Fail to reverse the inequality sign when multiplying/dividing by a negative.
An error-pattern drill: practice spotting and avoiding this specific SHSAT trap.
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Solve \(-3x < 12\). If you divide by \(-3\) and keep the sign, you get \(x < -4\). The right answer is \(x > -4\). When you multiply or divide both sides by a negative number, the inequality sign flips.
Why? Multiplying by a negative reverses the order of numbers: \(2 < 5\), but \(-2 > -5\). Students forget because with an equation, dividing by a negative needs no extra step.
Check habits:
Subtract 10: \(-2x ≤ 18\). Divide by -2 (negative), flipping the inequality: \(x ≥ -9\). (Forgetting to flip gives \(x ≤ -9\). Dividing by -2 first and then subtracting 10 gives \(x ≥ -24\). Dividing by 2 instead of -2 gives \(x ≤ 9\).)
Add \(2\) to all three parts: \(-4 \le -2x < 10\). Divide all three parts by \(-2\) and flip BOTH signs: \(2 \ge x > -5\), that is \(-5 < x \le 2\). Each endpoint keeps its own sign, so \(x = -4, -3, \dots, 2\). Their sum is \(-7\). (Not flipping the signs and then writing the ends in order, \(-5 \le x < 2\), gives a sum of \(-14\). Counting both \(-5\) and \(2\) as included gives a sum of \(-12\). Dividing by \(2\) instead of \(-2\) gives \(-2 \le x < 5\) and a sum of \(7\).)
First: \(-2x < 20\). Dividing by \(-2\) flips the sign: \(x > -10\). Second: \(3x \ge -39\), so \(x \ge -13\). Both must hold, and the first bound is the tighter one, so the least integer is \(-9\). (Forgetting to flip gives \(x < -10\); then only the second bound limits \(x\), which gives \(-13\). Including \(-10\), which the strict sign leaves out, gives \(-10\). Dividing by \(2\) instead of \(-2\) gives \(x > 10\) and \(11\).)