Solve the inequality. \( \left|\frac{1}{4} x-\frac{1}{3}\right| \leq \frac{1}{3} \) \( [?] \leq x \leq \square \)

To solve the inequality \( \left|\frac{1}{4} x-\frac{1}{3}\right| \leq \frac{1}{3} \), we start by rewriting it as two separate inequalities: 1. \( \frac{1}{4}x - \frac{1}{3} \leq \frac{1}{3} \) 2. \( \frac{1}{4}x - \frac{1}{3} \geq -\frac{1}{3} \) For the first inequality: \[ \frac{1}{4}x - \frac{1}{3} \leq \frac{1}{3} \] Adding \( \frac{1}{3} \) to both sides: \[ \frac{1}{4}x \leq \frac{2}{3} \] Multiplying by 4: \[ x \leq \frac{8}{3} \] For the second inequality: \[ \frac{1}{4}x - \frac{1}{3} \geq -\frac{1}{3} \] Adding \( \frac{1}{3} \) to both sides: \[ \frac{1}{4}x \geq 0 \] Multiplying by 4: \[ x \geq 0 \] Combining these results, we find: \[ 0 \leq x \leq \frac{8}{3} \] So, the final solution to the inequality is: \[ 0 \leq x \leq \frac{8}{3} \]