Sec 2 G2 Maths Bukit Timah | Negative Inequalities

Did you know that dividing an inequality by a negative number reverses its direction? For −2x<6, the correct solution is x>−3, not x<−3. If you test x=0, the original statement reads 0<6 and is true. That small substitution reveals why the sign reversal is necessary and why it makes sense on a number line.

A Secondary 2 G2 Mathematics tutor in Bukit Timah can help students distinguish this from ordinary equations. Adding or subtracting the same quantity preserves the order, and dividing by a positive number also preserves it. The special change occurs only when multiplication or division uses a negative number. The negative boundary itself does not trigger the reversal.

This guide builds from inequality symbols and negative-number order to one-step solutions, multi-step examples, brackets, whole-number contexts and a fast independent checking routine. The first goal is clear reasoning, then accurate arithmetic, rather than memorising an unexplained instruction to flip an arrow.

Quick answer: reverse only for a negative multiplier

Operation	What happens
Add or subtract same quantity on both sides	Order direction stays.
Multiply or divide by a strictly positive number	Order direction stays.
Multiply or divide by a strictly negative number	Reverse < to >, or ≤ to ≥, and vice versa.
Multiply or divide by zero	Cannot use as an equivalent inequality transformation.
Strict boundary < or >	Open endpoint; equality excluded.
Inclusive boundary ≤ or ≥	Closed endpoint; equality permitted.
A negative coefficient −2x<6	x>−3, checked by substitution.

Secondary 2 G2 Mathematics: which inequalities to practise

Singapore’s official G2 syllabus includes solving simple inequalities of the form ax≤b, ax≥b, ax<b and ax>b, with integer coefficients. The sign-reversal rule is therefore an important foundation. More involved bracketed or two-sided inequalities in this article are optional transfer challenges rather than a claim that every school examines them. Full Subject-Based Banding uses G2 as a Mathematics subject level; the learner’s current classwork should guide the immediate scope.

Official reference: MOE’s G2 Mathematics syllabus and one school’s 2026 Secondary 2 G2 topic outline.

Twenty-five core concepts behind the sign change

1. Inequality compares values rather than asserting equality

An equation says that two expressions have the same value at solutions. An inequality allows one side to be smaller or larger than the other. The symbol x<5 means every number smaller than five, not only x=4. Students often carry equation habits into inequality questions; the first change is to expect a set of solutions rather than one isolated number. A visual number line makes that set concrete.

Teaching move: Ask the learner to name three values satisfying x<5 and two failing it.

Avoid this: Do not report only x=4 when infinitely many values below five qualify.

2. Strict and inclusive symbols ask different questions

The symbol < means strictly less than, and ≤ means less than or equal to. The number three satisfies x≤3 but not x<3. Similarly, ≥ includes its boundary while > excludes it. These distinctions matter in questions about maximum capacity, whole-number ticket counts or required minimum scores. Before transforming an inequality, state whether equality at the boundary is permitted.

Teaching move: Place three test points, one below, one exactly at, and one above a threshold.

Avoid this: Do not shade a closed boundary for a strict inequality.

3. Adding the same value preserves order

If a<b, then a+3<b+3. A number line shows why: shifting both points right by three preserves their left-to-right order. Similarly, subtracting the same constant from both sides preserves the inequality direction. This is why x−5<2 becomes x<7 after adding five; it is not a special sign rule, simply the same translation on both values.

Teaching move: Write two numerical unequal values and add five to each before solving a letter example.

Avoid this: Do not reverse the inequality just because a positive constant moves across a displayed equals-like line.

4. Multiplying by a positive number preserves order

If 2<5, multiplying both by three gives 6<15. Scaling positive values, or even negative values, by a positive factor keeps their relative order. Therefore 3x≤12 becomes x≤4 after dividing by positive three. The multiplier must be strictly positive for this simple preservation rule. A zero multiplier destroys useful order information, since both sides would become zero.

Teaching move: Test a pair such as −4<2 and multiply by positive two.

Avoid this: Do not change < to > merely because a division operation occurred; the sign of the divisor matters.

5. Multiplying by a negative number reverses order

Starting with 2<5, multiply both values by −1: −2>−5. Their order reverses on the number line because reflection across zero turns rightmost numbers into leftmost ones. This is the reason behind the familiar inequality flip. In algebra, −2x<6 becomes x>−3 after dividing both sides by negative two. The reversal is mandatory, not optional notation.

Teaching move: Plot 2 and 5, reflect each point through zero, and read the reversed order.

Avoid this: Retaining < after division by −2 would give x<−3, which is false for a simple valid test value.

6. Division by a negative reverses exactly once for that operation

For −3x≥12, divide by −3 to get x≤−4. One sign reversal accompanies the negative divisor. If later multiplying or dividing by another negative, there may be another reversal. Rather than count minus symbols in a whole page, examine the sign of the single operation applied to both full sides at each step. This makes the method reliable in longer inequalities.

Teaching move: Label the dividing number above the step and write ‘negative: reverse’.

Avoid this: Do not flip repeatedly simply because a negative numeral remains somewhere in the expression.

7. A coefficient’s sign is different from the answer’s sign

In 2x<−6, the coefficient two is positive, so division by two preserves the symbol and gives x<−3, even though the resulting bound is negative. Conversely −2x<6 has a positive right-hand six but a negative coefficient, so the symbol reverses. Students should identify the number used as divisor, not merely look for any minus sign.

Teaching move: Compare the two examples side by side and circle the coefficient of x.

Avoid this: The sign of the final numerical boundary does not itself determine whether to reverse the symbol.

8. The boundary is often excluded even if the bound is exact

From −2x<6 we obtain x>−3. Substitute x=−3 into the original: −2(−3)=6, and 6<6 is false. This confirms why the point −3 remains excluded. A student can use the boundary test to check symbol strictness without trying many arbitrary values. For ≤ or ≥, the corresponding equality at the boundary would be allowed.

Teaching move: Test the boundary first and say whether the original comparison is true.

Avoid this: Do not turn strict < into inclusive ≥ after dividing by negative; the symbol becomes strict >.

9. One satisfying value and one failing value make a powerful check

For x>−3, x=0 should satisfy original −2x<6 because 0<6, while x=−4 should fail because 8<6 is false. Checking one value on each side of the boundary provides stronger evidence than checking a single convenient point. It exposes both incorrect inequality direction and incorrect boundary. Use exact integer points whenever possible.

Teaching move: Ask the pupil to choose their own one-true and one-false test input.

Avoid this: A test at only the boundary may not distinguish which side of that boundary should be shaded.

10. Number-line arrows represent unlimited possible solutions

If x≤4, mark a closed point at four and shade or draw an arrow left. If x>−3, mark an open point at −3 and arrow right. The line communicates a set, not an approximate graph. A horizontal number line must increase from left to right, even if all labelled values are negative. The boundary symbol should match the open or closed endpoint.

Teaching move: Draw x<2 and x≥2 on separate lines and compare their endpoints.

Avoid this: An arrow pointing right for x<2 reverses the intended set of solutions.

11. Negative values on the number line still have normal order

A number such as −8 lies left of −3 and is therefore smaller, while −1 lies right of −3 and is larger. This is why x>−3 includes −2,0 and100 but excludes −4. Students who think ‘eight is bigger than three’ may make the wrong comparison when the negatives are attached. Work with actual coordinates on a number line before adding algebra.

Teaching move: Mark −8,−3,−1 and0, then read inequalities in words.

Avoid this: Comparing only absolute values of negative boundaries produces incorrect directions.

12. Whole-number contexts may have discrete solutions

If x counts complete tickets and −2x<6 simplifies to x>−3, physical context may further restrict x to nonnegative integers. The algebraic solution set includes many real numbers, but a ticket-count problem has its own permitted domain. Do not replace the mathematical inequality with an arbitrary rounded boundary; intersect its solutions with the stated context. This distinction is important when interpreting ‘at least’ and ‘no more than’ constraints.

Teaching move: List real-number solutions first, then apply a nonnegative-integer count restriction.

Avoid this: A negative fraction cannot represent a count of complete people or objects simply because it satisfies the algebra.

13. A coefficient and constant can be removed in either sensible order

For −3x+5≤17, subtract five from both sides to obtain −3x≤12, then divide by −3 and reverse: x≥−4. A check at x=−4 gives 17≤17, true, while x=−5 gives20≤17, false. Separating the translation step from the negative division helps pupils identify exactly when reversal occurs.

Teaching move: Underline the subtraction step and the negative division step in different colours.

Avoid this: Reversing at subtraction and again at negative division would cancel the needed reversal incorrectly.

14. Brackets can hide the negative coefficient

The inequality −2(x−3)<10 expands to −2x+6<10, then −2x<4 and x>−2. It can also be solved by dividing both sides by −2 first: x−3>−5, then x>−2. Both approaches agree when the entire bracket and right-hand side are handled correctly. Showing two valid routes can build confidence that the sign flip is a property of the negative operation.

Teaching move: Compare expanding first with dividing first on the same inequality.

Avoid this: The minus before a bracket must affect every term; −2(x−3) is not −2x−6.

15. Fractional coefficients are handled with the same rule

For −x/2≥3, multiply both sides by positive two to get −x≥6; then divide by −1 and reverse to x≤−6. Alternatively multiply both sides by −2 in one step, reversing to x≤−6. Fractional notation does not change the logic of signed multiplication. The learner should choose a method that keeps each step transparent rather than guessing the direction.

Teaching move: Say whether the multiplier is positive or negative at each line.

Avoid this: A minus before a fraction is part of the coefficient and cannot be ignored when isolating x.

16. Two-sided inequalities can be simplified by collecting terms

For 5−2x> x−4, subtract x and then subtract five to obtain −3x>−9, so dividing by −3 gives x<3. The reversal happens only in the final negative division, not when collecting like terms through addition or subtraction. The boundary x=3 makes both sides equal at −1, so it is excluded for strict >. Testing x=0 confirms the correct side.

Teaching move: Move all x terms to one side with explicitly balanced subtraction.

Avoid this: Do not reverse the symbol when merely subtracting x from both sides.

17. Inequalities can be rearranged without negative division if chosen wisely

In 5−2x>x−4, add 2x to both sides instead, obtaining 5>3x−4, so 9>3x and 3>x, equivalent to x<3. This route avoids multiplying by a negative entirely. It is equally valid because only positive division is used. Pupils may use whichever method produces the clearest logical chain, while understanding the sign-reversal rule remains necessary in other problems.

Teaching move: Solve the same inequality by both routes, then compare final sets.

Avoid this: An unconventional line 3>x is valid and equivalent to x<3; reading its direction correctly matters.

18. The solution interval may not be a single number

For x≥−4, all real numbers from −4 upward satisfy the inequality. If a pupil gives x=0 alone, they have provided an example of a solution, not the full solution set. A number line or inequality statement communicates the whole set concisely. This is one of the central differences from solving linear equations, where a unique root may occur.

Teaching move: Ask for three satisfying inputs and one non-satisfying input after any algebraic answer.

Avoid this: One valid numerical example is not a complete solution to an inequality.

19. An inequality may become a contradiction

Consider x+2<x−3. Subtract x to get 2<−3, which is always false. There is no real solution. A learner who expects a boundary after every problem may invent one. Just as equations can be contradictory, inequality transformations can leave a statement about constants that is false for every input.

Teaching move: Test any two x-values to confirm that the left is always five more than the right.

Avoid this: Do not divide both sides by a missing x coefficient when the variable has cancelled.

20. An inequality may hold for every real number

Consider x+2>x−3. Subtract x to obtain 2>−3, which is always true. Every real x satisfies it. This is the counterpart of a contradiction. The student should state ‘all real numbers’ rather than claiming x=0 merely because no x remained. The original comparison shows why the same gap of five appears for every input.

Teaching move: Test x=−100 and x=100 to illustrate the constant difference.

Avoid this: A true constant statement after cancellation means all permitted inputs work, not just one.

21. Multiplying both sides by zero is not an equivalent transformation

If 2<5 and both sides are multiplied by zero, the result is 0=0, which does not retain the original strict comparison. Therefore an inequality-solving step that multiplies or divides by zero is not reversible. Numeric denominator clearing should use nonzero factors; variable multipliers require knowledge of their sign before reversing or preserving the relation. These are important boundaries of the ordinary rules.

Teaching move: Explain why zero times two and zero times five lose order information.

Avoid this: Do not divide by a possibly zero variable without a domain or sign condition.

22. An unknown-sign multiplier cannot be handled casually

Multiplying x<3 by another variable a cannot be done with one fixed inequality direction unless the sign of a is known. If a>0, direction stays; if a<0, it reverses; if a=0, the strict inequality becomes false under multiplication. This is a useful extension for pupils curious about why the standard rule always mentions positive and negative, rather than simply memorising one symbol-flip mnemonic.

Teaching move: Give three cases a=2, a=−2 and a=0 and examine their effects.

Avoid this: Do not apply the fixed-negative reversal rule to a multiplier whose sign is undetermined.

23. The boundary must be tested in the original expression

For −4x+7≥19, rearrangement gives −4x≥12, then x≤−3. The boundary x=−3 yields19≥19, true, so it belongs to the solution set. A point x=0 gives7≥19, false, while x=−4 gives23≥19, true. That establishes both the inclusive endpoint and shading toward the negative side.

Teaching move: Test the derived boundary before drawing the line.

Avoid this: Do not rely entirely on a memorised open/closed-circle pattern without checking the original symbol.

24. The word problem determines whether strictness makes sense

An instruction ‘no more than ten’ translates to x≤10, while ‘fewer than ten’ translates to x<10. Similarly ‘at least five’ is x≥5 and ‘more than five’ is x>5. The algebra can be identical except for endpoint inclusion, which makes the language significant. Ask the student to write the inequality from the sentence before solving for the unknown.

Teaching move: Create four quick phrases and ask the pupil to match each with its boundary symbol.

Avoid this: The difference between ‘at most’ and ‘less than’ cannot be ignored when deciding whether equality counts.

25. A final explanation should name both set and reason

The statement x>−3 is a full algebraic answer, but a useful explanation can add that the inequality reversed because both sides were divided by negative two, and the boundary −3 is excluded because the original sign was strict. That is a concise and meaningful reason. In timed school work, the level of written justification depends on instructions, but correct line-by-line arithmetic should remain traceable.

Teaching move: Ask the student to explain one reversal in no more than a sentence.

Avoid this: Writing an arrow in the wrong direction after correct algebra still presents a wrong solution set.

Thirty worked examples: signs, boundaries and number lines

Worked example 1: Divide by negative two

Question: Solve −2x<6.

Method: Divide by −2 and reverse: x>−3.

Check: x=0 gives 0<6 true; x=−4 gives8<6 false.

Common mistake: Do not keep < after negative division.

Changed question: Try −3x<12.

Worked example 2: Divide by negative three

Question: Solve −3x≥12.

Method: Divide by −3 and reverse: x≤−4.

Check: Boundary x=−4 gives12≥12; x=0 fails.

Common mistake: Keep the endpoint inclusive.

Changed question: Change ≥ to >.

Worked example 3: Negative bound with positive coefficient

Question: Solve 2x<−6.

Method: Divide by positive two: x<−3.

Check: x=−4 gives −8<−6 true.

Common mistake: A negative number on the right does not cause symbol flip.

Changed question: Try 4x≤−20.

Worked example 4: Negative x on both sides

Question: Solve 5−2x>x−4.

Method: Rearrange −3x>−9; divide by −3, reverse to x<3.

Check: x=0:5>−4 true; x=4:−3>0 false.

Common mistake: Only reverse on negative division.

Changed question: Solve instead by adding2x to both sides.

Worked example 5: One-step with inclusive sign

Question: Solve −4x≤20.

Method: Divide by −4, reverse: x≥−5.

Check: x=−5 gives20≤20 true.

Common mistake: An inclusive sign remains inclusive after reversal.

Changed question: Try −5x≥25.

Worked example 6: Subtraction first

Question: Solve −3x+5≤17.

Method: Subtract five: −3x≤12; divide by −3, reverse: x≥−4.

Check: x=−4 gives17≤17; x=−5 fails.

Common mistake: Do not reverse when subtracting five.

Changed question: Try −3x+5<17.

Worked example 7: Another constant

Question: Solve −4x+7≥19.

Method: Subtract seven: −4x≥12; divide by −4: x≤−3.

Check: Boundary −3 makes both sides19.

Common mistake: The arrow points left for x≤−3.

Changed question: Change ≥ to >.

Worked example 8: A bracket with negative outside

Question: Solve −2(x−3)<10.

Method: Expand −2x+6<10, so −2x<4, hence x>−2.

Check: x=0 gives6<10 true; x=−3 gives12<10 false.

Common mistake: −2(x−3)=−2x+6, not −2x−6.

Changed question: Try −3(x+1)≤6.

Worked example 9: Dividing first

Question: Solve −2(x−3)<10 without expanding first.

Method: Divide by −2 and reverse: x−3>−5; add three: x>−2.

Check: Matches expansion route.

Common mistake: Entire bracket, not one term, is divided.

Changed question: Compare both solutions.

Worked example 10: Fractional negative coefficient

Question: Solve −x/2≥3.

Method: Multiply by −2 and reverse: x≤−6.

Check: At x=−6, 3≥3 true; at x=0 false.

Common mistake: The negative multiplier triggers reversal.

Changed question: Try −x/3>4.

Worked example 11: Two terms with a minus

Question: Solve 7−5x≤2.

Method: Subtract seven: −5x≤−5. Divide by −5: x≥1.

Check: At x=1, 2≤2 true; x=0 gives7≤2 false.

Common mistake: Two negative numbers do not mean you can omit sign reversal.

Changed question: Try 7−5x<2.

Worked example 12: Inequality x/negative denominator

Question: Solve x/(−4)<2.

Method: Multiply both sides by −4, reverse: x>−8.

Check: At x=0:0<2 true; x=−10 gives2.5<2 false.

Common mistake: A denominator that is negative is still a negative multiplier.

Changed question: Change < to ≤.

Worked example 13: Inequality with minus fraction

Question: Solve −x/3+2>5.

Method: Subtract two: −x/3>3; multiply by −3 and reverse: x<−9.

Check: At x=−12,4+2=6>5 true.

Common mistake: Do not multiply by three and forget the negative sign.

Changed question: Try −x/3+2≥5.

Worked example 14: Fractional bound

Question: Solve −2x>5.

Method: Divide by −2: x<−5/2.

Check: x=−3 gives6>5; x=−2 gives4>5 false.

Common mistake: An exact boundary can be a fraction.

Changed question: Draw open point at −2.5.

Worked example 15: Inclusive fractional bound

Question: Solve −3x≤7.

Method: Divide by −3: x≥−7/3.

Check: At x=−7/3, left7≤7 true.

Common mistake: No requirement that boundary be integer.

Changed question: Identify a satisfying integer.

Worked example 16: Whole-number count domain

Question: A mathematical condition on a nonnegative integer n is −2n<6. Find feasible n.

Method: Algebra gives n>−3. Intersect with n∈{0,1,2,…}: all nonnegative integers satisfy.

Check: n=0 gives0<6.

Common mistake: Don’t report negative fractions as complete ticket counts.

Changed question: Add an upper limit n≤4.

Worked example 17: Strict capacity bound

Question: A model states −2x+14>6. Solve.

Method: Subtract14: −2x>−8; divide by −2: x<4.

Check: At x=4, equality6>6 false.

Common mistake: The endpoint four is open.

Changed question: If x is a nonnegative whole count, list0,1,2,3.

Worked example 18: Inclusive capacity bound

Question: Solve −2x+14≥6.

Method: Same algebra but inclusive: x≤4.

Check: At x=4, equality is accepted.

Common mistake: Changing strictness changes the endpoint.

Changed question: Compare number lines of two previous examples.

Worked example 19: An identity-type inequality

Question: Solve x+2>x−3.

Method: Subtract x:2>−3, true for all real x.

Check: Every substitution has left side five greater.

Common mistake: No unique boundary should be invented.

Changed question: Change > to <.

Worked example 20: A contradiction

Question: Solve x+2<x−3.

Method: Subtract x:2<−3, false, so no real solution.

Check: No x can reverse a fixed five-unit gap.

Common mistake: Do not say x=0.

Changed question: Explain the difference from identity.

Worked example 21: Inequality with x both sides

Question: Solve −2x+5≤3x−10.

Method: Subtract3x and5: −5x≤−15, divide by −5: x≥3.

Check: x=3 both sides −1; x=4 left−3, right2 true.

Common mistake: Check exact transformation of constants.

Changed question: Try reverse ≤ to <.

Worked example 22: A different two-sided inequality

Question: Solve 4−3x>2x+9.

Method: Rearrange −5x>5, divide negative: x<−1.

Check: x=−2 gives10>5 true; x=0 gives4>9 false.

Common mistake: Do not treat x<1 as equivalent.

Changed question: Try 4−3x≥2x+9.

Worked example 23: A double-negative term

Question: Solve −(x+2)≥−6.

Method: Expand −x−2≥−6, add two: −x≥−4; multiply by −1, reverse x≤4.

Check: x=4 gives−6≥−6 true.

Common mistake: The outside minus changes both x and2.

Changed question: Try −(x−3)>5.

Worked example 24: A negative multiplication from start

Question: Solve −5(x+1)>−10.

Method: Divide by −5 and reverse: x+1<2, so x<1.

Check: x=0 gives−5>−10 true; x=2 gives−15>−10 false.

Common mistake: The entire bracket must be divided.

Changed question: Expand first as a cross-check.

Worked example 25: Another fraction with brackets

Question: Solve −(x−1)/2≤4.

Method: Multiply by −2 and reverse: x−1≥−8, so x≥−7.

Check: At x=−7,−(−8)/2=4≤4.

Common mistake: The sign outside fraction affects whole numerator.

Changed question: Change ≤ to <.

Worked example 26: Boundary zero

Question: Solve −7x>0.

Method: Divide negative: x<0.

Check: Any x negative gives positive −7x, and zero fails strict >.

Common mistake: Zero is not always included as a boundary.

Changed question: Change > to ≥.

Worked example 27: Inequality word problem with fixed fee

Question: A toy model uses score S=20−3n, and S must remain at least8. Find permissible nonnegative integer n.

Method: 20−3n≥8 => −3n≥−12 => n≤4. With n≥0 integer, n=0,1,2,3,4.

Check: n=4 gives score8, n=5 gives5 below minimum.

Common mistake: State count domain and inclusive endpoint.

Changed question: Change threshold to strictly above8.

Worked example 28: Check a wrong solution set

Question: A learner claims −2x<6 gives x<−3. Find a counterexample.

Method: x=0 satisfies original0<6 but fails claimed x<−3. Therefore claimed direction is false; correct set x>−3.

Check: Use a satisfying test to disprove a wrong general solution.

Common mistake: Do not judge just by a copied final symbol.

Changed question: Test x=−4 for the correct set.

Worked example 29: Positive division does not reverse

Question: Solve 5x−8<−3.

Method: Add eight:5x<5; divide by positive five: x<1.

Check: x=0 gives−8<−3 true.

Common mistake: A negative constant does not trigger a flip.

Changed question: Compare with −5x−8<−3.

Worked example 30: Negative coefficient counterpart

Question: Solve −5x−8<−3.

Method: Add eight: −5x<5; divide by −5, reverse x>−1.

Check: x=0:−8<−3 true; x=−2:2<−3 false.

Common mistake: Contrast with positive-coefficient version.

Changed question: Draw both on one number line.

Ten precisely diagnosed inequality mistakes

1. The inequality sign is never flipped

Why it goes wrong: A student divides −2x<6 by −2 and writes x<−3. Reflect two numbers across zero to explain why order reverses, then test x=0.

Repair and test: Correct set x>−3 contains zero and excludes−4.

Independent evidence: A new negative-coefficient problem has proper direction without prompting.

2. Every negative number triggers a flip

Why it goes wrong: The equation 2x<−6 is incorrectly changed to x>−3 because the final bound is negative. Divide by positive two: direction stays x<−3.

Repair and test: Circle the actual multiplier or divisor, not the sign anywhere on the page.

Independent evidence: The student correctly contrasts positive and negative coefficients.

3. An open boundary becomes closed

Why it goes wrong: The problem says strict <, but the graph uses a filled circle at its boundary. Test the boundary in the original equation; equality fails for strict symbols.

Repair and test: Use x=−3 in −2x<6 and show6<6 is false.

Independent evidence: The learner draws endpoint inclusion correctly on a changed problem.

4. The number line points the wrong way

Why it goes wrong: The algebra gives x≥−4 but shading goes left because the boundary is negative. A number line still increases left-to-right; x≥−4 means all points at or right of−4.

Repair and test: Mark −5,−4,−3 and ask which satisfy.

Independent evidence: The arrow direction follows the actual order of numbers.

5. Sign reverses while subtracting a constant

Why it goes wrong: A student changes −3x+5≤17 into −3x≥12 after subtracting five. Subtraction by the same constant preserves order. Only dividing by −3 reverses.

Repair and test: Separate each operation into its own line and name its sign.

Independent evidence: A multi-step inequality has exactly the appropriate reversals.

6. Brackets are distributed incorrectly

Why it goes wrong: −2(x−3) becomes−2x−6 instead of−2x+6. The opposite sign on −3 creates a positive six.

Repair and test: Expand first, then compare with the divide-first solution.

Independent evidence: The next bracket inequality preserves every sign.

7. All real solutions reduced to one example

Why it goes wrong: The learner reports x=0 for x>−3 because zero works. Explain that the inequality solution is the full infinite set and a number-line interval or symbol states it.

Repair and test: List other qualifying positive and negative inputs.

Independent evidence: The final answer names a set, not just a point.

8. A valid algebraic root violates word-count context

Why it goes wrong: An inequality yields x≥−7/3, and a context requires nonnegative complete objects. Apply the extra n≥0 integer restriction after the algebra.

Repair and test: List feasible counts as a separate step.

Independent evidence: The next contextual solution includes both algebraic and physical domains.

9. A contradiction is treated as x=0

Why it goes wrong: After cancellation 2<−3 appears and the child assigns x=0. A false constant inequality has no solution because it cannot depend on x.

Repair and test: Test x=0 in the original to demonstrate failure.

Independent evidence: The learner classifies both true and false constant inequalities.

10. A possibly zero variable is used as a divisor

Why it goes wrong: A student divides both sides of an inequality by a variable a without knowing whether a is positive, negative or zero. The direction and validity cannot be decided unconditionally.

Repair and test: Examine three separate cases a>0, a<0 and a=0.

Independent evidence: No unconditional flip is applied to unknown-sign multipliers.