Why the inequality flips
Tonight's idea
\(a \lt b\) means \(a\) sits to the left of \(b\) on the number line. Every rule for inequalities asks what a move does to left and right.
Adding the same number to both sides slides both points the same way. Left stays left. Multiplying by a positive number stretches the line away from 0 (or squashes it toward 0). Left still stays left.
Multiplying by \(-1\) is different. It reflects every point through 0, to the same distance on the other side. The point that was on the right lands on the left.
Any negative multiplier is a reflection followed by a stretch, so it flips the order too. Dividing by a negative is the same flip.
Add or subtract anything: sign unchanged. Multiply or divide by a positive: unchanged. Multiply or divide by a negative: flip \(\lt\) to \(>\) (and \(\le\) to \(\ge\)).
You can also avoid the rule: if the \(x\) term is negative, move it to the other side, where it is positive, then finish. Both routes must agree, and checking one number settles any doubt.
Watch one: \(3 - 4x \le 11\)
Route one. Subtract 3: \(-4x \le 8\). Divide by \(-4\), a negative, so flip: \(x \ge -2\).
Route two, with no negatives. Add \(4x\) to both sides: \(3 \le 11 + 4x\). Subtract 11: \(-8 \le 4x\). Divide by 4: \(-2 \le x\). The same answer, read from the other side.
Check with numbers. \(x = 0\): \(3 \le 11\), true, and 0 is in the answer. \(x = -3\): \(3 + 12 = 15 \le 11\), false, and \(-3\) is not. The solution is every number from \(-2\) upwards, including \(-2\).
Your turn
- Solve each, and say whether you flipped: (a) \(5x - 7 \lt 13\); (b) \(-3x > 12\); (c) \(4 - x \ge 9\).
- Solve (a) \(2(x - 3) > 5x + 9\) and (b) \(-\dfrac{x + 1}{2} \le 4\). Check each answer with one number inside the solution and one outside.
- Suppose \(a \lt b\). For each statement, decide whether it must be true, and give the reason from the number line: (a) \(a + 5 \lt b + 5\); (b) \(3a \lt 3b\); (c) \(-2a \lt -2b\); (d) \(a - b \lt 0\).
- A student solves \(\dfrac{1}{x} \lt \dfrac{1}{2}\) by multiplying both sides by \(2x\) to get \(2 \lt x\). Test \(x = -1\) in the original. What went wrong, and what is the complete set of solutions?
Stretch: trapped on both sides
Solve \(-7 \lt 3 - 2x \le 5\) by doing each step to all three parts at once. Write the answer with \(x\) in the middle and the smaller number on the left. How many integers satisfy it?
Answers
- (a) \(5x \lt 20\), \(x \lt 4\); no flip. (b) Divide by \(-3\) and flip: \(x \lt -4\). (c) \(-x \ge 5\), so \(x \le -5\); flipped. (Or move \(x\) across: \(4 \ge 9 + x\), so \(x \le -5\).)
- (a) \(2x - 6 > 5x + 9\), so \(-15 > 3x\) and \(x \lt -5\). Check \(x = -6\): \(-18 > -21\), true; \(x = 0\): \(-6 > 9\), false. (b) Multiply by \(-2\) and flip: \(x + 1 \ge -8\), so \(x \ge -9\). Check \(x = 0\): \(-\tfrac12 \le 4\), true; \(x = -11\): \(5 \le 4\), false.
- (a) True: both slide 5 to the right. (b) True: stretching by 3 keeps left on the left. (c) False: multiplying by \(-2\) reflects, so \(-2a > -2b\). (d) True: subtract \(b\) from both sides of \(a \lt b\).
- \(x = -1\) gives \(-1 \lt \tfrac12\), true, yet \(-1\) is not bigger than 2. Multiplying by \(2x\) is multiplying by a negative whenever \(x \lt 0\), and then the sign should have flipped. Two cases. If \(x > 0\): \(2 \lt x\), so \(x > 2\). If \(x \lt 0\): the sign flips to \(2 > x\), which every negative \(x\) satisfies. Complete answer: \(x \lt 0\) or \(x > 2\). (\(x = 0\) is not allowed, since \(1/0\) is undefined.)
Stretch. Subtract 3 from all three parts: \(-10 \lt -2x \le 2\). Divide by \(-2\) and flip both signs: \(5 > x \ge -1\), that is, \(-1 \le x \lt 5\). The integers are \(-1, 0, 1, 2, 3, 4\): six of them.