequalities, this often means breaking down into two related simpler inequalities.

• And: We use the word 'and' when the solution must satisfy both inequalities simultaneously. For example, \(a \leq x \leq b\) indicates \(x\) must be between \(a\) and \(b\), inclusive.
• Or: If 'or' is used, then only one of the individual inequalities needs to be satisfied. This results in a union of solution intervals.

In the original exercise, the inequality \(|x-5| \leq 3\) becomes a compound inequality \(2 \leq x \leq 8\), where both parts \((x-5 \leq 3) \) and \(-(x-5) \leq 3) \) must be true at the same time. Hence, this is an 'and' compound inequality.

Solving Absolute Value Equations

Absolute value equations demand consideration of both positive and negative cases. The absolute value \( |x| \) indicates the distance of \( x \) from zero without regard to direction on the number line.
For simple expressions like \(|x-5| \leq 3\), this involves dealing with two scenarios

• Setting up the "positive case": \((x-5) \leq 3\), which directly affects the solution.
• Handling the "negative case": \(-(x-5) \leq 3\), requiring a shift in inequality direction after manipulation.

Keep in mind the steps:

• Transform the absolute value equation into two separate inequalities.
• Solve each inequality separately.
• Combine the solutions using the appropriate compound wording, either 'and' or 'or'.

With our task, handling \(|x-5| \leq 3\) meant identifying \(x \leq 8\) and \(x \geq 2\). Therefore, we combine them into the unified solution \(2 \leq x \leq 8\).