Solve and graph the following absolute value inequalities. Express the solution in interval notation.
(A) $\mid \frac{2x-5}{3}\mid \le \frac{1}{3}$

• A. $\left\lbrack-3,-2\right\rbrack$
• B. $\left(-3,-2\right)$
• C. $\left\lbrack2,3\right\rbrack$
• D. $\left(2,3\right)$

Answer: (C)

Verified step by step guidance

1. Start with the given inequality involving the absolute value: \(\left|\frac{2x - 5}{3}\right| \leq \frac{1}{3}\).
2. Recall that for any expression \(A\), the inequality \(|A| \leq B\) (where \(B \geq 0\)) is equivalent to \(-B \leq A \leq B\). Apply this to get: \(-\frac{1}{3} \leq \frac{2x - 5}{3} \leq \frac{1}{3}\).
3. Multiply all parts of the inequality by 3 to eliminate the denominator, remembering that multiplying by a positive number does not change the inequality signs: \(-1 \leq 2x - 5 \leq 1\).
4. Next, isolate the term with \(x\) by adding 5 to all parts: \(-1 + 5 \leq 2x \leq 1 + 5\), which simplifies to \$4 \leq 2x \leq 6$.
5. Finally, divide all parts by 2 to solve for \(x\): \(\frac{4}{2} \leq x \leq \frac{6}{2}\), giving the solution interval for \(x\). Express this solution in interval notation and then graph it on a number line.