Learn to solve G3 SEC K310 linear inequalities and, more importantly, interpret exact boundary conditions through forty-five original worked problems and real-world decision checks.
The decisive question is not merely what x equals
A linear equation identifies a value at which two expressions agree. A linear inequality identifies a set of acceptable values. In an examination problem about a budget, capacity, permitted time or minimum participation, the most important question is often not “What is x?” but “Which values of x satisfy every restriction?” You may calculate perfectly and still answer incorrectly if you forget that a person must be counted as a whole number, or if you include a strict boundary that the wording excludes. This guide develops the habits that prevent those mistakes.
Imagine a hall with room for twenty visitors. “At most twenty visitors” includes twenty; “fewer than twenty visitors” excludes twenty. The first is n≤20, the second n<20. In the second case, nineteen is the largest whole-number count. This is not just symbol practice. It is a method for making an accurate decision from stated conditions. The objective is to read the restriction, define the variable, write the inequality, solve it and interpret what is actually allowed.
Know exactly what SEC K310 assesses
For the first Singapore-Cambridge Secondary Education Certificate examinations in 2027, G3 Mathematics is identified as K310, with the previous O-Level code 4052 listed for reference. The official 2027 SEAB K310 syllabus includes solving linear inequalities in one variable and representing solutions on a number line. The two written papers are each 2 hours 15 minutes, each carry 90 marks and each contribute 50% of the assessment. Approved calculators may be used in both, while essential working still matters. Review the official G3 subject list for the current document.
This guide stays within one-variable inequalities while enriching the interpretation. We will sometimes combine an upper and a lower bound on the same variable; that is not a lesson in graphing feasible regions in two variables. The examples are original practice problems and do not claim to reproduce future examination questions. The transfer goal is broader: identify the quantity, retain every mathematical condition, and make a conclusion that can be checked.
Define the variable before writing an expression
Write a sentence: “Let n be the number of complete student teams.” This prevents the variable from silently changing midway. A team count is a non-negative integer. A measurement of length may take fractional values. A price measured to the nearest cent follows a different convention again. These domains change what a mathematical bound means in an application.
Suppose eight supervisors already occupy seats on a 46-seat bus, and every student team has three members. If n counts student teams, the occupancy is 8+3n, not n or 8+n. The seating condition is 8+3n≤46. You must solve for teams and give a whole-number answer. If n had instead been defined as the number of individual students, the model would be n+8≤46. Both can be mathematically valid models, but they answer different questions. The opening definition prevents that confusion.
Translate boundary words, not just familiar phrases
“No more than,” “at most,” “not exceeding” and “up to” usually mean ≤. “At least,” “not fewer than” and “a minimum of” usually mean ≥. “Fewer than” and “less than” mean . These are small words with large consequences. If at least twelve people are needed, a group of eleven fails even if it seems close. If fewer than twelve may enter, twelve itself fails.
Do not mechanically extract one word from a longer sentence. “No more than twelve extra visitors” counts additional people, whereas “no more than twelve people in total” counts everyone. A programme might also require two supervisors. Distinguish totals, additions, and quantities inside a category. Before using a symbol, test the boundary in ordinary English: Would the policy allow exactly twelve? The answer tells you whether to include equality.
Preserve the order when adding or subtracting
The order of two numbers does not change when both are shifted by the same amount. For x+7≤16, subtract seven on both sides to obtain x≤9. For 3x+8>29, subtract eight to obtain 3x>21, then divide by positive three to obtain x>7. All steps maintain the original strictness. If x counts items, the smallest admissible positive integer is eight; if x is a continuous measurement, values such as 7.1 may be possible.
Use substitution to check the proposed boundary. At x=7, 3x+8=29, so the strict >29 condition fails. At x=8, the left side is 32 and succeeds. Two substitutions can catch a mistaken comparison sign more reliably than another attempt to memorise a keyword. They also prepare you to defend a greatest or smallest integer answer.
Why negative multiplication reverses the direction
Consider the true statement 2−5. That reversal is the rule whenever an inequality is multiplied or divided by a negative number. It is not an arbitrary examination trick.
To solve 12−2x≥4, subtract twelve: −2x≥−8. Dividing by −2 gives x≤4. At x=4 the original left side equals four, so it qualifies; at x=5 the original left side equals two, so it fails. Return to Vol 0131 on inequality sign reversal if the operation itself remains uncertain. Here the point is to apply that foundation while preserving the meaning of the boundary in a practical decision.
Number lines make the solution set visible
An open dot means that an endpoint is excluded. A filled dot means that it is included. From x≥4, put a filled dot at four and shade towards larger values. From x<4, put an open dot at four and shade towards smaller values. For −2<x≤5, use an open dot at −2, a filled dot at five and shade the segment between them. The dots express strictness; the shading expresses the range.
If x is a real number, the shaded interval contains many non-integer values. If x counts whole tables, we select the whole numbers lying inside the same permitted interval. Do not confuse an interval drawing with a list of allowable discrete items. Likewise, do not write x=5 when the question asks you to solve x≤5. Five may be a permitted value or the greatest permitted integer, but it is not the whole mathematical solution set.
Integer restrictions change how you choose an answer
Suppose n≤12.8 and n counts complete teams. The maximum is twelve, not thirteen. Rounding 12.8 to the nearest integer would produce a forbidden result. Instead choose the largest integer that satisfies the original inequality. If n>6.2, the smallest integer is seven. The same seven is also the smallest integer satisfying n≥6.2, yet the corresponding real-number sets differ because one includes 6.2 and the other does not.
For measurements, do not invent whole-number restrictions. A travel time less than twelve minutes may be 11.9 minutes, and there is no greatest real number below twelve. A price recorded in whole cents is discrete, but a theoretical real-valued amount is continuous. Read how the quantity is defined and recorded. The correct interpretation follows from the domain rather than from a convenient rounding habit.
Two limits mean their intersection, not their union
A workshop needs at least four participants but permits at most nine. It imposes n≥4 and n≤9. Together these give 4≤n≤9. If another rule says fewer than seven, the permitted whole numbers become four, five and six. You must satisfy every condition at the same time.
A budget might permit twelve visitors while seating permits ten. Ten is the actual maximum, not twelve. Conversely, a programme requiring at least eighteen visitors cannot run in a room holding only fifteen. In that case the correct result is that no feasible size exists. Mathematics is allowed to reveal impossibility. Do not manufacture an answer because a word problem appears to expect a number. The conclusion is only valid when all its conditions hold.
Fixed costs belong once; per-unit costs repeat
A visit costs $90 to organise plus $14 per participant. If the total budget is no more than $300, let n count people and write 90+14n≤300. Subtract the fixed fee to get 14n≤210, so n≤15. The maximum whole-number group is fifteen. If at least eight are needed, the permitted range is eight through fifteen. A mistake such as 14(90+n) confuses a fixed fee with a charge that repeats for every participant.
When reading an unfamiliar situation, make two brief notes: what stays constant and what changes per unit. A hall reservation might remain constant while drinks cost per guest. A base fare might remain constant while distance adds a charge per kilometre. Translating the story accurately often requires more thought than the subsequent algebra, so do not rush the modelling stage.
Brackets and fractions must be simplified carefully
For 4(2x−1)≤3x+11, expand to 8x−4≤3x+11. Subtract 3x and add four, giving 5x≤15, so x≤3. At x=3 both original sides equal twenty, which checks the inclusive endpoint. For (2x+3)/5<3, multiply by positive five to obtain 2x+3<15, then x<6. The positivity of the multiplier matters.
Do not casually multiply inequalities by variable expressions whose sign is unknown. This guide uses known positive denominators when removing fractions. Likewise, when simplifying 5−(2x−3), distribute the negative sign to produce 8−2x rather than 2−2x. Most apparent “hard” problems in this area fail through ordinary arithmetic and notation errors. Clean intermediate lines protect the underlying idea.
When x cancels, interpret the statement
An inequality does not always end in x≤a or x>a. Solve 2(x+3)>2x+7 and the x terms cancel, leaving 6>7. This is false for every real x, so the solution set is empty. Solve 2(x+3)<2x+7 and the result is 6<7, true for every real x. If the problem has a narrower domain, such as non-negative integers, that separate restriction still applies.
Do not write x=0 just because the x terms vanish. Cancellation reveals whether the original comparison is always true or always false. This is a particularly useful check in real-world modelling: incompatible conditions can emerge from algebra, and they must be reported rather than disguised.
What the final sentence needs to say
The instruction “solve” usually asks for all mathematical solutions. “Find the largest whole number” asks for one admissible integer. “State all possible numbers of groups” asks for a discrete range or list. “Is the proposed plan feasible?” asks for a yes-or-no conclusion with a numerical reason. Read this command before writing the final line.
Compare “x≤12,” “twelve groups,” and “twelve is the maximum number of complete groups.” They are not interchangeable. The first is a solution set, the second an unspecified count, and the third a complete answer to a maximum question. Write the requested units. When necessary, demonstrate that the next larger whole number violates the condition. That is how you show that the maximum is not merely one example that works.
Graph intersections and inequality intervals are different questions
In Vol 0234, the emphasis was solving equalities and interpreting a point where lines intersect. This volume concerns which numbers satisfy upper and lower limits on one variable. A capacity restriction can have many admissible counts even though the break-even point in another problem might be a single value. Preserve that distinction when a question uses everyday words such as feasible, minimum and maximum.
The PSLE Learning Guide supports earlier comparisons and ratio ideas. The Secondary 4 Mathematics notation guide extends the discussion of symbols. Use the right owner for the right weakness rather than repeating a full introductory lesson in every new practice session.
A consistent error ledger shortens revision
Record a specific cause beside each mistake: translated “fewer than” as ≤; counted supervisors as student teams; rounded a fractional upper bound upwards; forgot to reverse the sign after dividing by −3; used a filled dot for a strict endpoint; answered with one value when a set was requested; or checked seating but not budget. Each label points to a distinct repair.
Practise one changed example immediately after correcting a mistake. If the error concerns strictness, keep the numbers but change “no more than” to “fewer than.” If it concerns integers, change a continuous distance to a number of packets. If it concerns multiple conditions, add a minimum requirement after solving a maximum. This kind of variation tests understanding. Repeating ten identical calculations may feel productive while leaving the original misunderstanding untouched.
An end-to-end example with three independent rules
A science club has a hall with 96 seats, including fourteen seats reserved for supervisors. Every student group contains six learners. At least five groups must take part. A materials budget of $350 must retain at least $110 after purchasing $20 of materials for each group. Let g count complete student groups. Seating gives 14+6g≤96, hence g≤13⅔, which means at most thirteen complete groups. The budget gives 350−20g≥110, hence −20g≥−240 and g≤12 after dividing by negative twenty. The participation rule gives g≥5. Combining all three, 5≤g≤12 with g an integer.
Twelve groups use 72 student seats plus fourteen supervisor seats, or 86 seats, and they spend $240 on materials, leaving exactly $110. Thirteen groups would still fit in the hall but would leave only $90 after materials and therefore fail the budget. The maximum under all rules is twelve. Notice how checking only one constraint would produce a plausible but wrong answer.
A suggested practice sequence, not a predicted paper
Start with ten boundary phrases and translate each into an inequality before doing any algebra. Next solve five expressions, including one with a negative coefficient and one in which x cancels. Then tackle three practical scenarios: a budget, seating and time. Finish by substituting the final integer into each original restriction and testing the nearest excluded integer. A calm twenty-five-minute rehearsal of this kind is more useful than rushing through many questions without examining the mistakes.
Return several days later with changed numbers and slightly different wording. Can you still distinguish an inclusive boundary from a strict one? Can you say why a fractional value may represent time but not people? Can you explain which of two constraints actually controls the maximum? If so, the skill is transferring. That is the objective of a learner’s guide, rather than collecting a long list of tricks.
Forty-five original worked problems with boundary checks
Worked problem 1: A no-more-than rule
Question: A hall permits no more than 24 participants.
Worked solution: Let n be the count. n≤24. The maximum whole-number count is twenty-four.
Interpretation check: Boundary check: twenty-four is permitted because equality is allowed; twenty-five is excluded.
Worked problem 2: A strict cap
Question: A hall permits fewer than 24 participants.
Worked solution: n<24. If n is a non-negative integer, the maximum is twenty-three.
Interpretation check: Twenty-four fails. Strictness changes the answer even when the stated number is unchanged.
Worked problem 3: A minimum team
Question: A project needs at least nine volunteers.
Worked solution: n≥9. Nine is the minimum whole-number count.
Interpretation check: Larger counts may also work unless a second condition imposes a maximum.
Worked problem 4: An examination threshold
Question: A qualifying result must be more than 18 whole points.
Worked solution: s>18. The smallest qualifying integer score is nineteen.
Interpretation check: Eighteen does not qualify; do not silently turn > into ≥.
Worked problem 5: One addition
Question: Solve x+7≤16.
Worked solution: Subtract seven on both sides and obtain x≤9.
Interpretation check: The order does not change when both sides are shifted equally.
Worked problem 6: One subtraction
Question: Solve 3x−4≥11.
Worked solution: Add four, giving 3x≥15, and divide by positive three to obtain x≥5.
Interpretation check: At x=5 the left side is eleven; the boundary belongs.
Worked problem 7: Negative multiplication
Question: Solve −2x≤−14.
Worked solution: Divide by negative two and reverse the sign: x≥7.
Interpretation check: At x=7 equality holds, while x=6 fails.
Worked problem 8: Negative coefficient in context
Question: Solve 15−3x>0.
Worked solution: Subtract fifteen, divide by negative three and reverse the sign: x<5.
Interpretation check: At x=5 the original expression is zero, which is not positive.
Worked problem 9: A bracketed comparison
Question: Solve 4(2x−1)≤3x+11.
Worked solution: Expand: 8x−4≤3x+11. Rearrange: 5x≤15, so x≤3.
Interpretation check: Substitution of x=3 gives twenty on both sides.
Worked problem 10: A positive fraction
Question: Solve (2x+3)/5<3.
Worked solution: Multiply by positive five: 2x+3<15; hence x<6.
Interpretation check: The endpoint six makes equality, which fails strict <.
Worked problem 11: Two sides at once
Question: Solve 5≤2x+1<13.
Worked solution: Subtract one from each part and divide by positive two: 2≤x<6.
Interpretation check: Include two; exclude six. The whole-number choices are 2, 3, 4, 5.
Worked problem 12: A negative endpoint
Question: Solve −3<2x+1≤9.
Worked solution: Subtract one: −4<2x≤8. Divide by two: −2<x≤4.
Interpretation check: The left endpoint is open and the right endpoint closed.
Worked problem 13: Three-person teams
Question: A 46-seat bus has eight supervisors, and each student team has three members.
Worked solution: 8+3t≤46 gives t≤38/3. The maximum integer team count is twelve.
Interpretation check: Twelve teams occupy 44 seats including staff; thirteen would need 47.
Worked problem 14: Delivery and packets
Question: A $96 budget includes a $12 fixed delivery fee and packets costing $7 each.
Worked solution: 12+7p≤96 gives p≤12. At most twelve packets can be purchased.
Interpretation check: Thirteen packets would cost $103 and fail.
Worked problem 15: Trip booking fee
Question: A school pays $90 fixed plus $14 per participant within a $300 budget.
Worked solution: 90+14n≤300 gives n≤15. The maximum complete participant count is fifteen.
Interpretation check: Fifteen uses exactly $300; the fee is not multiplied by the participant count.
Worked problem 16: A price after discount
Question: An original price p receives a $5 discount and must become strictly below $18.
Worked solution: p−5<18, so p<23.
Interpretation check: An original price of $23 produces exactly $18 and fails.
Worked problem 17: Three consecutive integers
Question: Three non-negative consecutive integers have sum no more than 72. Let n be the first.
Worked solution: n+(n+1)+(n+2)≤72 gives n≤23. The greatest triple is 23, 24, 25.
Interpretation check: Its sum is 72; starting at 24 would exceed the cap.
Worked problem 18: A no-solution case
Question: Solve 2(x+3)>2x+7.
Worked solution: The inequality reduces to 6>7, which is false. No real x satisfies it.
Interpretation check: Cancellation of x does not mean x=0.
Worked problem 19: An all-values case
Question: Solve 2(x+3)<2x+7.
Worked solution: The inequality reduces to 6<7, always true. Every real x works.
Interpretation check: Any separate domain restriction still remains in a word problem.
Worked problem 20: Minimum and maximum
Question: An activity needs at least six students and allows at most ten.
Worked solution: Use n≥6 and n≤10, giving 6≤n≤10.
Interpretation check: This is the overlap of the two conditions, not a choice of one.
Worked problem 21: A conflicting pair
Question: An event requires at least eighteen participants but has only fifteen places.
Worked solution: The conditions n≥18 and n≤15 have no common value.
Interpretation check: Report that the plan cannot satisfy both restrictions.
Worked problem 22: Seat groups
Question: A 50-seat hall reserves twelve seats for demonstrators; each learner group has four members.
Worked solution: 12+4g≤50 gives g≤9.5, so at most nine complete groups fit.
Interpretation check: Ten groups would require 52 seats, exceeding capacity.
Worked problem 23: Retained cash
Question: A club has $350 and must keep $110 after purchasing $20 kits.
Worked solution: 350−20k≥110 gives k≤12. Maximum whole kits: twelve.
Interpretation check: Thirteen kits would leave only $90; dividing by negative twenty reverses the sign.
Worked problem 24: Continuous time
Question: Setup takes at least fifteen minutes and must finish within forty-two minutes.
Worked solution: If t is the setup duration, 15≤t≤42.
Interpretation check: Fractional minutes are possible unless the question says whole minutes.
Worked problem 25: Late arrival
Question: A learner can enter no more than twelve minutes after a session begins.
Worked solution: Let t be minutes late, with 0≤t≤12.
Interpretation check: Defining t as minutes late excludes early arrival; another definition might include it.
Worked problem 26: One open dot
Question: Represent x<4 on a number line.
Worked solution: Place an open dot at four and shade left towards smaller numbers.
Interpretation check: A filled dot would incorrectly allow x=4.
Worked problem 27: One closed dot
Question: Represent x≥4 on a number line.
Worked solution: Place a filled dot at four and shade right towards larger numbers.
Interpretation check: A filled dot communicates that equality belongs.
Worked problem 28: A mixed interval
Question: Represent −1<x≤4 on a number line.
Worked solution: Open at −1, closed at four and shade only the segment between.
Interpretation check: If x counts non-negative whole items, allowed counts are 0 through 4.
Worked problem 29: Whole-number lower limit
Question: Find the smallest integer n satisfying 5n−2>18.
Worked solution: Rearrange to n>4. The smallest integer is five.
Interpretation check: At n=4 the left side is eighteen, which does not exceed eighteen.
Worked problem 30: A fractional upper limit
Question: Find the greatest integer n satisfying n≤12.8.
Worked solution: The greatest admissible integer is twelve.
Interpretation check: Thirteen would violate the inequality. Nearest-integer rounding is not the method.
Worked problem 31: A real distance limit
Question: A path must be shorter than 12.8 km.
Worked solution: The permissible distances satisfy d<12.8; there is no greatest real permissible d.
Interpretation check: Whole-kilometre rounding is irrelevant unless explicitly requested.
Worked problem 32: A family group limit
Question: A 35-seat hall holds seven helpers and each family group occupies four seats.
Worked solution: 7+4f≤35 gives f≤7. Up to seven complete family groups fit.
Interpretation check: Seven groups use 28 seats plus seven helpers, exactly filling the hall.
Worked problem 33: Two upper constraints
Question: A 60-seat hall already has ten supervisors. Each table has five pupils and policy permits at most eight tables.
Worked solution: Seating gives t≤10; policy gives t≤8. The maximum is eight tables.
Interpretation check: Meeting the seating limit alone is insufficient.
Worked problem 34: A higher bound after division
Question: Solve 2−5x<17.
Worked solution: Subtract two and divide by −5, reversing the sign: x>−3.
Interpretation check: At x=−3 the expression equals seventeen and fails the strict comparison.
Worked problem 35: The variable disappears
Question: Solve 7x−7x≤3.
Worked solution: The statement is 0≤3, which is always true. Any real x satisfies this condition.
Interpretation check: The absence of x does not force x to be zero.
Worked problem 36: Subtract across a chain
Question: Solve 2≤x+4<11.
Worked solution: Subtract four throughout: −2≤x<7.
Interpretation check: The left endpoint belongs; seven is excluded.
Worked problem 37: Keep empty seats empty
Question: A room has 80 seats. Fifteen must remain empty, eight are used by staff, and family groups have five people.
Worked solution: 15+8+5g≤80 gives g≤57/5=11.4. The maximum is eleven complete groups.
Interpretation check: Twelve groups would need 83 seats counting empty reservations and staff.
Worked problem 38: Choose the right answer type
Question: Suppose the solution is 6≤n≤10 and n counts students.
Worked solution: For the maximum, give ten. For all possible counts, list 6,7,8,9,10.
Interpretation check: The mathematical result has not changed; the requested answer type has.
Worked problem 39: Minimum duration expressed in seconds
Question: A safety procedure must last at least 2.5 minutes.
Worked solution: In seconds, t≥150. If t is minutes, t≥2.5.
Interpretation check: Do not compare 2.5 minutes directly with a number stated in seconds without conversion.
Worked problem 40: No negative quantities
Question: A pure algebra result is x>−3, but x represents the number of packets.
Worked solution: Apply the separate domain: non-negative integers x=0,1,2,3,… qualify.
Interpretation check: The original x>−3 is correct algebra, but an item count is additionally restricted.
Worked problem 41: Combined attendance and budget
Question: At least five groups are needed. Materials cost $20 per group and no more than $240 may be spent.
Worked solution: g≥5 and 20g≤240 imply integer group counts 5 to 12 inclusive.
Interpretation check: If asked for all feasible counts, giving only twelve is incomplete.
Worked problem 42: One last threshold
Question: A reduced capacity rule says fewer than twelve teams may attend, while a sponsor needs at least eight.
Worked solution: Use 8≤t<12. For whole teams, the possibilities are 8,9,10,11.
Interpretation check: The bound at twelve is strict, so twelve is not allowed even though funding may favour it.
Worked problem 43: Two different units
Question: An activity has a $200 maximum budget; a bus has forty seats. Each pupil costs $5 and six seats are for staff.
Worked solution: The budget permits n≤40, seating permits n≤34. The combined maximum is 34 pupils.
Interpretation check: Compare the permitted pupil counts after converting the dollar and seat conditions into the same variable.
Worked problem 44: A missing third restriction
Question: A hall permits 25 people, four teachers must attend, and the school allows at most eighteen pupils.
Worked solution: Seating permits n≤21; school policy permits n≤18. The maximum is eighteen pupils.
Interpretation check: A correct plan must satisfy all relevant conditions, even when a less obvious policy is tighter.
Worked problem 45: Maximum is not the only choice
Question: A range gives 4≤n≤9 and n is a whole count.
Worked solution: Every count from four to nine is admissible; nine is only the greatest one.
Interpretation check: Read whether the question wants all values or a single maximum.
Worked problem 46: A strict price comparison
Question: A coupon reduces a price by $6 and the resulting price must be more than $30.
Worked solution: With original price p, p−6>30, so p>36.
Interpretation check: At p=$36 the reduced price is exactly $30, which fails the strict rule.
Final rehearsal: defend the interval, not the guess
Choose any five scenarios from different categories: a strict lower limit, an inclusive upper limit, an expression with a negative coefficient, a fixed-cost budget and a compound constraint. For each, define the variable and say whether it is discrete or continuous. Solve the inequality with all sign changes visible. Test a boundary value and a rejected neighbour. Then state the answer in the noun and unit requested, not simply as an unexplained symbol. If you can justify why the excluded choice fails, you have a result that stands up to scrutiny.
Keep the learning sequence coherent. Earlier volumes handled the sign-reversal rule and equations; this one focuses on which values are actually permitted. The next Science volume examines the human eye’s response when a viewed object moves from near to far. Although the subject changes, the same examination discipline applies: name the relevant mechanism, identify precisely what changes, and describe the consequence without adding an unsupported assumption. A short, defensible explanation is better than a long guess dressed up in technical language.
