Singapore School Mathematics Operating Manual · Chapter 16
A mathematical calculation can produce a candidate that the problem does not permit.
The candidate may violate a denominator restriction, fall outside a stated interval, represent a negative length, produce a non-integer count, break a triangle condition, exceed a budget, or satisfy one equation while violating another.
These are not secondary details. They define the feasible set: the collection of values or configurations that satisfy every active constraint.
This chapter develops a practical constraint discipline for school Mathematics. The central habit is: solve the mathematics, then intersect the result with what the problem actually allows.
What is a constraint? · Intersect conditions · Geometry · Integer constraints · Optimisation · Practice · Worked answers
1. A constraint limits the admissible state
If x is a positive integer, then x belongs to {1, 2, 3, …}. That is a domain constraint.
If x + y = 10, that is a relationship constraint.
If x ≥ 0 and y ≥ 0, those are inequality constraints.
If a rectangle must fit inside a fixed perimeter, the side lengths are constrained by that perimeter equation.
If a logarithm appears, its argument must be positive. If a denominator appears, it must be non-zero.
The feasible set is the intersection of all such conditions.
2. Satisfying one condition is not enough
Suppose x must satisfy x² = 25 and x > 0.
The equation gives candidates x = 5 and x = −5.
The positivity constraint removes −5.
The feasible solution set is therefore {5}.
The algebraic solution set and the admissible solution set are not always identical.
3. Constraints can come from the problem statement
“x is a whole number”, “0 ≤ t ≤ 10”, “the angle is acute”, “the radius is positive”, “the budget is at most $200” and “the point lies inside the triangle” are all constraints.
They should be extracted before or during solution, not discovered only after an impossible answer appears.
A useful first-pass annotation is to circle or list every phrase that restricts what values are allowed.
4. Constraints can also come from the mathematics itself
For 1/(x − 3), x ≠ 3 even if the question never states that restriction separately.
For √(2x − 1) over the real numbers, 2x − 1 ≥ 0, so x ≥ 1/2.
For log(x + 4), x > −4.
These restrictions arise from the definitions of the operations involved.
They belong to the problem whether or not they are printed in a highlighted box.
5. Algebraic candidates should be filtered through the original domain
Solve √(x + 2) = x.
The right side must be non-negative, so x ≥ 0.
Squaring gives x + 2 = x², hence x² − x − 2 = 0.
Factorising gives (x − 2)(x + 1) = 0, so candidates are x = 2 and x = −1.
The feasible-domain filter removes −1. Direct checking confirms x = 2.
The extra candidate was created by a one-way transformation. The domain condition recovers control.
6. Multiple inequalities create a feasible interval
Suppose x ≥ 2 and x < 7.
The feasible set is 2 ≤ x < 7.
If x is also an integer, the admissible values become 2, 3, 4, 5 and 6.
The continuous interval and discrete domain combine through intersection.
This is why every active condition matters.
7. Simultaneous equations define a feasible intersection
The equations x + y = 10 and x − y = 2 each describe a line.
The solution is the point satisfying both lines simultaneously.
Adding gives 2x = 12, so x = 6 and y = 4.
Geometrically, the feasible set is the intersection point (6, 4).
If the lines were parallel and distinct, the feasible set would be empty.
8. Inconsistent constraints produce no feasible solution
Suppose x + y = 8 and x + y = 10 are both required.
No pair can satisfy both.
The correct conclusion is not to average them or choose the more convenient equation. The feasible set is empty.
This connects to the earlier chapter on Data Sufficiency, Unique Answers and Counterexamples: a problem can be underdetermined, uniquely determined, multiply determined or inconsistent.
9. Too few constraints can leave many feasible solutions
If x + y = 10 is the only condition over real numbers, infinitely many pairs are feasible.
(1, 9), (4, 6), (10, 0) and many others satisfy it.
If the problem asks for one unique pair, more information is needed.
A feasible set can be large without being wrong. Uniqueness is a separate property.
10. Geometry imposes feasibility conditions beyond equations
Three positive lengths form a non-degenerate triangle only if the largest side is less than the sum of the other two.
Lengths 3, 4 and 8 cannot form such a triangle because 8 ≥ 3 + 4.
Even if a separate algebraic equation involving those lengths is satisfied, the geometric configuration remains infeasible.
A valid geometry answer must satisfy both algebraic and spatial constraints.
11. Angle constraints filter trigonometric candidates
If sin θ = 1/2 and θ is an acute angle, the only admissible solution is 30°.
If the domain is 0° ≤ θ ≤ 360°, then 30° and 150° are both admissible.
The trigonometric equation generates candidates; the angle-domain constraint determines which survive.
The same equation can therefore have different feasible solution sets under different domains.
12. Coordinate geometry can combine line, circle and region constraints
Suppose a point P lies on the circle x² + y² = 25 and in the first quadrant.
The circle alone permits infinitely many points.
The first-quadrant condition adds x > 0 and y > 0.
If P also lies on y = x, then 2x² = 25 and x = ±5/√2.
The first-quadrant constraint selects x = 5/√2 and y = 5/√2.
13. Physical interpretation often adds sign constraints
An equation for a length may yield x = −3 or x = 7.
If x represents a physical length in an ordinary school geometry model, the negative candidate is inadmissible.
But do not turn “negative means impossible” into a universal rule. Negative values can be perfectly meaningful for coordinates, temperature changes, displacement, debt models or signed rates.
Interpretation determines the constraint.
14. Counts require integer feasibility
Suppose 100 pupils are divided into groups of size x, with x required to be a positive integer divisor of 100.
The equation 100/x = number of groups can be evaluated for any positive real x, but only certain x-values describe complete equal groups.
Feasible group sizes include 1, 2, 4, 5, 10, 20, 25, 50 and 100.
A decimal such as x = 7.5 may be algebraically usable in a ratio but not admissible as a whole-pupil group size.
15. Capacity decisions use ceiling-type feasibility
If a bus holds at most 40 passengers and 93 passengers must all travel, 93/40 = 2.325.
The feasible bus count must be a whole number and must provide capacity at least 93.
Two buses provide only 80 places. Three provide 120.
Therefore the minimum feasible count is 3.
Ordinary nearest-integer rounding is not the governing rule. Feasibility is.
16. Complete-group questions use a different feasible target
With 93 pupils and groups of exactly 40, the number of complete groups is 2 with 13 pupils left over.
The arithmetic quotient is the same as in the bus problem, but the target constraint differs.
One problem requires covering all pupils. The other counts only fully populated groups.
Whole-number decisions must follow the problem’s feasibility definition.
17. Budget constraints create upper feasible limits
Suppose each item costs $7 and the budget is at most $50.
If n is the whole-number item count, 7n ≤ 50.
Thus n ≤ 50/7 ≈ 7.14.
The largest feasible integer is 7.
Eight items violate the budget even though 8 is numerically close to the quotient.
18. Time constraints can remove otherwise valid routes
Suppose a journey model requires arrival within 90 minutes.
Several speed choices may produce valid travel times mathematically, but only those at or below 90 minutes belong to the feasible set.
Operational constraints can therefore sit on top of algebraic relationships.
In real settings, additional safety and legal constraints would also matter; a school model should not be mistaken for complete operational advice.
19. Optimisation means choosing the best feasible candidate
An objective function does not operate alone.
Suppose profit P = 8x + 5y, subject to x ≥ 0, y ≥ 0 and x + y ≤ 10.
The objective asks what to maximise. The inequalities define which points are permitted.
A point with huge objective value but outside the feasible region is irrelevant to the constrained problem.
Optimisation means best among admissible choices.
20. Linear constraints create feasible regions
For x ≥ 0, y ≥ 0 and x + y ≤ 10, the feasible region is the triangle bounded by the coordinate axes and line x + y = 10.
Every point inside or on the relevant boundaries satisfies all three inequalities.
A point such as (7, 5) satisfies non-negativity but violates x + y ≤ 10.
Checking constraints one at a time reveals which boundary excludes the point.
21. Boundary points can be feasible or infeasible depending on inequality signs
x + y ≤ 10 includes the line x + y = 10.
x + y < 10 excludes it.
The feasible region changes only on the boundary, but that boundary can contain the optimum.
This is why strict and inclusive inequalities matter in constrained problems.
22. A maximum may fail to exist on an open feasible set
Suppose 0 < x < 1 and the objective is to maximise x.
Values can get arbitrarily close to 1, but 1 is not feasible.
There is no maximum value.
The supremum is 1, but it is not attained.
Feasibility of the boundary determines whether an optimum is actually achieved.
23. Parameter changes can change the feasible set itself
Consider x + y = 10, with x and y positive integers, and x − y = k.
If k = 2, x = 6 and y = 4, feasible.
If k = 3, the real solution is x = 6.5 and y = 3.5, which violates the integer constraints.
The equations remain solvable over the reals, but the integer feasible set becomes empty.
This connects to Parameter Sensitivity and Stability.
24. A candidate can satisfy the transformed problem but fail the original one
Squaring, multiplying by denominators, cancelling factors and applying inverse functions can alter visible constraints.
For this reason, the original problem should remain the final feasibility test.
A candidate is admissible only if it satisfies the original equation, original domain and original contextual restrictions.
This is the intersection point between constraint reasoning and Reversible Steps and Extraneous Solutions.
25. Approximate data can create uncertain feasibility
Suppose an item must be shorter than 10 cm, and its length is reported as 9.9 cm to the nearest 0.2 cm under a stated rounding model.
The true interval may cross the 10 cm threshold depending on the exact reporting convention.
In that case, the central recorded value is feasible while some compatible exact values may not be.
The correct conclusion is not automatically “fits”. It may be “not guaranteed”.
The Rounded Data chapter develops this fully.
26. Constraints should be classified by source
Useful categories include:
Domain constraints — such as positive integer or real interval.
Definition constraints — such as non-zero denominator or positive logarithm argument.
Relationship constraints — equations, inequalities or geometric conditions.
Context constraints — budget, capacity, time, count or physical interpretation.
Decision constraints — such as “minimum whole number that guarantees enough”.
Classifying them helps prevent one condition from silently disappearing.
27. Redundant constraints need not change the feasible set
Suppose x ≥ 5 and x ≥ 2 are both stated.
The second condition adds nothing because every x satisfying x ≥ 5 already satisfies x ≥ 2.
The feasible set remains x ≥ 5.
A redundant constraint can still be true and useful for checking, but it does not narrow the set further.
28. Contradictory constraints empty the feasible set
x ≥ 5 and x < 3 cannot both hold.
No algebraic manipulation should force a value through that contradiction.
Recognising an empty feasible set early can prevent wasted calculation.
In word problems, contradiction may indicate impossible stated conditions or a modelling error.
29. Feasibility comes before optimality
A candidate cannot be “best” if it is not allowed.
In a geometry optimisation, a side length that maximises a formula but makes another side negative is infeasible.
In an integer problem, a continuous optimum at x = 4.7 is not an admissible decision if x must be whole.
In a budget problem, the highest-revenue plan is irrelevant if it exceeds the budget.
The correct order is: define feasible choices, then compare them.
30. Feasibility checks are powerful error diagnostics
If a probability exceeds 1, the candidate violates a universal feasibility condition.
If a length is negative, inspect the interpretation and equation setup.
If a point claimed to lie on a circle fails the circle equation, the coordinate pair is infeasible.
If a timetable solution uses a time outside the stated interval, the branch is invalid.
These checks can locate errors faster than repeating the entire method.
31. A feasible set can be represented in several ways
It may be a list of integers, an interval, a shaded region, a set of coordinate points, a collection of graph branches or a family described by a parameter.
Choosing the right representation can make the constraints visible.
For two-variable inequalities, a graph may be clearer than prose.
For a small integer set, a list may be clearer than interval notation.
For an algebraic family, a parameter form may be best.
32. A practical feasibility audit
Before accepting a final candidate, ask:
Does it satisfy the original equation? Is it inside the domain? Does it satisfy every inequality? Does it respect whole-number or sign conditions? Does it fit the geometry? Does it obey the context? If the task is optimisation, is it actually feasible before I compare objective values?
This audit should become automatic in problems where constraints are dense.
33. Independent practice
1. Solve x² = 36 subject to x > 0.
2. Solve √(x + 5) = x and identify the feasible solution.
3. List the integer x satisfying 1 ≤ x < 6.
4. Determine whether x + y = 4 and x + y = 7 have any common solution.
5. Can side lengths 5, 6 and 12 form a non-degenerate triangle?
6. Solve sin θ = 1/2 for acute θ.
7. A bus holds 35 passengers. Find the minimum feasible number of buses for 91 passengers.
8. An item costs $9 and the budget is $70. Find the largest feasible whole-number item count.
9. Describe the feasible region for x ≥ 0, y ≥ 0 and x + y ≤ 6.
10. Explain why a continuous optimum of x = 7.6 is not automatically the answer when x must be an integer.
11. Give an example of a redundant pair of inequalities.
12. Give an example of contradictory inequalities.
34. Worked answers
1. x = ±6 from the equation, then x > 0 selects x = 6.
2. The right side requires x ≥ 0. Squaring gives x + 5 = x², so x² − x − 5 = 0. The roots are (1 ± √21)/2. Only (1 + √21)/2 is non-negative and can satisfy the original equation. Direct substitution confirms it.
3. 1, 2, 3, 4, 5.
4. No. A single pair cannot have sum both 4 and 7. The feasible set is empty.
5. No. The largest side 12 is greater than 5 + 6 = 11, so the triangle inequality fails.
6. θ = 30°.
7. 91/35 = 2.6. Two buses hold only 70, so 3 buses are required.
8. 9n ≤ 70 gives n ≤ 7.77…, so the largest feasible whole-number count is 7.
9. The triangular region in the first quadrant on or below the line x + y = 6, including the axes and boundary line.
10. 7.6 is not admissible. Evaluate nearby feasible integers, commonly 7 and 8, together with the original objective and constraints.
11. x ≥ 5 and x ≥ 2. The second adds no restriction beyond the first.
12. x ≥ 5 and x < 3.
35. Continue through the School Mathematics manual
Use Necessary and Sufficient Conditions to distinguish filters from decisive tests, Notation, Scope and Quantifiers to preserve each restriction, and Local and Global Conclusions to decide whether a feasible conclusion applies at one point or across the complete domain.
Return to the BTT Mathematics Hub for Batch 04.