Q.Solve $|3x - 1| \le 5$ for real $x$.

[!TLDR] Expand $|3x-1|\le 5$ as $-5\le 3x-1\le 5$, then isolate $x$. [!ANSWER] $x \in \left[-\dfrac{4}{3}, 2\right]$.

$|3x-1|\le 5$ means $-5 \le 3x-1 \le 5$ (slack version of Rule A, applied to the expression $3x-1$). Add $1$ to all three parts: $-4 \le 3x \le 6$. Divide all three parts by the positive number $3$: $-\dfrac{4}{3} \le x \le 2$. Both endpoints are included since the original inequality is slack ($\le$). [!ANSWER] $x \in \left[-\dfrac{4}{3}, 2\right]$.

Expand the modulus into a compound inequality on the inner expression $3x-1$, then add 1 and divide by 3 across all three parts.

Dividing by 3 correctly keeps the direction (3 is positive), but students sometimes flip the inequality here out of habit from Rule 3 — only division by a negative number requires a flip.