Secondary Mathematics Diagnostic Casebook: Find the First Wrong Step

A page of red crosses tells you which answers were wrong. It does not yet tell you what to teach next. Two students can produce the same answer for different reasons. One may have misunderstood the question; another may have represented it correctly and then changed a sign. Giving both students the same extra worksheet does not investigate that difference.

This casebook provides original mathematical tasks, fictional student working, follow-up questions and fully explained repairs. Its purpose is practical: inspect one line, choose a question that distinguishes plausible explanations, and decide which mathematical relationship needs attention. The cases are teaching examples, not actual student records, a standardised diagnostic test, or evidence of an improvement rate.

Use only cases whose underlying topics have already been taught in the student’s school course. School year alone does not establish coverage. In particular, do not use an untaught upper-secondary topic to diagnose a lower-secondary learner as weak.

Start with the work, not the label

Before showing an answer, ask the student to complete one selected task independently. Keep the original attempt. Ask what each symbol represents and which line they would check first. Record any prompt you provide: “Read the question again”, “What is the original amount?” and “Divide both sides by negative three” are not equivalent amounts of help.

The general framework for separating conceptual, representation, procedural and transfer difficulties is explained in eduKateSG’s existing diagnosis guide. This casebook supplies additional worked material rather than a replacement taxonomy.

A follow-up question should distinguish explanations. “Do another one” is often too vague. A numerical substitution can test whether two expressions agree. Giving an equation after a word problem can help separate construction of the model from manipulation of that model. Neither observation, by itself, establishes a permanent trait or a complete explanation of the student’s performance.

Case 1: The negative sign outside a bracket

Task. Simplify −3(2x − 5) + 4x.

Fictional attempt. The student writes −6x − 15 + 4x, followed by −2x − 15.

The first incorrect transition is the expansion. Multiplying −3 by −5 gives +15, not −15. Combining −6x and +4x in the next line is correct, given the student’s preceding expression. A useful review therefore does not describe every line as wrong. It locates the first change that failed to preserve the value.

Correct working: −3(2x − 5) + 4x = −6x + 15 + 4x = 15 − 2x.

Ask three short follow-ups, without assuming their answers. What is (−3)(−5)? What is −3(2 − 5)? What does −3 multiply in the original bracket? A correct isolated product with incorrect expansion suggests checking distribution or written execution. An incorrect isolated product makes signed multiplication a plausible earlier repair target. These are provisional interpretations of the observed work, not diagnoses from a single response.

At x = 2, the original expression is −3(4 − 5) + 8 = 11. The correct simplified expression gives 11; the student’s expression gives −19. One disagreement refutes the proposed identity. Agreement at one substitution, however, would not prove that two expressions are identical for every x.

Repair task. Expand −2(3y − 4), explaining both products before combining anything else. The answer is −6y + 8. Then simplify −2(3y − 4) + 5y = 8 − y. Check whether the student carries the positive constant into the complete expression without a sign prompt.

Case 2: A division that applies to only part of a numerator

Task. Simplify (6x + 9)/3.

Fictional attempt. The student writes 2x + 9.

The numerator is the entire quantity 6x + 9. Division by 3 applies to both terms: (6x + 9)/3 = 6x/3 + 9/3 = 2x + 3. The answer is not obtained by dividing whichever term happens to contain a letter.

Ask the student to compare (6x + 9)/3 with 6x + 9/3. These expressions have different grouping. At x = 1, the first is 5 and the second is 9. That comparison makes the role of the fraction bar visible without relying on a slogan about order of operations.

Next ask for 15/3 and (6 + 9)/3. A learner who can do both may need help treating the algebraic numerator as a grouped quantity. A learner who cannot perform the numerical division needs a different starting point. Keep the interpretation tied to the actual follow-up response.

Repair task. Simplify (8a − 12)/4. Write it as 8a/4 − 12/4 to obtain 2a − 3. Then compare it with 8a − 12/4 = 8a − 3. Ask the student to state which part of each expression the denominator divides.

Case 3: A cancelled factor does not restore an excluded value

Prerequisite. Use this case only after factorisation and algebraic fractions have been taught.

Task. Simplify (x² − 4)/(x − 2), stating any restriction.

Fictional attempt. The student correctly obtains x + 2 but then writes, “At x = 2, the original fraction equals 4.”

Factorising gives (x − 2)(x + 2)/(x − 2). Cancelling the common factor is valid when x − 2 is nonzero. The simplified expression is therefore x + 2 with x ≠ 2. The polynomial x + 2 itself can be evaluated at 2, but the original fraction cannot: its denominator is zero there.

Ask the student what the denominator is at x = 2. Then ask whether simplifying the notation changes which inputs were permitted in the original expression. This tests the domain condition rather than repeating another factorisation exercise.

Repair task. Simplify (t² − 9)/(t − 3). The result is t + 3, with t ≠ 3. At t = 4, both permitted forms give 7. At t = 3, the original fraction remains undefined. Do not describe it as zero, infinity, or an ordinary fraction with a removable arithmetic inconvenience.

The useful learning outcome is not just a correct cancellation. It is a correct cancellation that carries its condition into subsequent use. In a later graphing course, the distinction can matter even when two displayed formulas appear almost identical.

Case 4: Choosing the correct percentage base

Task. A quantity increases from 80 units to 92 units. Find the percentage increase.

Fictional attempt. The student calculates 12/92 × 100%.

The increase is 12 units. The question asks how large that increase is relative to the original 80 units, so the calculation is 12/80 × 100% = 15%. Dividing by 92 answers a different comparison: the increase as a fraction of the final amount.

Ask, “Which quantity represents 100% before the increase?” Then ask the student to write the relationship 92 = 80 × 1.15. The two representations should agree about the reference quantity. Avoid accepting a memorised denominator rule without checking what the denominator means.

Now change the question. A quantity falls from 92 units to 80 units. The decrease is still 12 units, but the percentage decrease is 12/92 × 100%, approximately 13.04%. This is not a contradiction. The starting value has changed. Equal absolute changes need not produce equal percentage changes.

Repair task. An item costs $72 after a 20% discount. Its original price P satisfies 0.8P = 72, so P = $90. Multiplying 72 by 1.2 gives $86.40 and does not reverse the discount. Ask the student to check the proposed original price by applying the stated discount to it. All prices here are fictional mathematical examples, not current offers.

Case 5: Solving an equation versus constructing it

Task. A club charges a fixed joining fee of $18 and $3 for each session attended. A member pays $42 in total. How many sessions are included?

Fictional attempt. The student writes 18n + 3 = 42.

The variable needs a meaning: let n be the number of sessions. The total session charge is 3n dollars, while the joining fee is paid once. The model is 18 + 3n = 42, giving 3n = 24 and n = 8.

A discriminating follow-up is to supply the correct equation without its answer and ask the student to solve it. If they obtain 8 independently, the evidence points more strongly toward representation of the story than execution of that equation. If they cannot solve it, representation and equation-solving may both need inspection. Do not pretend that one successful supplied equation eliminates every procedural difficulty.

Ask what the formula predicts for zero sessions and one session. The totals should be $18 and $21. These boundary checks expose the reversed roles of fixed and repeated charges in 18n + 3.

Repair task. A fictional service charges a fixed $12 and $4 per booking. A total charge is $44. Write 12 + 4b = 44, so b = 8. Then ask the student to create a sentence for 12b + 4 instead. It would describe $12 per booking and a fixed $4 charge. Both equations can make sense; they model different rules.

Case 6: Why the inequality reverses

Task. Solve −3x < 12.

Fictional attempt. The student divides by −3 and writes x < −4.

Dividing by a negative number reverses the inequality, so the correct solution is x > −4. A check can make the contradiction visible. The value x = 0 satisfies the original inequality because 0 < 12. It must therefore appear in the solution set. The student’s proposed x < −4 excludes it.

Ask the student to begin with a true numerical comparison, such as 2 < 5, and multiply both sides by −1. The results satisfy −2 > −5. The reversal is a property of order, not an arbitrary mark that a teacher asks students to remember.

Repair task. Solve 7 − 2y ≥ 15. Subtracting 7 gives −2y ≥ 8. Dividing by −2 gives y ≤ −4. The boundary value −4 works because the original relation includes equality. A value such as −5 also works; a value such as 0 does not.

Record separately whether the learner handled subtraction correctly, reversed the direction, and retained the inclusive boundary. Three distinct decisions should not disappear beneath the single comment “inequalities wrong”.

Case 7: Average speed is not generally the average of speeds

Task. A traveller covers 60 km at 30 km/h and then another 60 km at 60 km/h, with no other elapsed time included. Find the average speed for the whole journey.

Fictional attempt. The student writes (30 + 60)/2 = 45 km/h.

Average speed is total distance divided by total time. The first section takes 60/30 = 2 hours. The second takes 60/60 = 1 hour. The complete journey is 120 km in 3 hours, so the average speed is 40 km/h.

Ask the student what quantity the denominator 2 in the attempted average represents. There are two journey sections, but they do not last the same amount of time. An unweighted arithmetic average of the two speeds therefore does not perform the required distance-over-time calculation.

Repair task. Now suppose the traveller spends one hour at 30 km/h and one hour at 60 km/h. The distances are 30 km and 60 km, giving 90 km in 2 hours, or 45 km/h. The arithmetic average works in this altered case because the durations are equal. A strong explanation distinguishes the conditions rather than declaring that averaging two speeds is always forbidden.

Case 8: A line’s gradient is not its height

Task. A straight line passes through (2, 7) and (5, 13). Find its gradient and equation.

Fictional attempt. The student divides 13 by 5 to obtain a gradient of 2.6.

The gradient is the change in vertical coordinate divided by the change in horizontal coordinate: (13 − 7)/(5 − 2) = 6/3 = 2. Substituting (2, 7) into y = 2x + c gives 7 = 4 + c, so c = 3 and the equation is y = 2x + 3.

Ask what assumption would make y/x equal to the gradient for a nonzero x on a line. It would require the line to pass through the origin. The given points do not justify that assumption. The distinction matters whenever a graph includes a starting quantity or fixed charge.

Repair task. A line passes through (1, 5) and (4, 11). Its gradient is 6/3 = 2, and its equation is y = 2x + 3. The coordinates changed but the underlying line did not. Ask the student to verify both points and explain why using a different pair of points on this line preserves the gradient.

Turn the case into a bounded teaching decision

A useful record contains the original line, the competing explanations, the follow-up response, the chosen repair and the next independent task. For Case 1, that might read: “Expanded the negative constant incorrectly; isolated negative product correct; original expression corrected only after a distribution prompt; next task will test bracket expansion without that prompt.” This is more informative than “careless with negatives”.

Do not turn the eight cases into an unvalidated total score with pass bands. Success here says something about the attempted tasks under the recorded conditions. It does not establish placement into G1, G2 or G3, predict an examination grade, or prove readiness for Additional Mathematics.

Likewise, completing a corrected example immediately after watching it is not the same observation as completing a fresh task later without help. Retain those different conditions in the record. The existing Mistake Ledger provides the broader correction route; this casebook gives examples of what a precise entry can contain.

A small independent exit set

After teaching only the relevant cases, select a few fresh questions rather than automatically giving the entire set. Ask the student to show the governing relationship as well as the answer.

A. Simplify −5(2p − 3) + 8p. Answer: 15 − 2p.

B. A quantity rises from 50 to 65. Find the percentage increase. Answer: 15/50 × 100% = 30%.

C. A fictional charge is $14 plus $6 per visit. The total is $44. Find the number of visits. Answer: 14 + 6v = 44, giving v = 5.

D. Solve 9 − 4z ≤ 1. Answer: −4z ≤ −8, so z ≥ 2.

E. Find the gradient through (1, 4) and (3, 10). Answer: (10 − 4)/(3 − 1) = 3.

For each selected question, record independent, self-corrected, prompted, or not completed, together with the actual help given. These are descriptive labels, not a calibrated scale. A student who catches their own error has produced a different piece of evidence from one who changes an answer after being told the exact operation.

Return to the correct owner

For topic teaching and worked stage guides, continue through the Secondary Mathematics capability map. For the wider school-stage route, use the Mathematics Learning Hub. For a separate A-Math question, use the Additional Mathematics Hub rather than assuming that a general mathematics result establishes A-Math readiness.

The next teaching step should be narrow enough to explain and concrete enough to check. A better diagnosis does not make a larger claim about the child. It makes a more useful claim about the next piece of mathematics to investigate.

Scope and source notes

All tasks, fictional attempts and worked answers in this casebook were prepared for this draft. They are not copied examination questions, actual tuition outcomes or a validated assessment instrument. The linked eduKate guides retain their existing explanation and navigation roles.

For current examination scope, consult SEAB’s SEC information and the official syllabus for the student’s subject level and examination year. SEC begins in 2027; a 2026 candidate must not be relabelled merely because a newer syllabus page exists. Reviewed 6 September 2026. This casebook does not claim comprehensive coverage of any one syllabus.