Twenty-eight additional inequality problems: bounds, choices and later algebra

The first twenty-four checks were about reading comparisons and identifying allowed values. These later questions make the same ideas useful in stories and simple algebra. Problems that require solving an inequality with x, distributing a negative coefficient or describing an interval of real solutions are labelled extensions where they exceed a learner’s current school level. The important first principle remains whether the statement includes or excludes the boundary.

In every case, test at least one value claimed to satisfy the inequality and one value that should fail. A correct inequality is not established merely because a rearrangement looks like an equation. The direction of the comparison and the permitted domain must remain consistent.

Additional worked inequality 25 · a minimum count

Question. A fictional display needs at least 12 books. Write the condition for whole-number count n.

Solution. n ≥ 12.

Reason. Exactly twelve is allowed.

Independent check. n=12 passes, n=11 fails.

Additional worked inequality 26 · a maximum count

Question. A fictional box holds at most 20 tokens. Express its count n.

Solution. 0 ≤ n ≤ 20 for a whole-object count.

Reason. The physical count cannot be negative, and 20 is the inclusive capacity.

Independent check. n=20 qualifies; n=21 does not.

Additional worked inequality 27 · strictly older

Question. An invented activity is for ages strictly greater than 12. Write age condition a.

Solution. a > 12.

Reason. At exactly twelve the strict condition is not met.

Independent check. a=13 passes and a=12 fails.

Additional worked inequality 28 · permitted temperature model

Question. A fictional reading is below −2 units. Write condition t.

Solution. t < −2.

Reason. Numbers farther left than −2 are smaller.

Independent check. t=−3 passes; t=−1 fails.

Additional worked inequality 29 · real interval

Question. Represent x between −1 and 4, both included.

Solution. −1 ≤ x ≤ 4.

Reason. Two closed boundary points with the region between them.

Independent check. x=−1 and x=4 both qualify.

Additional worked inequality 30 · real interval with open top

Question. Represent values at least 2 and strictly less than 7.

Solution. 2 ≤ x < 7.

Reason. Include lower endpoint, exclude upper.

Independent check. x=2 passes; x=7 fails.

Additional worked inequality 31 · count positive integers

Question. How many positive integers satisfy x < 6?

Solution. 5: 1,2,3,4,5.

Reason. Positive excludes zero and negative values.

Independent check. Each is strictly below six.

Additional worked inequality 32 · nonnegative integers

Question. List nonnegative integers less than four.

Solution. 0,1,2,3.

Reason. Nonnegative includes zero.

Independent check. Four is excluded by the strict upper bound.

Additional worked inequality 33 · compare negative reciprocals

Question. Which is greater: −2/3 or −3/4?

Solution. −2/3.

Reason. −2/3≈−0.667 lies to the right of −0.75.

Independent check. −2/3 > −3/4.

Additional worked inequality 34 · fraction comparison

Question. Compare 7/10 and 2/3.

Solution. 7/10 > 2/3.

Reason. Convert to thirtieths: 21/30 compared with 20/30.

Independent check. Difference is 1/30.

Additional worked inequality 35 · a false inclusive claim

Question. Is x=5 allowed by x < 5?

Solution. No.

Reason. The strict sign excludes the boundary.

Independent check. 5 < 5 is false.

Additional worked inequality 36 · two inclusive conditions

Question. Find integers that satisfy x ≥ −2 and x ≤ 2.

Solution. −2,−1,0,1,2.

Reason. Take the intersection of two inclusive rays.

Independent check. Each candidate lies between −2 and 2.

Additional worked inequality 37 · no whole-number solution

Question. Find whole numbers satisfying 0 < x < 1.

Solution. None.

Reason. No integer lies strictly between zero and one.

Independent check. Real numbers such as 0.5 would qualify under a different domain.

Additional worked inequality 38 · one integer solution

Question. Find integers satisfying 2 < x < 4.

Solution. x=3.

Reason. Three is the sole integer between the bounds.

Independent check. x=2 and x=4 are excluded.

Additional worked inequality 39 · use an integer domain

Question. Find all integers satisfying −1 < x ≤ 2.

Solution. 0,1,2.

Reason. −1 is excluded, +2 included.

Independent check. Each choice satisfies both conditions.

Additional worked inequality 40 · a comparison becomes equation at boundary

Question. If x ≥ 3, does x=3 qualify?

Solution. Yes.

Reason. The ≥ symbol includes equality at three.

Independent check. 3 ≥ 3 is true.

Additional worked inequality 41 · first algebraic inequality (extension)

Question. Solve x+2 > 7.

Solution. x > 5.

Reason. Subtract two from both sides without changing order.

Independent check. x=6 passes; x=5 fails.

Additional worked inequality 42 · another positive shift (extension)

Question. Solve x−4 ≤ 9.

Solution. x ≤ 13.

Reason. Add four to both sides.

Independent check. x=13 passes; x=14 fails.

Additional worked inequality 43 · positive coefficient (extension)

Question. Solve 3x ≥ 12.

Solution. x ≥ 4.

Reason. Dividing by positive three preserves the inequality direction.

Independent check. x=4 passes, x=3 fails.

Additional worked inequality 44 · two-step inequality (extension)

Question. Solve 4x+3 < 19.

Solution. x < 4.

Reason. Remove three and divide by positive four.

Independent check. x=3 passes; x=4 fails.

Additional worked inequality 45 · inclusive two-step (extension)

Question. Solve 2x−5 ≤ 7.

Solution. x ≤ 6.

Reason. Add five, divide by positive two.

Independent check. x=6 passes and x=7 fails.

Additional worked inequality 46 · divide by negative (extension)

Question. Solve −2x < 8.

Solution. x > −4.

Reason. Dividing by a negative reverses inequality direction.

Independent check. x=0 gives 0<8 true; x=−5 gives 10<8 false.

Additional worked inequality 47 · another negative coefficient (extension)

Question. Solve −3x ≥ 12.

Solution. x ≤ −4.

Reason. Division by −3 reverses ≥ into ≤.

Independent check. x=−4 gives 12≥12; x=0 fails.

Additional worked inequality 48 · signed constant (extension)

Question. Solve 5−2x ≥ 1.

Solution. x ≤ 2.

Reason. Subtract five, giving −2x≥−4, then divide by −2 reversing direction.

Independent check. x=2 gives 1≥1, x=3 gives −1≥1 false.

Additional worked inequality 49 · another signed expression (extension)

Question. Solve 7−3x > −2.

Solution. x < 3.

Reason. Subtract seven: −3x>−9. Divide by −3 and reverse.

Independent check. x=2 gives 1>−2 true; x=4 gives −5>−2 false.

Additional worked inequality 50 · mixed two-sided inequality (extension)

Question. Solve 2 ≤ x+3 < 7.

Solution. −1 ≤ x < 4.

Reason. Subtract three from every part.

Independent check. x=−1 and x=3 qualify; x=4 fails.

Additional worked inequality 51 · interval contains infinitely many real values

Question. How many real numbers satisfy 0 < x < 1?

Solution. Infinitely many.

Reason. All fractional and decimal real numbers strictly inside qualify.

Independent check. For example 1/2, 1/3 and 0.9 qualify.

Additional worked inequality 52 · changed inequality direction

Question. A pupil claims multiplying x < 3 by −1 gives −x < −3. Correct it.

Solution. −x > −3.

Reason. Multiplication by a negative reverses the order.

Independent check. For x=2, 2<3 and −2>−3 are both true.

Why multiplying an inequality by a negative reverses the sign

Start with the true comparison 2 < 3. Multiply both values by positive two and obtain 4 < 6, preserving the order. Now multiply both by negative one: −2 and −3. Their signed order is −2 > −3. The sign reverses because multiplication by a negative reflects the values across zero; what was farther right becomes farther left relative to the other number.

This fact explains why solving −2x < 8 requires x > −4. A pupil who divides both sides by −2 but keeps the same symbol obtains x < −4, a false solution set. To expose the mistake without another long derivation, test x=0 in the original expression: −2(0)=0 is less than eight, so zero should qualify. The incorrect x < −4 would reject zero.

The rule must be connected to number-line order before it becomes an algebraic shortcut. If a learner cannot yet compare −3 and −8, they are unlikely to understand why negative multiplication changes the direction. Return to signed positions and simple numerical examples, then progress to expressions with x when the school’s syllabus permits.

Boundary points: open and closed circles communicate exact meaning

Inequality	Number-line endpoint	Shaded direction	A boundary test
x > 2	Open at 2	Right	x=2 fails
x ≥ 2	Closed at 2	Right	x=2 passes
x < 2	Open at 2	Left	x=2 fails
x ≤ 2	Closed at 2	Left	x=2 passes
−1 ≤ x < 4	Closed at −1; open at 4	Between	−1 passes, 4 fails

A small filled circle and a hollow circle might look like drawing details, but they encode whether the endpoint belongs to the solution set. A student may correctly solve an inequality symbolically but lose the boundary when translating it to a diagram. A simple final check is to substitute exactly the boundary value into the original statement.

For integer-only problems, a plotted set of isolated points can be clearer than a continuous shaded region. For example, 2 < n ≤ 5 with integer n gives the three values 3,4,5. If the variable measures a continuous length, real numbers such as 2.5 also qualify. The mathematical notation should be interpreted with its stated domain.

Seven diagnostic errors and the first repair

Visible mistake	First unstable idea	Repair prompt
−8 > −3 because eight is bigger	Magnitude confused with signed position	Place both on the number line
x > 4 with a filled dot on four	Strictness of endpoint lost	Does four pass when substituted?
‘At most’ written x ≥ c	Language interpreted backwards	Which side of the boundary is allowed?
Only lists x=4 as solution for x ≥ 4	Inequality mistaken for equation	Try x=5 and x=6
Divides by −2 without reversing sign	Negative multiplication and order disconnected	Compare 2<3 with −2>−3
Lists only integers when x is real	Domain ignored	Does x=3.5 satisfy the bounds?
Treats two conditions as alternatives	Logical ‘and’ confused with ‘or’	Which candidate satisfies both?

Correcting an inequality requires two linked decisions: the mathematical transformation and the interpretation of allowed values. A tutor should listen to the pupil’s explanation of the first wrong step. A child who solves correctly but plots the wrong endpoint needs representation practice, whereas a child who draws boundaries correctly but reverses a signed comparison needs number-line work.

A ten-minute parent diagnostic

Write the four signs <, >, ≤ and ≥ and ask the child to read each aloud. Then present −5 and −2 and ask which is greater. After that, show x > 3 and x ≥ 3 and ask whether three itself is allowed. If the child can explain the difference without a prompt, move to writing ‘at least seven’ and ‘fewer than five’ in symbols.

Finish with a domain change: for integer n, list numbers satisfying −2 ≤ n < 2, giving −2,−1,0,1. Then ask what changes when x represents a real measurement rather than an integer count. The child should realise many non-integer values also satisfy the bounds. This is evidence of an actual mathematical concept, not simply remembering how to draw an arrow.

Once a misconception is repaired, choose a new boundary and fresh values. The pupil should explain which endpoint is included, shade in the correct direction and test one passing and one failing value. A copied diagram is a useful start, but an independent new case after a few days is a better sign that the meaning has settled.

A flexible six-week comparison and inequalities route

Week	Focus	What to see independently
1	Signed-number ordering	Compares negative values without looking only at digits
2	Strict and inclusive inequality signs	Reads symbols and draws valid boundaries
3	Translate ordinary-language constraints	Writes at least, at most and between correctly
4	Integer and real-valued solutions	Names domain and tests candidate values
5	Simple algebraic inequality transformations if appropriate	Maintains equality restrictions and order
6	Mixed comparison and number-line questions	Identifies first wrong step and corrects unaided

This plan is illustrative, not a claim that every G1, G2 or G3 Secondary 1 classroom in 2028 teaches all listed material at the same time. The official G2/G3 Secondary One numbers strand includes comparison signs; deeper linear-inequality solving is an extension to match the actual school sequence. A pupil still uncertain with negative numbers should receive more time there before algebraic bounds are introduced.

A productive small-group lesson

Three pupils might each get a number-line item wrong for different reasons. One compares negative numbers by unsigned magnitude, one uses the correct direction but a filled boundary for a strict inequality, and another draws the right diagram but cannot express ‘at most’ in symbols. A tutor should identify those causes separately and give each pupil a different changed-context task.

A three-student 90-minute lesson may combine short shared explanations with close inspection of working, independent practice and delayed retrieval. The goal is that each child can decide which values qualify without needing a teacher to keep naming the rule. A small class can make that observation easier, but it does not guarantee examination results by itself.

Frequently asked questions

Is x > 5 the same as x ≥ 5?

No. The first excludes five; the second includes it. Both permit values greater than five.

Why does the sign reverse when dividing by a negative?

Negative multiplication reverses the order of numbers on the number line. Test 2<3 and then compare −2 with −3 to see the change.

What does an open circle mean on a number-line solution?

It means the boundary value is not included. A closed point means it is included.

Can a single inequality have infinitely many answers?

Yes. For real numbers, x>3 includes infinitely many values. If the domain is restricted to whole numbers, the admissible values change accordingly.

Are two-sided algebraic inequalities compulsory at the start of Secondary 1?

Not for every subject level. This guide includes them as later-stage extensions; the applicable MOE syllabus and school planning determine the scope.

How should an inequality solution be checked?

Test one value that the solution permits and one that it rejects, preferably including a boundary value. Both tests should agree with the original statement.

Sources and related Punggol Mathematics

• MOE G2/G3 Mathematics syllabus — Secondary One numerical inequality symbols and related progression.
• MOE Full Subject-Based Banding — subject-level arrangements.
• Punggol PSLE 2027 to Secondary 1 Mathematics 2028 hub.
• Signed integers and number lines — prerequisite number sense.
• Linear equations and balance — a different relationship from inequalities.
• eduKate Secondary 1 Mathematics teaching framework — diagnostic learning and small-group support.

Inequality mastery is recognising a set of valid values, not guessing a single number. Read the boundary carefully, preserve the correct order of signed quantities, use the right endpoint and explain which candidates satisfy the condition.

Five complete boundary checks: from words to valid values

One — a maximum is inclusive. Imagine a fictional display with a maximum capacity of twenty items. A whole-number item count n must satisfy 0 ≤ n ≤ 20. With nineteen items, the display has room for one more. With twenty items, the capacity is full but the numerical condition remains satisfied. With twenty-one, it fails. The phrase ‘at most twenty’ includes twenty. Change it to ‘fewer than twenty’ and the count must be strictly below twenty. An open or closed boundary is therefore a statement about permission, not a drawing preference.

Ask a pupil to substitute n=20 into both conditions. For n ≤ 20, the statement is true. For n < 20, it is false. A child may know that the arrow points left but still draw a filled point for the strict condition. Testing the boundary helps identify exactly where the interpretation changed. The same logic works for any meaningful upper limit once its unit and permitted domain have been specified.

Two — negative values are ordered on a line. Suppose a hypothetical score must be greater than negative two. The condition is x > −2. The value −1 qualifies, zero qualifies and positive three qualifies. The value −2 itself does not. Negative five is smaller than negative two, so it also fails. A pupil who compares only the unsigned numerals may think −5 is larger because five is larger than two. Return to the number line and ask which point lies farther right.

Move the boundary from −2 to −5 and repeat. The condition x > −5 permits −4 and −3 even though both are negative. This demonstrates that an inequality is about relative position, not a requirement that all allowed values be positive. The open endpoint remains at −5 because exactly −5 is excluded. Try one passing and one failing value to check the shading.

Three — the domain changes the solution list. For 3 ≤ n ≤ 7, the permissible whole-number counts are 3, 4, 5, 6 and 7. But if x represents a real-valued measurement, the same numerical bounds include fractions and decimals such as 3.25, 4.5 and 6.99. The symbolic comparisons have not changed. What changed is the defined kind of number that x may represent. Before listing a finite number of answers, the child should ask whether the problem explicitly restricts x to integers.

The condition 0 < x < 1 makes the difference especially clear: there are no whole-number solutions, but infinitely many real-number solutions. A pupil who writes ‘no solution’ without reading the domain is making a modelling error, not an arithmetic error. An accurately drawn continuous number line between open endpoints at zero and one can communicate the real-number solution set.

Four — two conditions must both be satisfied. The statement x ≥ 2 and x < 5 means 2 ≤ x < 5. With integer x, the solutions are 2, 3 and 4. The number five fails despite satisfying the first condition, because it does not satisfy the second. The number one fails the first. Both tests matter. Ask the pupil to explain why the overlapping region between the two constraints is the answer.

By contrast, x < 0 and x > 3 cannot both hold for the same real number. No number can lie to the left of zero and simultaneously to the right of three. This does not mean either separate inequality is wrong. The contradiction appears only because the word ‘and’ requires both conditions together. The idea prepares students for later set intersections without requiring formal set notation prematurely.

Five — a negative factor reverses numerical order. The valid comparison 2 < 3 becomes −2 > −3 after multiplication by negative one. This concrete example explains why solving −2x < 8 produces x > −4 rather than x < −4. Test x=0: the original left-hand value is zero, which is below eight, so zero must be permitted. The wrong expression x < −4 would reject it. Try x=−5: the original product is ten, so that value must be excluded.

Do not teach the sign reversal as a chant detached from signed-number order. A pupil who cannot reliably compare −8 with −3 needs that number-line foundation first. A student ready for the algebraic extension can check the transformation with two numerical candidates, one that satisfies the original and another that fails it. Equivalent inequalities must permit exactly the same values in the given domain.

A ten-minute home transfer assessment

Ask the child to write ‘at least six but less than ten’ as 6 ≤ x < 10, list its integer solutions 6, 7, 8 and 9, and draw a closed left endpoint at six and an open right endpoint at ten. Then change the wording to ‘more than six and no more than ten’, giving 6 < x ≤ 10 and integer values 7, 8, 9 and 10. The two cases differ only in endpoint inclusion, but that difference controls which numbers are allowed.

Finish with negative bounds, such as −3 ≤ x < 2. The integer solutions are −3, −2, −1, 0 and 1. Ask for an example of a real solution that is not an integer, such as −2.5, and a value outside the set, such as two. The child should explain the result from the sign and boundary rather than depend on a tutor naming the rule. Return to a changed condition a few days later to check whether the skill has persisted.