Q.Solve $-\dfrac{x}{3} + 2 > 5$ for real $x$, showing every step; represent the solution on the number line.

[!TLDR] Isolate the $-x/3$ term, then multiply by $-3$ — a negative number — which reverses the inequality's direction. [!ANSWER] $x < -9$, i.e. $x \in (-\infty, -9)$.

Start with $-\dfrac{x}{3} + 2 > 5$. Subtract $2$ from both sides (Rule 1, no flip): $-\dfrac{x}{3} > 3$. Now multiply both sides by $-3$, a negative number — by Rule 3 the inequality direction must reverse: multiplying the left side gives $(-3)\left(-\dfrac{x}{3}\right) = x$, and multiplying the right side gives $(-3)(3) = -9$, and the $>$ flips to $<$: $x < -9$. Check with $x = -10$: $-(-10)/3 + 2 = 10/3 + 2 \approx 5.33 > 5$ ✓; and at the boundary $x=-9$: $-(-9)/3+2 = 3+2 = 5$, which is not strictly greater than $5$, confirming $-9$ is correctly excluded. On the number line: an open circle at $-9$ with an arrow extending left. [!ANSWER] $x < -9$, i.e. $x \in (-\infty, -9)$.

Isolate the term with $x$ first using addition/subtraction, then multiply by the negative coefficient $-3$ and flip the inequality sign.

The most common error here is multiplying by $-3$ but forgetting to reverse the inequality sign, which would wrongly give $x > -9$.