The learning goal
Name the operation at each line: dividing −3x≥6 by −3 gives x≤−2, not x≥−2. Substituting −3, −2 and −1 into the original inequality gives true, true and false. Those checks expose a direction error, but a few tested points do not prove an entire solution set; equivalent algebra establishes its full range.
Look at the operation, not the presence of a minus sign
Adding the same number and multiplying by a negative do different things to order. Start with two easy values: 1<4. Adding −2 to both gives −1<2, with order preserved. Multiplying both by −2 gives −2>−8, with order reversed.
OpenStax's inequality properties distinguish addition from multiplication, including ≤ and ≥. A minus sign alone is therefore not the trigger; identify whether both sides were actually multiplied or divided by a known negative number. A negative exponent is a separate issue, covered by the base, parentheses and exponent guide.
Three operations from the same starting comparison, 1<4
Add −2 to both sides
−1<2. Both positions move by the same amount, preserving their order.
Multiply both sides by 2
2<8. A known positive factor preserves order.
Multiply both sides by −2
−2>−8. The negative factor reverses order.
A complete example: reverse at division by −3
The numbers, two solution routes and practice problems here are original editorial exercises over real x, not a test of student outcomes. To solve −3(x−2)≥12, first expand the left side as −3x+6. The −2 inside the parentheses must also multiply −3, giving +6. This rewrites one expression; it does not multiply both sides of the inequality by −3.
Subtract 6 from both sides to obtain −3x≥6, preserving the comparison. Only when both sides are divided by the known negative number −3 do you obtain x≤−2. Labeling the operation beside each line makes the change checkable instead of guessing from the visible signs.
| Current inequality | Operation | Reason for its direction |
|---|---|---|
| −3(x−2)≥12 | Given problem over real x | No transformation yet |
| −3x+6≥12 | Distribute on the left | Rewrite the same left-hand expression |
| −3x≥6 | Subtract 6 from both sides | Equal addition or subtraction preserves order |
| x≤−2 | Divide both sides by −3 | The divisor is negative, so ≥ reverses to ≤ |
Use a positive divisor instead and reconcile the answer
You can avoid a final negative divisor by collecting the x term on the right. Still write the same operation on both sides. This is another sequence of equivalent steps, not a shortcut that removes the need to check.
From −3x+6≥12, add 3x to both sides to get 6≥12+3x. Subtract 12 to get −6≥3x, then divide by positive 3 to obtain −2≥x. Rewriting that statement with x first gives x≤−2. Exchanging the positions of the compared objects also exchanges the written symbol; it expresses the same comparison rather than another multiplication or division.
Do the two written directions express the same comparison?
−2 ≥ x
x ≤ −2
Substitute the boundary and both sides to expose a direction error
Equality is allowed in the original problem, so −2 is included. The solution is x≤−2, or (−∞,−2] in interval notation: a filled point at −2 and a ray extending left. Infinity is not an endpoint value you can include.
Evaluate the original left side, −3(x−2), rather than test only the final line that may already be wrong. A point on each side and the boundary help diagnose direction and endpoint mistakes. They do not exhaust all real numbers; the equivalent derivation above establishes the full range.
| Position and x | Left-hand calculation | Original comparison | What the check shows |
|---|---|---|---|
| Left: −3 | −3×(−3−2)=15 | 15≥12 is true | There is a valid value to the left |
| Boundary: −2 | −3×(−2−2)=12 | 12≥12 is true | Include the endpoint: ≤ and a filled point |
| Right: −1 | −3×(−1−2)=9 | 9≥12 is false | This value to the right must not be included |
The first error is in the division step, not the final number
−3x≥6, therefore x≥−2.
Divide both sides by −3 and reverse the comparison: x≤−2.
Change a condition and identify which rule changes
Solve these three problems before looking at the answers. Write each operation, the complete range and whether the boundary is included. A contains negative numbers without a negative divisor; B changes ≥ to >; C requires an operation on all three parts. They check unnecessary reversal, strictness and changing only one of two comparisons.
Solve over the real numbers and name the operation on both sides at each step.
A. x−5 < −2
B. −3(x−2) > 12
C. −4 < −2x ≤ 6
For each problem, check at least one boundary value and explain why it is included or excluded.| Problem | Derivation | Complete answer | Boundary and sample check |
|---|---|---|---|
| A | x−5<−2; add 5 to both sides, giving x<3 | x<3, or (−∞,3) | At x=3, −2<−2 is false; at x=2, −3<−2 is true |
| B | −3x+6>12; −3x>6; divide by −3 to obtain x<−2 | x<−2, or (−∞,−2) | At x=−2, 12>12 is false: use an open endpoint; at x=−3, 15>12 is true |
| C | Divide all three parts by −2: 2>x≥−3; then reorder from least to greatest | −3≤x<2, or [−3,2) | At x=−3, −4<6≤6 is true; at x=2 the left strict comparison fails; at x=0, −4<0≤6 is true |
Know a letter divisor's sign before dividing
Do not apply the known-negative rule when the sign is unknown. In another original example, ax≥6, a>0 gives x≥6/a, a<0 gives x≤6/a, and a=0 makes the original statement 0≥6, which is false and has no solution. Separate those conditions before dividing by a, rather than divide first and add a possible reversal afterward.
This boundary concerns the sign of a fixed coefficient; it is not a universal recipe for variable denominators or quadratic inequalities. Use the first-invalid-step error record to keep the original problem, actual operation, failing line, full range and one changed-condition retry. The review goal is to explain why order reverses; no learning improvement was measured here.
Check four things before putting the answer away
- I can identify the exact multiplication or division by a known negative number, rather than just mention a minus sign.
- My ≤, <, ≥ or > preserves the original problem's equality condition.
- I substituted a boundary and values on both sides into the original problem, while recognizing that these points do not replace a full derivation.
- I did not divide by a value whose sign is unknown or that might be zero.
