Solve each inequality. Give the solution set in interval notation. | (2/3)x + 1/2 | ≤ 1/6

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Start by understanding that the inequality involves an absolute value expression: \(\left| \frac{2}{3}x + \frac{1}{2} \right| \leq \frac{1}{6}\). The absolute value inequality \(|A| \leq B\) means that \(-B \leq A \leq B\).

Rewrite the inequality without the absolute value by setting up a compound inequality: \(-\frac{1}{6} \leq \frac{2}{3}x + \frac{1}{2} \leq \frac{1}{6}\).

Next, solve the left part of the compound inequality: \(-\frac{1}{6} \leq \frac{2}{3}x + \frac{1}{2}\). Subtract \(\frac{1}{2}\) from both sides to isolate the term with \(x\).

Then, solve the right part of the compound inequality: \(\frac{2}{3}x + \frac{1}{2} \leq \frac{1}{6}\). Similarly, subtract \(\frac{1}{2}\) from both sides to isolate the term with \(x\).

After isolating \(\frac{2}{3}x\) in both inequalities, multiply all parts of the compound inequality by the reciprocal of \(\frac{2}{3}\), which is \(\frac{3}{2}\), to solve for \(x\). Remember to keep the inequality signs consistent since you are multiplying by a positive number. Finally, express the solution set in interval notation.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Absolute Value Inequalities

An absolute value inequality involves expressions within absolute value bars, representing distance from zero. To solve |A| ≤ B, where B ≥ 0, rewrite it as a compound inequality: -B ≤ A ≤ B. This approach helps isolate the variable and find the range of values satisfying the inequality.

Solving Linear Inequalities

Solving linear inequalities requires isolating the variable by performing inverse operations, similar to solving equations. When multiplying or dividing by a negative number, the inequality sign reverses. The solution is often expressed as an interval or union of intervals.

Interval Notation

Interval notation is a concise way to represent sets of numbers between two endpoints. Use parentheses () for values not included and brackets [] for values included. For example, [a, b] includes both endpoints, while (a, b) excludes them, clearly showing the solution set.