Singapore School Mathematics Operating Manual · Chapter 5
Many school solutions look like a chain of equal signs or implications. Yet not every step in that chain has the same logical strength. Some operations preserve exactly the same set of solutions in both directions. Others create extra possibilities, delete valid possibilities, or require conditions that must be checked separately.
This chapter develops one of the most important pieces of mathematical control: knowing whether a transformation is reversible. A reversible step allows us to move from one statement to another and back again without changing the solution set. A one-way step may still be useful, but its output must be checked against the original problem.
The examples range from Primary-style inverse operations to Secondary algebra and later exponential, logarithmic and calculus contexts. The deeper principle is the same throughout: when a transformation changes the information carried by the problem, the learner must account for what was lost or introduced.
Equivalence · Squaring and roots · Division and cancellation · Domains and restrictions · Practice · Worked answers
1. An equation is not merely a calculation
Consider 3x + 5 = 20. Subtracting 5 from both sides gives 3x = 15. Dividing both sides by 3 gives x = 5. Each move is reversible because adding 5 and multiplying by 3 recover the previous equation. The equations have exactly the same real solution set.
This is stronger than saying that the arithmetic is correct. It says the transformed statement is logically equivalent to the previous one under the stated domain. The solution x = 5 works in every equation in the chain, and no other real x has been added or removed.
A useful symbol distinction is between “=” inside algebraic expressions and “if and only if” between equivalent statements. Students do not need to write formal logic on every line, but they should know what the chain is claiming. A sequence of transformations should not silently change the set being solved.
For simple linear equations, standard balance operations are usually reversible when their inverse is always defined. Add the same quantity to both sides; subtract the same quantity; multiply by the same non-zero constant; divide by the same non-zero constant. These operations preserve equivalence.
The complication begins when the operation itself is not one-to-one or when its inverse is not always defined.
2. Inverse operations in Primary Mathematics already teach reversibility
Suppose a child thinks of a number, multiplies it by 4, then adds 7 to obtain 31. The forward path is “multiply by 4, add 7”. The reverse path is “subtract 7, divide by 4”. This works because each operation has a well-defined inverse in the given number system.
31 − 7 = 24, and 24 ÷ 4 = 6. Checking forward gives 6 × 4 + 7 = 31. The reverse method is not a trick. It works because the two operations can be undone in reverse order.
Now compare “square a number and get 25”. Squaring does not have a single-valued inverse over the real numbers in the same sense. Both 5 and −5 square to 25. If the original problem says the number is positive, the domain selects 5. Without that restriction, reversing the square requires two branches.
This is why “do the opposite operation” is a useful beginner idea but not a universal algebra law. Mature school Mathematics asks a second question: does the proposed inverse recover every original possibility?
3. Squaring can introduce extraneous solutions
Solve √(x + 1) = x − 1 over the real numbers. Before squaring, note that the square-root side is non-negative, so x − 1 must also be non-negative. Hence x ≥ 1.
Squaring gives x + 1 = (x − 1)² = x² − 2x + 1. Rearranging gives x² − 3x = 0, so x(x − 3) = 0. The squared equation has candidate solutions x = 0 and x = 3.
But x = 0 violates the earlier requirement x ≥ 1. Direct substitution into the original equation also shows the failure: √1 = 1, while x − 1 = −1. Thus x = 0 is an extraneous solution created by the squaring step.
The value x = 3 works: √4 = 2 and 3 − 1 = 2. Therefore the original equation has the single real solution x = 3.
Why did squaring create an extra solution? Because a = b implies a² = b², but the reverse statement is weaker: a² = b² implies a = b or a = −b. Squaring forgets the sign relationship between the two sides.
The repair is not “never square”. Squaring is often exactly the right operation. The repair is to treat the resulting values as candidates and check them against the original equation and its domain.
4. Taking square roots can delete a branch if done carelessly
Solve x² = 49. A common incorrect step is “take square root: x = 7”. This loses the valid solution x = −7.
The correct reasoning is x² = 49, so x = 7 or x = −7. Equivalently, x² − 49 = 0 gives (x − 7)(x + 7) = 0.
It is important to distinguish this from evaluating √49. The symbol √49 denotes the non-negative square root, which is 7. The equation x² = 49 asks for every real x whose square is 49. Those are different mathematical objects.
The positive-square-root notation is single-valued; the inverse relation of squaring over the reals has two branches except at zero. Confusing the two produces systematic missing solutions.
5. Absolute value creates another two-branch inverse
Solve |2x − 1| = 7. Absolute value discards sign, so the expression inside can be 7 or −7.
Case 1: 2x − 1 = 7, giving x = 4. Case 2: 2x − 1 = −7, giving x = −3. Both satisfy the original equation.
Writing only 2x − 1 = 7 assumes the inside is positive without justification. Squaring both sides would also lead to the same pair if followed carefully, but direct case splitting exposes the structure more clearly.
The deeper idea is that a many-to-one transformation cannot generally be reversed by one path. Absolute value maps 7 and −7 to the same output. Recovering the input therefore requires multiple possible branches.
6. Dividing by an expression can delete a solution
Solve x(x − 4) = 0. One tempting move is to divide both sides by x and obtain x − 4 = 0, hence x = 4. But the original equation also has x = 0.
The problem is that dividing by x assumes x ≠ 0. If x = 0, the division is not defined. The step has silently excluded one of the original cases.
The correct zero-product reasoning is x = 0 or x − 4 = 0, giving x = 0 or x = 4.
More generally, cancellation by a variable expression is safe only after the possibility that the expression equals zero has been handled. A factor can be cancelled from a fraction under a non-zero condition; it cannot simply be erased from an equation when that factor itself may create a solution.
7. Cancelling factors in fractions requires domain tracking
Consider the rational expression (x² − 9)/(x − 3). Factorising gives (x − 3)(x + 3)/(x − 3). For x ≠ 3, this simplifies to x + 3.
But the original expression is undefined at x = 3. The simplified formula x + 3 is defined there. Thus the two formulas are equal on the domain x ≠ 3, but they are not identical as unrestricted functions.
This distinction matters in graphing and solving. The graph of the original rational function is the line y = x + 3 with a hole at x = 3. If a student simplifies and forgets the restriction, that missing point disappears from the model.
Cancellation preserves values where the cancelled factor is non-zero. It does not repair the original denominator at the excluded value.
8. Multiplying by an expression can also introduce hidden conditions
Solve 1/(x − 2) = 3. The original domain excludes x = 2. Multiplying by x − 2 gives 1 = 3(x − 2), so 1 = 3x − 6 and x = 7/3.
The candidate 7/3 is not excluded and checks in the original equation. This multiplication is safe because the original equation already requires x − 2 ≠ 0. The domain condition was present before the transformation.
Now imagine starting from an equation such as x/(x − 2) = 0. Multiplying by x − 2 gives x = 0, which is valid. But the transformed equation by itself no longer displays the exclusion x ≠ 2. The restriction must be carried forward separately.
Cross-multiplication is therefore not a magic pattern. It is multiplication by denominators under non-zero conditions. When those conditions matter, they belong in the solution.
9. Logarithms are reversible only inside their domain
Solve log₂(x − 1) = 3. The logarithm requires x − 1 > 0, so x > 1. Converting to exponential form gives x − 1 = 2³ = 8, hence x = 9. The value lies inside the domain.
The transformation log₂(A) = 3 to A = 8 is reversible when A > 0. The positivity condition is not an extra optional check; it is part of what it means for the logarithm to exist over the reals.
Consider log₂(x − 1) + log₂(x − 3) = 3. The domain requires x > 3. Combining logs gives log₂[(x − 1)(x − 3)] = 3, so (x − 1)(x − 3) = 8. Expanding gives x² − 4x − 5 = 0, with algebraic roots x = 5 and x = −1.
Only x = 5 lies in the domain x > 3. The root −1 belongs to the polynomial equation created after the logarithms were combined and exponentiated, but not to the original logarithmic equation.
10. Exponentials preserve order differently depending on the base
For base b > 1, the function bˣ is strictly increasing. Therefore bᵃ = bᵇ implies a = b, and bᵃ < bᵇ implies a < b. These steps are reversible because the exponential function is one-to-one on the real numbers.
For 0 < b < 1, equality remains one-to-one, but inequalities reverse direction because the function is strictly decreasing. For example, (1/2)ˣ < (1/2)³ implies x > 3.
This is another example of why an algebraic-looking manipulation needs structural knowledge. The same symbolic base pattern can preserve equality while changing the direction of an inequality depending on the function’s monotonicity.
11. Multiplying an inequality by an unknown-sign expression is dangerous
Suppose we want to solve (x − 1)/(x + 2) > 0. Multiplying both sides by x + 2 without knowing its sign can produce the wrong inequality direction.
The correct approach is to locate the critical values x = 1 and x = −2, then analyse signs across the intervals. For x < −2, numerator and denominator are both negative, so the quotient is positive. For −2 < x < 1, the signs differ, so the quotient is negative. For x > 1, both are positive.
Thus the solution is x < −2 or x > 1, with x = −2 excluded because the expression is undefined and x = 1 excluded because the inequality is strict.
Alternatively, one can multiply by (x + 2)², which is positive wherever the original expression is defined. That preserves inequality direction while retaining the condition x ≠ −2. Choosing a transformation whose sign is controlled is a form of mathematical engineering.
12. Trigonometric equations also require complete inverse branches
Solve sin θ = 1/2 for 0° ≤ θ ≤ 360°. A calculator inverse sine gives 30°, but this is not the complete solution set. Sine is also 1/2 in the second quadrant, giving 150°.
The inverse-sine function returns a principal value. Solving the equation requires every angle in the specified domain that produces the target sine value.
The same distinction appears for cosine and tangent. A calculator’s inverse function is a way to recover one representative angle. The periodic trigonometric function has many inputs with the same output, so the full equation requires the appropriate symmetry and periodicity.
Again, a many-to-one function cannot be reversed by a single branch unless the domain has first been restricted enough to make the function one-to-one.
13. Differentiation loses information; integration restores a family
If f(x) = x² + 5, then f′(x) = 2x. If g(x) = x² − 17, then g′(x) = 2x as well. Differentiation has removed the constant difference between the functions.
Therefore, if f′(x) = 2x, we cannot conclude f(x) = x². The general antiderivative is f(x) = x² + C, where C is an arbitrary constant.
An initial condition can recover the missing information. If f(3) = 14, then 9 + C = 14 and C = 5.
This is a sophisticated version of the same operating-manual principle. The derivative transformation is not one-to-one over all differentiable functions because every vertical shift has the same derivative. Integration produces a family until an additional condition selects one member.
14. Equivalent expressions are not always equivalent solution steps
The identity (x − 2)(x + 3) = x² + x − 6 holds for every real x. Replacing one side with the other inside an equation preserves the equation exactly because the expressions are equal everywhere on the domain.
By contrast, replacing x with √(x²) is not valid for all real x, because √(x²) = |x|, not x. The two expressions agree only when x ≥ 0.
A transformation can therefore fail even when every symbol looks familiar. The learner must know the identity being used and the conditions under which it is true.
This is why formula recall should include domain and structure, not only surface shape.
15. A compact audit for every non-obvious transformation
Before accepting a transformed equation, ask four questions. Was the operation defined for every current candidate? Is the operation one-to-one on the relevant domain? Could the step have introduced extra candidates? Could it have removed a case?
If the answer to either of the last two questions is yes, mark the result as a candidate set rather than the final solution set. Then substitute back into the original statement or reapply the original restrictions.
This audit is especially useful after squaring, taking roots, cancelling variable factors, multiplying inequalities by variable expressions, using inverse trigonometric functions, combining logarithms, and integrating after differentiation.
The aim is not to make every line slower. It is to identify the small number of transformations where logical control matters most.
16. Independent practice
1. Solve x² = 16 over the real numbers.
2. Solve √(x + 6) = x.
3. Solve |3x + 2| = 8.
4. Solve x(x − 5) = 0 without losing a solution.
5. Simplify (x² − 4)/(x − 2), stating the domain restriction.
6. Solve 2/(x − 1) = 1 and state the excluded value.
7. Solve log₃(x − 2) = 2.
8. Solve log₂(x) + log₂(x − 2) = 3.
9. Solve (x − 4)/(x + 1) ≥ 0.
10. Solve cos θ = 1/2 for 0° ≤ θ ≤ 360°.
11. If f′(x) = 6x and f(2) = 17, find f(x).
12. Explain why replacing x by √(x²) can change an expression when x is negative.
17. Worked answers
1. x = 4 or x = −4. Both square to 16.
2. The square-root side is non-negative, so x ≥ 0. Squaring gives x + 6 = x², hence x² − x − 6 = 0 and (x − 3)(x + 2) = 0. Candidates are 3 and −2. The domain excludes −2. Check x = 3: √9 = 3. Therefore x = 3.
3. 3x + 2 = 8 or 3x + 2 = −8. Thus x = 2 or x = −10/3.
4. Zero-product reasoning gives x = 0 or x − 5 = 0. Hence x = 0 or 5. Dividing by x would incorrectly delete the zero solution.
5. Factor to obtain (x − 2)(x + 2)/(x − 2) = x + 2, valid for x ≠ 2. The simplified expression carries the original exclusion.
6. The original domain excludes x = 1. Multiplying by x − 1 gives 2 = x − 1, so x = 3, which is valid.
7. Domain x > 2. Exponential form gives x − 2 = 9, hence x = 11.
8. Domain x > 2. Combine logs: log₂[x(x − 2)] = 3, so x² − 2x = 8. Rearranging gives x² − 2x − 8 = 0, hence (x − 4)(x + 2) = 0. Candidates are 4 and −2. Only x = 4 lies in the domain.
9. Critical values are x = 4 and x = −1. The quotient is positive on x < −1 and x > 4, zero at x = 4, and undefined at x = −1. Therefore x < −1 or x ≥ 4.
10. θ = 60° or 300°. The inverse cosine principal value alone is incomplete.
11. Integrating gives f(x) = 3x² + C. Since f(2) = 12 + C = 17, C = 5. Therefore f(x) = 3x² + 5.
12. √(x²) means the non-negative number whose square is x², so it equals |x|. If x = −4, then √(x²) = √16 = 4, not −4. The replacement is valid only when x ≥ 0.
18. Continue through the operating manual
Use Case Splitting and Piecewise Reasoning when one mathematical situation naturally breaks into branches. Use Boundary Cases, Endpoints and Extremes when validity changes at equality, zero or an endpoint. Use Discrete and Continuous Models when the type of quantity changes the allowed answers.
Return to the BTT Mathematics Hub for the full Batch 02 route.
When Algebra Changes the Solution Set: Dividing, Squaring and Domain Restrictions
An algebraic line can look cleaner than the line before it and still be logically weaker. That is the hidden difficulty behind extraneous roots, lost solutions and forgotten domain restrictions. The visible work is symbol manipulation; the deeper work is controlling which values are still allowed after each transformation.
This section turns that control into a repeatable decision system. The question is not merely, “Can I perform this operation?” It is, “What does this operation do to the solution set?”
1. Equality of expressions is not the same as equivalence of equations
If two expressions are equal for every value in their common domain, one can replace the other without changing meaning. For example, 2(x+3) and 2x+6 are equal for every real number x. Expanding or factorising between those forms is reversible.
An equation transformation needs a stronger audit. Starting from A=B, an operation may preserve exactly the same solutions, preserve only some solutions, or create additional candidates. The symbols alone do not announce which of these has happened.
Use three labels:
- Equivalent step: the old and new statements have exactly the same solution set.
- Forward-valid step: every old solution satisfies the new statement, but the new statement may have extra candidates.
- Case-restricting step: the operation is valid only after a condition has been declared; without the condition, solutions may be lost.
2. The arrow test
Write the logic as arrows before trusting the algebra. If statement S implies statement T, write S ⇒ T. If both implications hold, write S ⇔ T. A chain of equations in a school solution often visually suggests ⇔ even when the mathematics only justifies ⇒.
For example:
√(x+6)=x ⇒ x+6=x².
Squaring is forward-valid: every solution of the original equation satisfies the squared equation. The converse is not automatic. The squared equation may admit a value that does not satisfy the original square-root equation.
By contrast:
3x+5=20 ⇔ 3x=15 ⇔ x=5.
Subtracting five and dividing by the non-zero constant three are reversible. The arrows can safely point both ways.
3. Division by a variable expression is a case split disguised as a shortcut
Consider x(x-4)=0. A learner sees the factor x on the left and divides by x:
x-4=0, so x=4.
The solution x=0 has disappeared. Division by x was valid only under the condition x≠0. The correct logical structure is:
- Case 1: x=0, which satisfies the original equation.
- Case 2: x≠0, so division by x is allowed and gives x=4.
Whenever the divisor contains the unknown, ask whether it can be zero. If yes, either preserve the zero case before dividing or use a method such as zero-product reasoning that does not discard it.
4. A cancellation can carry an excluded value
Take
(x²-9)/(x-3)=0.
Factor the numerator:
(x-3)(x+3)/(x-3)=0.
It is tempting to cancel x-3 and solve x+3=0. That does produce the valid solution x=-3. But the simplified expression must retain the original restriction x≠3. At x=3, the original expression is undefined even though the simplified formula x+3 has a numerical value.
The cancellation changes the visible formula; it does not rewrite history. The original domain still governs the problem.
5. Multiplying by a variable expression can also change what must be checked
Solve
2/(x-1)=x, with x≠1.
Multiplying by x-1 gives:
2=x(x-1), hence x²-x-2=0, so (x-2)(x+1)=0.
The candidates are x=2 and x=-1. Both lie in the original domain and both satisfy the original equation, so both survive.
In this example multiplication did not introduce an extraneous root. The important point is that the original excluded value remains excluded and every candidate is checked against the original statement.
6. Squaring forgets sign
The map t → t² sends both 3 and -3 to 9. It is many-to-one on the real numbers. That is why squaring can create extra candidates: two different inputs can become indistinguishable after the transformation.
Consider:
√(x+2)=x-2.
The left side is non-negative, so the original equation already requires x-2≥0, or x≥2. Squaring gives
x+2=(x-2)²=x²-4x+4,
so
x²-5x+2=0.
The quadratic formula produces two real candidates. But only candidates with x≥2 are even eligible for the original equation, and every surviving candidate must still be substituted into the unsquared equation. The domain check is not decoration at the end; it is part of the logic from the start.
7. Squaring both sides can be safe on a restricted domain
Squaring is one-to-one on the non-negative reals. If both sides are already known to be non-negative, squaring can be reversible. For example, if a≥0 and b≥0, then a=b if and only if a²=b².
This is why domain information can turn a dangerous operation into a controlled one. Mathematics becomes safer when the learner states the relevant domain before performing the transformation.
8. Taking a square root can lose the negative branch
From x²=25, writing x=5 loses a solution. The correct conclusion is x=±5.
The principal square-root symbol √25 denotes the non-negative square root, which is 5. Solving x²=25 is a different job: find every real number whose square is 25. The equation therefore has two solutions.
This is a useful contrast: squaring can introduce candidates; taking a principal square root as though it were a two-sided inverse can delete candidates.
9. Absolute value exposes the hidden branch explicitly
If |2x-1|=7, the expression inside the absolute value may equal 7 or -7. The two branches are:
2x-1=7 or 2x-1=-7, giving x=4 or x=-3.
A learner who simply “removes the absolute value signs” keeps only one branch. The error is not arithmetic. It is loss of a case.
10. Logarithms carry domain conditions before any manipulation
Solve:
log₂(x-1)+log₂(x-3)=3.
The original logarithms require x-1>0 and x-3>0, so x>3. Combining gives
log₂[(x-1)(x-3)]=3,
hence
(x-1)(x-3)=8.
Expanding gives x²-4x-5=0, so x=5 or x=-1. The algebraic equation has two roots; the original logarithmic equation admits only x=5. The domain restriction performs the final selection.
11. Rational inequalities cannot be handled like ordinary equations
Consider
(x-2)/(x+1) ≥ 0.
Multiplying both sides by x+1 without knowing its sign is unsafe because the inequality direction depends on whether the multiplier is positive or negative. The safer route is to identify critical values x=2 and x=-1, partition the number line, and determine the sign of the quotient on each interval.
The endpoint x=2 is included because it makes the quotient zero. The value x=-1 is excluded because the expression is undefined. The solution is x<-1 or x≥2.
12. Solving after substitution: preserve the substitution domain
Suppose a substitution u=x² is used in a biquadratic equation. If solving the transformed quadratic gives u=-4, that value may be algebraically valid for the quadratic in u but impossible under u=x² over the reals. The substitution introduced a new variable with its own inherited condition u≥0.
Every substitution carries a return ticket. After solving in the new variable, map the candidates back and reapply the conditions that came with the substitution.
13. Inverse trigonometric functions return a principal value, not the whole equation
If sin θ=1/2 on 0°≤θ≤360°, a calculator may return 30°. That is one angle, not the full solution set. The sine function is not one-to-one across the whole interval. The second solution is 150°.
The inverse key solves a restricted inverse problem. The trigonometric equation asks for every input in the stated domain. The learner must restore the branches that the inverse function does not display.
14. The domain ledger
For equations with several restrictions, keep a short ledger beside the working. It prevents conditions from disappearing during simplification.
| Source of restriction | Condition to record |
|---|---|
| denominator g(x) | g(x)≠0 |
| even root √g(x) | g(x)≥0 |
| logarithm log g(x) | g(x)>0 |
| inverse trigonometric branch | declared output interval and original angle domain |
| context | length, count, time, probability or other physical restrictions |
| substitution | conditions inherited from the substituted expression |
The ledger is especially useful in long problems where the algebra later looks ordinary and the original restriction is easy to forget.
15. A worked decision case: divide or factor?
Solve x²-7x=0.
Two routes appear:
- Divide by x, giving x-7=0.
- Factor x(x-7)=0 and use the zero-product property.
The second route is safer because it preserves both branches automatically. The first route is valid only in the case x≠0; it needs a separate x=0 case before division. The method decision is therefore not about which route is shorter on paper. It is about which route preserves the solution set with the least logical overhead.
16. A worked decision case: square now or isolate first?
Solve √(2x+3)+1=x.
Squaring immediately produces a more complicated expression and makes sign conditions harder to see. Isolate the root first:
√(2x+3)=x-1.
Now the right side must be non-negative, so x≥1. Squaring gives 2x+3=(x-1)². The later candidates can be filtered using x≥1 and checked in the original equation.
The method choice exposes the condition before the non-reversible operation. That makes the later audit easier.
17. A worked decision case: cross-multiply or keep the denominator visible?
Solve (x+2)/(x-4)=3.
First record x≠4. Cross-multiplication is shorthand for multiplying both sides by the non-zero denominator on the declared domain:
x+2=3(x-4).
Then x+2=3x-12, so 14=2x and x=7. The candidate satisfies the restriction and the original equation.
The important step is not the cross-multiplication pattern. It is the declaration that the denominator cannot be zero.
18. A worked decision case: is an observed root enough?
Suppose a graph appears to cross the axis near x=2. That gives a candidate. It does not prove there are no other roots outside the visible window, no repeated root that only touches the axis, and no numerical display artefact. Return to the equation, factor where possible, use exact reasoning, or apply the appropriate theorem and domain analysis.
This connects solution-set control to the Functions and Graphs owner: a graph can suggest where to look, but algebra and domain reasoning establish what the solution set actually is.
19. The first invalid step protocol
When reviewing a worked solution, do not correct the final answer first. Find the earliest line where the new statement is no longer equivalent to or safely implied by the previous one.
- Write the domain of the original problem.
- Compare each pair of consecutive lines.
- Name the operation used.
- Ask whether it is reversible on the current domain.
- If not, label what may have been introduced or lost.
- Repair from that line, not from the end.
- Check final candidates in the original statement.
20. Five planted transformations
Case A. From x(x-6)=0, a solution divides by x and reports x=6. The first invalid move is unqualified division by a factor that may be zero. Repair: split the zero case or use zero-product reasoning. Final solutions: x=0,6.
Case B. From √(x+5)=x-1, a solution squares and accepts both quadratic roots without checking. The squaring step is forward-valid, not automatically equivalent. Repair: apply the non-negative condition and substitute candidates into the original.
Case C. From (x²-1)/(x-1)=3, a solution cancels x-1 and later accepts x=1. The simplification itself is valid only on the original domain x≠1. Repair: retain the exclusion.
Case D. From x²=9, a solution writes x=√9=3. The first invalid step treats the principal square root as the complete inverse of squaring. Repair: x=±3.
Case E. From 1/(x-2)>0, a solution multiplies by x-2 and writes 1>0, concluding all x≠2 work. The sign of x-2 was not controlled. Repair: analyse intervals; the solution is x>2.
21. Fresh transfer set
For each problem, identify the domain first, mark every non-reversible or condition-dependent transformation, solve, and check the final set against the original statement.
- x(x+3)=0.
- √(x+10)=x.
- (x²-16)/(x-4)=5.
- log₃(x-1)+log₃(x-4)=2.
- |2x+5|=9.
- (x+1)/(x-2)≤0.
- sin θ=√3/2 for 0°≤θ≤360°.
- Let u=x² in x⁴-5x²+4=0; restore every real x.
22. Fresh transfer answers and checks
1. x=0 or x=-3. Do not divide by x before preserving the zero case.
2. The original requires x≥0. Squaring gives x+10=x², so x²-x-10=0. Any real candidates must satisfy both the non-negative condition and the unsquared equation.
3. Original domain x≠4. Factor and cancel only with that restriction retained; solve the simplified equation, then reject any excluded value.
4. Domain requires x>4. Combine the logs, solve the resulting quadratic, and keep only roots in the domain.
5. Split into 2x+5=9 or 2x+5=-9.
6. Critical values are -1 and 2; 2 is excluded. Use a sign chart rather than uncontrolled multiplication.
7. The solutions are 60° and 120°.
8. The transformed equation is u²-5u+4=0, so u=1 or u=4. Because u=x², the real solutions are x=±1,±2.
23. Tutor feedback: name the logical error, not only the algebraic symptom
Feedback such as “check your working” is too broad when the real problem is transformation control. Use specific language:
- You divided by an expression before protecting its zero case.
- You squared both sides and treated candidates as confirmed solutions.
- You simplified the formula but dropped the original excluded value.
- You used a principal inverse value as though it represented every solution in the domain.
- You multiplied an inequality by an expression whose sign was unknown.
- You solved the transformed problem but did not return through the substitution.
24. A compact release test
The learner is ready to move on when they can do more than obtain correct roots. On an unfamiliar problem they should be able to:
- state the original domain;
- distinguish reversible and one-way steps;
- explain why squaring may introduce candidates;
- explain why division by a variable factor may lose a case;
- preserve exclusions through cancellation;
- restore branches after inverse or substitution methods;
- check every candidate against the original statement;
- solve a changed problem without being told which danger is present.
25. Continue through BTT
Use Equations, Balance and Checking for the broader secondary-equation route. Use Equivalence, Canonical Forms and Choosing Useful Representations when the question is whether two forms express the same object or relationship. Use How to Check Maths Answers for independent verification and planted-error checking.
The durable rule is simple: every algebraic transformation has a logical cost. Strong Mathematics keeps the receipt.
Domain Repair Clinic: Five More Cases Where the Algebra Is Locally Correct but the Answer Set Is Wrong
The most difficult transformation errors are often not calculation errors. Every visible manipulation may be legal inside a restricted case, yet the final answer can still be incomplete because the restriction was never restored. These five cases train the habit of carrying domain information alongside the algebra.
Case 1 — division hides the zero branch
Solve x(x+5)=2x. A tempting route is to divide both sides by x, giving x+5=2 and x=-3. The result is valid for the case x≠0, but the division has silently excluded x=0.
A safer route is to collect terms:
x(x+5)-2x=0
x(x+3)=0.
Therefore x=0 or x=-3. If division is used, the solution must explicitly split into the cases x=0 and x≠0.
Transfer question: when does division simplify a problem without losing information? Answer: when the divisor is known to be non-zero on the active domain, or when the zero case has already been handled separately.
Case 2 — an even power erases sign information twice
Solve (x-1)²=16. A learner writes x-1=4 and obtains x=5. The missing solution is x=-3.
The equation says that the square of x-1 is 16. Both 4 and -4 have square 16, so:
x-1=4 or x-1=-4.
This is not an “extra ± rule” added by convention. It restores the two inputs that the squaring operation maps to the same positive output.
Case 3 — a logarithmic equation produces an algebraic candidate outside the log domain
Solve log₂(x+2)=log₂(6-x).
Before equating arguments, the original logarithms require:
x+2>0 and 6-x>0, so -2<x<6.
Because the logarithm is one-to-one on positive inputs, equality of the logs implies x+2=6-x. Thus 2x=4 and x=2. The candidate lies in the domain and is valid.
Now change the problem to log₂(x+2)=log₂(x-8). Equating arguments gives the impossible statement 2=-8, so there is no solution. Domain work still matters: both arguments would require x>8, but the structural equality never occurs.
Case 4 — substitution creates a transformed domain that must be remembered
Suppose u=√x in an equation. The substitution automatically carries u≥0 and x≥0. If the transformed equation later gives u=-3, that value may solve the algebra in u but cannot correspond to √x over the real numbers.
The transformed problem is not free of the original structure. Every substitution creates a new variable plus a relationship that controls which transformed values are admissible.
Case 5 — context is part of the domain
A model gives t²-9t+20=0 for the time, in minutes, at which an event occurs after observation begins. Factorisation gives (t-4)(t-5)=0, so t=4 or 5. Both are mathematically admissible if the observation interval includes them.
Now suppose the problem states that measurements were recorded only for 0≤t≤4.5. The algebra has not changed, but the contextual domain has. Only t=4 belongs to the stated interval.
Domain restrictions can therefore come from algebra, functions, geometry, units or the real situation being modelled.
The solution-set ledger
For difficult problems, keep three short columns:
| Stage | Candidate set | Restriction still active |
|---|---|---|
| original statement | unknown | write the full domain |
| after transformation | new candidates may appear | carry every earlier restriction |
| after solving | list all algebraic candidates | filter by domain and context |
| final check | confirmed solutions only | substitute into the original statement |
This separates two jobs that students often collapse: generating candidates and certifying solutions.
Fresh release problems
- Solve x(x-2)=3x without losing the zero branch.
- Solve (2x+1)²=25 and explain why two branches appear.
- Solve log₃(x-1)=log₃(7-x), stating the domain first.
- Let u=x² in x⁴-13x²+36=0; recover every real solution.
- A quadratic model gives possible lengths -4 and 7. Explain why only one can represent a physical length.
Release answers
1. x(x-5)=0, so x=0 or 5. 2. 2x+1=±5, so x=2 or -3. 3. Domain is 1<x<7; equating arguments gives x-1=7-x, so x=4. 4. u²-13u+36=(u-4)(u-9), so u=4 or 9 and therefore x=±2,±3. 5. A physical length is positive in the stated model, so 7 is admissible and -4 is rejected by context.
The final habit
Do not ask only whether the algebra produced an answer. Ask which set the answer belongs to, which transformations produced it, and whether it still satisfies the original problem. That is the difference between manipulating equations and controlling solution sets.