Problem 62

Question

Solve the inequality. $$ \left|x+\frac{1}{2}\right| \leq 2 $$

Step-by-Step Solution

Verified
Answer
\(-\frac{5}{2} \leq x \leq \frac{3}{2}\)
1Step 1: Understand the Inequality
The given inequality is \( \left| x + \frac{1}{2} \right| \leq 2 \). This inequality denotes that the absolute value of \( x + \frac{1}{2} \) is less than or equal to 2. It means the expression inside the absolute value has to satisfy \( -2 \leq x + \frac{1}{2} \leq 2 \).
2Step 2: Solve the First Part of the Compound Inequality
To solve \( -2 \leq x + \frac{1}{2} \), subtract \( \frac{1}{2} \) from both sides: \(-2 - \frac{1}{2} \leq x \). This simplifies to \( -\frac{5}{2} \leq x \), since \(-2 - \frac{1}{2} = -\frac{5}{2}\).
3Step 3: Solve the Second Part of the Compound Inequality
Next, solve \( x + \frac{1}{2} \leq 2 \). Subtract \( \frac{1}{2} \) from both sides: \( x \leq 2 - \frac{1}{2} \), which simplifies to \( x \leq \frac{3}{2} \).
4Step 4: Combine the Results
Combine the results from Step 2 and Step 3 to get the solution to the original inequality: \( -\frac{5}{2} \leq x \leq \frac{3}{2} \). This means \( x \) is in the interval \( \left[ -\frac{5}{2}, \frac{3}{2} \right] \).

Key Concepts

Compound InequalitiesInterval NotationSolving Inequalities
Compound Inequalities
In mathematics, compound inequalities involve two or more inequalities combined into one statement by using the words "and" or "or." A compound inequality is instrumental for expressing a range of possible solutions within a single expression. In our exercise, we encounter a compound inequality when solving a problem involving absolute values.
This occurs because the absolute value inequality \( \left| x + \frac{1}{2} \right| \leq 2 \) translates to two simultaneous inequalities:
  • \( -2 \leq x + \frac{1}{2} \)
  • \( x + \frac{1}{2} \leq 2 \)
The 'and' between these two means each must be true to satisfy the entire inequality. This leads to finding the overlap or "intersection" of solutions that fulfill both parts simultaneously. In simple terms, compound inequalities help define a range where multiple conditions are true.
Understanding compound inequalities is crucial, as it forms the basis for solving more complex expressions in algebra.
Interval Notation
Interval notation provides a concise way of writing the solution set of inequalities. It helps in clearly representing the range of values for a solution. Let's see how it's used in our context:
After finding that \( -\frac{5}{2} \leq x \leq \frac{3}{2} \), we can express this compound inequality using interval notation as \( \left[ -\frac{5}{2}, \frac{3}{2} \right] \).
  • The square brackets \([ ]\) indicate that both endpoints, \( x = -\frac{5}{2} \) and \( x = \frac{3}{2} \), are included in the solution, known as a closed interval.
  • If the endpoints were not included, parentheses \(( )\) would be used instead, which indicate open intervals.
Interval notation is powerful as it allows mathematicians to effectively and succinctly communicate complex solution sets without listing every individual number within an interval. It is a vital skill to master for interpreting and solving inequalities.
Solving Inequalities
Solving inequalities involves finding the set of values that make an inequality true. In our exercise, we focused on an absolute value inequality, where the goal is to solve for \( x \).
The key steps include:
  • Removing the absolute value by considering both the positive and negative scenarios (compound inequalities).
  • Solving each resulting inequality separately.
  • Combining solutions to provide a full picture of possible values.
For instance, having \( \left| x + \frac{1}{2} \right| \leq 2 \) required us to:
  • Consider both \( -2 \leq x + \frac{1}{2} \) and \( x + \frac{1}{2} \leq 2 \).
  • Solve them: \( -2 - \frac{1}{2} \leq x \)and \( x \leq 2 - \frac{1}{2} \), leading to \( -\frac{5}{2} \leq x \leq \frac{3}{2} \).
  • Use logical reasoning to merge these solutions.
Solving such inequalities systematically improves problem-solving skills and builds confidence in handling various algebraic challenges. Mastering these methods is essential for progressing in mathematical studies.