What you're inspecting
This is the s1s3-algebra generator unit — 12 templates covering HK junior secondary Unit 8 (Linear equations, one unknown) and Unit 9 (Linear equations, two unknowns). The probe below draws 96 fresh questions (8 per template) and shows you each one rendered as multiple choice with trap-analysed distractors, the canonical solution steps the autograder will use, and a preview of how the same stem would be presented as fill-in or long-answer.
Templates in this unit
12
7 for Unit 8 · 5 for Unit 9
Samples in this probe
96
8 per template, freshly generated on page build
Difficulty mix observed
advanced: 17 basic: 12 intermediate: 67
across all 96 sampled questions
Pipeline coverage
✓ generator
✓ verifier
✓ distractors
✓ solutions
✓ Chinese
✓ verifier
✓ distractors
✓ solutions
✓ Chinese
all 12 IDs present in every map
Eval gate
PASS
L1 (deterministic) + L2 (LLM judge)
see
see
eval-unit.jsS1–S3 curriculum coverage
2 / 31 units
19 / 311 hours = 6.1% of junior-secondary classroom time
This unit covers Units 8 + 9 (Number & Algebra)
This unit covers Units 8 + 9 (Number & Algebra)
S1–S3 scope · what this 12-template unit is, in context
The HK junior secondary maths curriculum has 31 units across 3 strands totaling 311 teaching hours and 125 declared objectives. This generator unit covers 2 of those 31 units (Units 8 + 9, both in Number & Algebra), which represent 19 of 311 hours (6.1%) of junior secondary classroom time.
| Unit | Name | Hours | Strand | Generators |
|---|---|---|---|---|
| 1 | Basic computation | 8 | Number & Algebra | — |
| 2 | Directed numbers | 9 | Number & Algebra | — |
| 3 | Approximate values & estimation | 6 | Number & Algebra | — |
| 4 | Rational & irrational numbers | 7 | Number & Algebra | — |
| 5 | Using percentages | 15 | Number & Algebra | — |
| 6 | Rates, ratios & proportions | 8 | Number & Algebra | — |
| 7 | Algebraic expressions | 7 | Number & Algebra | — |
| 8 | Linear equations in one unknown in scope | 7 | Number & Algebra | 7 templates |
| 9 | Linear equations in two unknowns in scope | 12 | Number & Algebra | 5 templates |
| 10 | Laws of integral indices | 11 | Number & Algebra | — |
| 11 | Polynomials | 15 | Number & Algebra | — |
| 12 | Identities | 8 | Number & Algebra | — |
| 13 | Formulae | 9 | Number & Algebra | — |
| 14 | Linear inequalities in one unknown | 6 | Number & Algebra | — |
| 15 | Errors in measurement | 6 | Measures, Shape & Space | — |
| 16 | Arc lengths & areas of sectors | 8 | Measures, Shape & Space | — |
| 17 | 3-D figures | 5 | Measures, Shape & Space | — |
| 18 | Mensuration | 15 | Measures, Shape & Space | — |
| 19 | Angles & parallel lines | 11 | Measures, Shape & Space | — |
| 20 | Polygons | 8 | Measures, Shape & Space | — |
| 21 | Congruent triangles | 14 | Measures, Shape & Space | — |
| 22 | Similar triangles | 9 | Measures, Shape & Space | — |
| 23 | Quadrilaterals | 13 | Measures, Shape & Space | — |
| 24 | Centres of triangles | 8 | Measures, Shape & Space | — |
| 25 | Pythagoras' theorem | 6 | Measures, Shape & Space | — |
| 26 | Rectangular coordinate system | 19 | Measures, Shape & Space | — |
| 27 | Trigonometry | 18 | Measures, Shape & Space | — |
| 28 | Organisation of data | 4 | Data Handling | — |
| 29 | Presentation of data | 17 | Data Handling | — |
| 30 | Measures of central tendency | 10 | Data Handling | — |
| 31 | Probability | 12 | Data Handling | — |
Coverage by sub-objective (Units 8 + 9)
Each curriculum objective listed here is broken out separately. The "Generators serving this objective" column answers your question — usually it's one generator per shape; multiple generators per objective when the curriculum objective spans different equation forms (e.g. "Solve linear equations" has 5 different shapes).
Unit 8
7 hours curriculum · 96 reference questions on disk
Linear equations in one unknown
◐
Solve linear equations
5 generators serving this objective
s1s3-eq-axb-c— Linear: ax + b = c 1vs1s3-eq-axb-cx-d— Variables both sides (+): ax + b = cx + d 1vs1s3-eq-ax-b-cx-d— Variables both sides (−): ax − b = cx + d 1vs1s3-eq-fraction— Fractional form: x/a ± b = c 3vs1s3-eq-bracket-both— Brackets both sides: a(x+b) = c(x+d) 1v
Shapes seen in reference bank, NOT yet a generator (1):
- Pure "solve only" with mixed bracket + fraction (multi-step)
○
Formulate from problem situations
0 generators serving this objective
Shapes seen in reference bank, NOT yet a generator (3):
- "Set up an equation only" — answer is the equation, not the value (common in EdCity prompts)
- Multi-clause word relations: 3+ entities (e.g. "Ming, May, Helen distributed candies, …")
- Three-person relational chains
◐
Solve word problems
2 generators serving this objective
Shapes seen in reference bank, NOT yet a generator (3):
- Consecutive integers / consecutive even numbers
- "k times age t years ago" structure (e.g. "Ceci is 3× her age 10 years ago")
- Fraction-of-population bus problems (passengers got off, then got on)
Unit 9
12 hours curriculum · 201 reference questions on disk
Linear equations in two unknowns
○
Concept and graphs
0 generators serving this objective
Shapes seen in reference bank, NOT yet a generator (2):
- "Which graph could represent ax + by + c = 0?" (graph-MC)
- Read intercept / slope from a given line
○
Solve simultaneous equations (graphical)
0 generators serving this objective
Shapes seen in reference bank, NOT yet a generator (2):
- Plot two lines, identify intersection as solution
- Match a 2×2 system to its intersection picture
◐
Solve simultaneous equations (substitution, elimination)
4 generators serving this objective
s1s3-simul-sub— Simultaneous: substitution method 1vs1s3-simul-elim-add— Simultaneous: elimination by addition 1vs1s3-simul-elim-sub— Simultaneous: elimination by subtraction 1vs1s3-simul-fraction-coeff— Simultaneous: fractional coefficients 4v
Shapes seen in reference bank, NOT yet a generator (2):
- Method-step MC ("If we eliminate y first, which is the first step?")
- Pick which equation to substitute into (process-aware MC)
◐
Formulate and solve word problems
1 generator serving this objective
Shapes seen in reference bank, NOT yet a generator (4):
- Chickens-and-pigs (heads + feet) as equation-pair-selection
- Exam scoring (+/− marks) systems
- Digit / place-value two-unknown stories
- "Pick 2 equations from a list" formulation MC
Reference question bank · how much room for more generators
Total EdCity items on disk
11,826
across all grades, all subjects
workdir/math/corpus/edcity-star/canonical/S1–S3 items
3,146
grade_canon = S1-S3
For Units 8 + 9 (this unit)
297
96 for Unit 8 · 201 for Unit 9
vs 12 generators currently — clear room to expand
vs 12 generators currently — clear room to expand
Variety potential — answer to "are we at the limit?"
No, not at the limit. The 12 generators cover the most common solve-and-give-the-number patterns, but the reference bank shows entire shape categories that aren't yet generators:
- Unit 8 — "set up an equation only" (answer is the equation, not the value); 3-person relational chains; consecutive-integer / consecutive-even patterns; "k times age t years ago"; fraction-of-population bus stories.
- Unit 9 — entire graphical pathway (no graphs at all yet); method-step MC ("if we eliminate y first, which is the first step?"); "pick 2 equations from this list"; chickens-pigs as equation-pair-selection; exam-score (+/−) systems; digit / place-value two-unknown stories.
Each bullet is a candidate new generator. Conservative estimate: +8 to +12 additional generators for Units 8+9 alone if we systematically work through the EdCity reference shapes — roughly doubling the variety surface.
What the pipeline does to each question
1
generate
Template draws random parameters, builds stem, computes answer, sets difficulty + skillsTested.
→
2
verify
Independent verifier in
verify_answers.js re-derives answer from raw data. Catches generator drift.→
3
add MC options
Trap-analysed distractors from
traps.js modeling specific student misconceptions; fallback distractors when traps are insufficient.→
4
build solution
Step-by-step working with intermediate math, plus a verification line.
→
5
add Chinese
Translates stem and solution to 繁體中文 via
chinese.js.→
6
seed Supabase
Enriched JSON →
db/seed-supabase.js → math_questions table.Question type readiness
The HK curriculum declares MC, fill-in, and word-problem types for both units. Here's where each format sits today:
| Format | Status | Notes |
|---|---|---|
| Multiple choice | ✓ Production-ready | 4 options per question, error-modeled distractors with explicit trap analysis. Fallback distractors when traps insufficient. |
| Fill-in / Short answer | ✓ Ready (drop options) | Same generator output — just hide options, use q.answer as expected. UI mockup shown in each card under "Other formats". |
| Long answer / Show working | ⚠ Solution steps ready; UI not built | Canonical step-by-step exists per question. Frontend doesn't yet collect or grade student-written working. |
| Word problem | ✓ Covered | 4 of the 12 templates are word problems (age, price, two-unknowns, fraction-coeff with 5 contexts). |
| Multi-part / Structured (a, b, c) | ○ Future work | No partA/partB scaffolding in current generator. Would need new template shape. |
| Graphical (Unit 9 obj) | ⚠ Curriculum gap | Unit 9 lists "Concept and graphs" + "graphical solution of simultaneous equations" — current templates are algebraic-only. Needs diagram-pipeline integration. |
Atomic knowledge wiring · NEW (2026-04-23)
Up until now, the generators emitted skillsTested tags as free-text strings with no canonical decomposition. As of this run, every generator's tags are anchored to a structured competency map per unit, and db/seed.js auto-loads them at seed time.
| Unit | Competency map | Atomic skills defined | Tags anchored | Coverage gaps documented |
|---|---|---|---|---|
| U8 | knowledge/competency-map/s1s3-u8-linear-eq-one-unknown.json | 8 (transpose, collect, brackets, fraction-coeff, var-both-sides, word-problem, …) | ✓ all 5 | 3 (formulate-only, negative coefficients, fractional answers) |
| U9 | knowledge/competency-map/s1s3-u9-linear-eq-two-unknowns.json | 11 (substitution, elimination, fraction-coeff, word, recognition, point-on-line, line-graph, intersection-read, formulate, …) | ✓ all 5 | 5 (graphical intersection-read, line-recognition, elimination-multiply-first, equation-pair-selection, formulate-only) |
Both files include a difficultySignal block recording the per-unit EdCity teacher calibration (suggested_mark + suggested_time) so future generators can target a known difficulty band rather than guessing.
Difficulty tagging · DECIDED 2026-04-24 → (B) dual-tag
Eric chose Option (B) — dual-tag schema
"Add 'marks + expected seconds' as the real difficulty under the hood. Keep basic/intermediate/advanced as a derived bucket for any UI that already expects it." — Eric, 2026-04-24 email reply.
Each generated question now carries difficulty: { mark, estTimeSec, bucket, levers{...} }. bucket is computed from (mark, estTimeSec) by a fixed table; nothing existing breaks. Full schema in the dimensions doc below.
Background: difficulty-audit-2026-04-23.md (problem framing).
Difficulty dimensions · the 9 levers + new template plan (NEW 2026-04-24)
After Eric chose (B), next question: "How do we fully understand the difficulty dimension so we can come up with new templates and/or flags?" Answer: 9 levers extracted from 296 sampled EdCity items, mapped against our 12 templates. Most templates lock 7-8 of the 9 levers — that's why every basic feels the same.
The 9 levers
| # | Lever | Settings | +sec per step | EdCity evidence |
|---|---|---|---|---|
| 1 | Solver step count | 1 / 2 / 3 / 4 / 5+ steps | +15s | 1-step −5x=60 30s; 4-step 5[3(1−x)+x]−12=0 120s |
| 2 | Coefficient magnitude | small (1-9) / medium (10-25) / large (26-50) | +10-15s | 5x=60 mental vs 12x+22=58 written |
| 3 | Sign mix | all-pos / one-neg / mixed-neg | +10-20s | −5x=60, 6−x=−9 |
| 4 | Answer form | pos-int / neg-int / proper fraction / improper | +15-45s | EdCity 90s items often have fractional answer |
| 5 | Bracket / fraction structure | none / single / both-sides / nested / frac-coeff | +20-45s | 9(x+1)=6(x+2) 90s; nested 120s |
| 6 | Variable position | one-side / both-sides / both-with-brackets | +15-30s | U8 60s vs 90s split |
| 7 | Formulation depth | none / single-eq / multi-entity 2-eq | +60-120s | EdCity 120-150s word problems |
| 8 | Graphical layer (U9) | none / read-1 / read-intersection / line-recognition / point-on-line | +20-45s | 30s "find intersection from labelled graph" MC |
| 9 | Presentation | free / MC / multi-MC / conceptual-MC | +15-30s | Conceptual MC "which equation results from eliminating x?" 90s |
Lever × template coverage today
✅ = template randomises this lever per generation. ❌ = locked at one setting. Out of 108 cells, only 15 are randomised today — ~86% of the difficulty space is locked.
| Template | step | coeff | sign | ans form | brackets | var pos | formulate | graph | present |
|---|---|---|---|---|---|---|---|---|---|
s1s3-eq-axb-c | ❌ | ✅ | ❌ | ❌ | ❌ | ❌ | ❌ | ❌ | ❌ |
s1s3-eq-axb-cx-d | ❌ | ✅ | ❌ | ❌ | ❌ | ❌ | ❌ | ❌ | ❌ |
s1s3-eq-ax-b-cx-d | ❌ | ✅ | ✅ | ❌ | ❌ | ❌ | ❌ | ❌ | ❌ |
s1s3-eq-bracket-both | ❌ | ✅ | ❌ | ❌ | ❌ | ❌ | ❌ | ❌ | ❌ |
s1s3-eq-fraction | ✅ | ✅ | ❌ | ❌ | ❌ | ✅ | ❌ | ❌ | ❌ |
s1s3-eq-word-age | ❌ | ✅ | ❌ | ❌ | ❌ | n/a | ✅ | ❌ | ❌ |
s1s3-eq-word-price | ❌ | ✅ | ❌ | ❌ | ❌ | n/a | ✅ | ❌ | ❌ |
s1s3-simul-sub | ❌ | ✅ | ❌ | ❌ | ❌ | n/a | ❌ | ❌ | ❌ |
s1s3-simul-elim-add | ❌ | ✅ | ❌ | ❌ | ❌ | n/a | ❌ | ❌ | ❌ |
s1s3-simul-elim-sub | ❌ | ✅ | ❌ | ❌ | ❌ | n/a | ❌ | ❌ | ❌ |
s1s3-simul-fraction-coeff | ❌ | ✅ | ❌ | ❌ | ❌ | n/a | ❌ | ❌ | ❌ |
s1s3-simul-word-two-unknowns | ❌ | ✅ | ❌ | ❌ | ❌ | n/a | ✅ | ❌ | ❌ |
Phase A — promote 4 levers to in-template flags
Lift coefficient magnitude, sign mix, answer form, and presentation from "locked" to "randomised per generation". Same code paths, widened ranges. Projected per-template spread:
| Template | Today's spread | Projected after Phase A | Wider by |
|---|---|---|---|
s1s3-eq-axb-c | 30-50s | 30-90s | 3× |
s1s3-eq-axb-cx-d | 50-70s | 50-105s | 2.7× |
s1s3-eq-bracket-both | flat 90s | 75-120s | ∞ |
s1s3-simul-elim-add | flat 75s | 60-110s | ∞ |
Phase B — 11 new templates that target currently-blank bands
Some bands are unreachable even with all flags maxed (e.g. no template has graphical layer ≠ none). Sorted by EdCity prevalence (= teacher-validated demand):
| # | New template | Mark | Time | EdCity n | Why |
|---|---|---|---|---|---|
| 1 | s1s3-substitute-point-find-k | 1 | 60s | ~80 | Single biggest U9 family — 38% of U9 EdCity volume; zero coverage today |
| 2 | s1s3-graphical-intersection-read-MC | 1 | 30s | ~14 | Closes the U9 graphical hole |
| 3 | s1s3-formulate-only-MC | 1 | 45s | ~15 | Tests formulation independent of execution |
| 4 | s1s3-line-recognise-points-MS-MC | 1 | 60s | ~10 | Conceptual property check |
| 5 | s1s3-simul-conceptual-elim-MC | 1 | 90s | ~10 | Tests procedural understanding without computing |
| 6 | s1s3-simul-elim-need-multiply | 2 | 100s | ~12 | Bridge between basic elim and fraction-coeff |
| 7 | s1s3-eq-one-step-transposition | 1 | 30s | ~16 | The 30s warm-up band, currently empty |
| 8 | s1s3-eq-bracket-nested | 1 | 120s | ~16 | The 120s deep-bracket band |
| 9 | s1s3-graphical-find-line-eq | 1 | 45s | ~6 | Inverse direction of substitute-point |
| 10 | s1s3-eq-word-distance-time | 2 | 150s | ~6 | Top-band motion problem |
| 11 | s1s3-eq-word-mixture-percent | 2 | 150s | ~5 | Same band, different surface |
Open decisions for Eric
| # | Decision | Default recommendation |
|---|---|---|
| D1 | Phase A first, then Phase B, or in parallel? | Sequential — Phase A is 1 day, validates calibration before B (3-4 days) |
| D2 | "Conceptual MC" as a flag everywhere, or only specific templates? | Only on templates with clear intermediate "next move" (elim-* and bracket-both) |
| D3 | Phase B priority order — close U9 graphical hole first, or build #1 (substitute-point-find-k)? | #1 first (highest prevalence + simplest), then #2 (closes hole), then #3-#5 |
| D4 | Word-problem templates: allow large coefficients? | No — keep ages/prices realistic; use sign + formulation lever |
Full doc with EdCity sample evidence per band: workdir/math/audits/difficulty-dimensions-2026-04-24.md
How to read the per-template tabs
- Top of each tab: what the template tests, the input shape (parameter ranges), the curriculum unit + subtopic it maps to, and the difficulty mix in this 8-sample probe.
- Each card: one freshly generated question. Three toggle views — MC options (correct + traps with reasons), canonical solution steps, and previews of fill-in / long-answer formats from the same stem.
- Data pills: the random parameters that produced this specific question — useful for spotting any narrow ranges.
Unit 8
Solve linear equations
Curriculum objective in Linear equations in one unknown. 5 generators currently serve this objective.
⚠ Reference shapes seen on disk but NOT yet served by these 5 generators (1 candidates):
- Pure "solve only" with mixed bracket + fraction (multi-step)
Variety dashboard · all 5 generators side-by-side
Compare structural variant counts, observed difficulty mix, and a representative stem from each. Click a row's button to jump to the full sample set below.
| Generator | Variants | Difficulty mix (8 samples) | First sample stem | |
|---|---|---|---|---|
ax + b = cs1s3-eq-axb-c | 1v Single algebraic shape; variety from coefficient sampling only. | basic: 8 | Solve the equation: 4x + 4 = 64. | |
ax + b = cx + ds1s3-eq-axb-cx-d | 1v Single algebraic shape; variety from parameter sampling only. | intermediate: 8 | Solve the equation: 12x + 2 = 8x + 42. | |
ax − b = cx + ds1s3-eq-ax-b-cx-d | 1v Search loop + fixed fallback (same exam shape). | intermediate: 8 | Solve the equation: 3x − 3 = x + 7. | |
Fraction coeff.s1s3-eq-fraction | 3v 3 structural branches (basic / intermediate / advanced). | advanced: 1 basic: 4 intermediate: 3 | Solve the equation: x/2 − 2 = 5. | |
Brackets both sidess1s3-eq-bracket-both | 1v Retry + fallback. Single algebraic shape. | advanced: 8 | Solve the equation: 9(x + 1) = 6(x + 2). |
Drill into one generator · 8 samples each
s1s3-eq-axb-cUnit 8 · Solve linear equations
Linear: ax + b = c
The fundamental case. Isolate x by undoing addition then multiplication.
Variety profile: 1 internal structural variant · Single algebraic shape; variety from coefficient sampling only.
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1basicanswer:
15Solve the equation: 4x + 4 = 64.
a=4b=4c=64x=15variant=axb-c
A.
60
→ Forgot to divide by coefficient of x after isolating x term
B.
17
→ Transposition sign error — added b instead of subtracting
C.
16
→ Ignored the constant b — divided c by a directly
D.
15 ✓ correct
1.
Subtract 4 from both sides.
4x = 64 − 4 = 60
2.
Divide both sides by 4.
x = 60 ÷ 4 = 15
4(15) + 4 = 64. ✓
Fill-in / Short Answer
Solve the equation: 4x + 4 = 64.
Answer: ________
Expected:
15Long answer / Show working
Solve the equation: 4x + 4 = 64.
Student writes solution steps here. Auto-grader compares final answer (
15) and rewards intermediate steps from the canonical solution above.#2basicanswer:
7Solve the equation: 11x + 17 = 94.
a=11b=17c=94x=7variant=axb-c
A.
10
→ Transposition sign error — added b instead of subtracting
B.
7 ✓ correct
C.
77
→ Forgot to divide by coefficient of x after isolating x term
D.
9
→ Ignored the constant b — divided c by a directly
1.
Subtract 17 from both sides.
11x = 94 − 17 = 77
2.
Divide both sides by 11.
x = 77 ÷ 11 = 7
11(7) + 17 = 94. ✓
Fill-in / Short Answer
Solve the equation: 11x + 17 = 94.
Answer: ________
Expected:
7Long answer / Show working
Solve the equation: 11x + 17 = 94.
Student writes solution steps here. Auto-grader compares final answer (
7) and rewards intermediate steps from the canonical solution above.#3basicanswer:
15Solve the equation: 10x + 3 = 153.
a=10b=3c=153x=15variant=axb-c
A.
15 ✓ correct
B.
150
→ Forgot to divide by coefficient of x after isolating x term
C.
16
→ Transposition sign error — added b instead of subtracting
D.
14
→ Off-by-one in root counting
1.
Subtract 3 from both sides.
10x = 153 − 3 = 150
2.
Divide both sides by 10.
x = 150 ÷ 10 = 15
10(15) + 3 = 153. ✓
Fill-in / Short Answer
Solve the equation: 10x + 3 = 153.
Answer: ________
Expected:
15Long answer / Show working
Solve the equation: 10x + 3 = 153.
Student writes solution steps here. Auto-grader compares final answer (
15) and rewards intermediate steps from the canonical solution above.#4basicanswer:
3Solve the equation: 12x + 22 = 58.
a=12b=22c=58x=3variant=axb-c
A.
5
→ Ignored the constant b — divided c by a directly
B.
36
→ Forgot to divide by coefficient of x after isolating x term
C.
7
→ Transposition sign error — added b instead of subtracting
D.
3 ✓ correct
1.
Subtract 22 from both sides.
12x = 58 − 22 = 36
2.
Divide both sides by 12.
x = 36 ÷ 12 = 3
12(3) + 22 = 58. ✓
Fill-in / Short Answer
Solve the equation: 12x + 22 = 58.
Answer: ________
Expected:
3Long answer / Show working
Solve the equation: 12x + 22 = 58.
Student writes solution steps here. Auto-grader compares final answer (
3) and rewards intermediate steps from the canonical solution above.#5basicanswer:
6Solve the equation: 9x + 12 = 66.
a=9b=12c=66x=6variant=axb-c
A.
6 ✓ correct
B.
54
→ Forgot to divide by coefficient of x after isolating x term
C.
9
→ Transposition sign error — added b instead of subtracting
D.
7
→ Ignored the constant b — divided c by a directly
1.
Subtract 12 from both sides.
9x = 66 − 12 = 54
2.
Divide both sides by 9.
x = 54 ÷ 9 = 6
9(6) + 12 = 66. ✓
Fill-in / Short Answer
Solve the equation: 9x + 12 = 66.
Answer: ________
Expected:
6Long answer / Show working
Solve the equation: 9x + 12 = 66.
Student writes solution steps here. Auto-grader compares final answer (
6) and rewards intermediate steps from the canonical solution above.#6basicanswer:
11Solve the equation: 7x + 20 = 97.
a=7b=20c=97x=11variant=axb-c
A.
77
→ Forgot to divide by coefficient of x after isolating x term
B.
17
→ Transposition sign error — added b instead of subtracting
C.
11 ✓ correct
D.
14
→ Ignored the constant b — divided c by a directly
1.
Subtract 20 from both sides.
7x = 97 − 20 = 77
2.
Divide both sides by 7.
x = 77 ÷ 7 = 11
7(11) + 20 = 97. ✓
Fill-in / Short Answer
Solve the equation: 7x + 20 = 97.
Answer: ________
Expected:
11Long answer / Show working
Solve the equation: 7x + 20 = 97.
Student writes solution steps here. Auto-grader compares final answer (
11) and rewards intermediate steps from the canonical solution above.#7basicanswer:
14Solve the equation: 5x + 18 = 88.
a=5b=18c=88x=14variant=axb-c
A.
70
→ Forgot to divide by coefficient of x after isolating x term
B.
14 ✓ correct
C.
21
→ Transposition sign error — added b instead of subtracting
D.
18
→ Ignored the constant b — divided c by a directly
1.
Subtract 18 from both sides.
5x = 88 − 18 = 70
2.
Divide both sides by 5.
x = 70 ÷ 5 = 14
5(14) + 18 = 88. ✓
Fill-in / Short Answer
Solve the equation: 5x + 18 = 88.
Answer: ________
Expected:
14Long answer / Show working
Solve the equation: 5x + 18 = 88.
Student writes solution steps here. Auto-grader compares final answer (
14) and rewards intermediate steps from the canonical solution above.#8basicanswer:
3Solve the equation: 2x + 9 = 15.
a=2b=9c=15x=3variant=axb-c
A.
8
→ Ignored the constant b — divided c by a directly
B.
3 ✓ correct
C.
12
→ Transposition sign error — added b instead of subtracting
D.
6
→ Forgot to divide by coefficient of x after isolating x term
1.
Subtract 9 from both sides.
2x = 15 − 9 = 6
2.
Divide both sides by 2.
x = 6 ÷ 2 = 3
2(3) + 9 = 15. ✓
Fill-in / Short Answer
Solve the equation: 2x + 9 = 15.
Answer: ________
Expected:
3Long answer / Show working
Solve the equation: 2x + 9 = 15.
Student writes solution steps here. Auto-grader compares final answer (
3) and rewards intermediate steps from the canonical solution above.s1s3-eq-axb-cx-dUnit 8 · Solve linear equations
Variables both sides (+): ax + b = cx + d
Move x-terms to one side, constants to the other. Tests sign-tracking.
Variety profile: 1 internal structural variant · Single algebraic shape; variety from parameter sampling only.
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1intermediateanswer:
10Solve the equation: 12x + 2 = 8x + 42.
a=12b=2c=8d=42x=10variant=axb-cxd
A.
10 ✓ correct
B.
-10
→ Subtracted x-coefficients in wrong order (denominator sign error)
C.
2
→ Added all terms instead of moving variables to one side
D.
40
→ Collected constants correctly but forgot to divide by (a−c)
1.
Bring x-terms to LHS, constants to RHS.
12x − 8x = 42 − 2
2.
Simplify.
4x = 40
3.
Divide both sides by 4.
x = 40 ÷ 4 = 10
12(10) + 2 = 122 = 8(10) + 42. ✓
Fill-in / Short Answer
Solve the equation: 12x + 2 = 8x + 42.
Answer: ________
Expected:
10Long answer / Show working
Solve the equation: 12x + 2 = 8x + 42.
Student writes solution steps here. Auto-grader compares final answer (
10) and rewards intermediate steps from the canonical solution above.#2intermediateanswer:
2Solve the equation: 12x + 2 = 8x + 10.
a=12b=2c=8d=10x=2variant=axb-cxd
A.
1
→ Added all terms instead of moving variables to one side
B.
2 ✓ correct
C.
8
→ Collected constants correctly but forgot to divide by (a−c)
D.
-2
→ Subtracted x-coefficients in wrong order (denominator sign error)
1.
Bring x-terms to LHS, constants to RHS.
12x − 8x = 10 − 2
2.
Simplify.
4x = 8
3.
Divide both sides by 4.
x = 8 ÷ 4 = 2
12(2) + 2 = 26 = 8(2) + 10. ✓
Fill-in / Short Answer
Solve the equation: 12x + 2 = 8x + 10.
Answer: ________
Expected:
2Long answer / Show working
Solve the equation: 12x + 2 = 8x + 10.
Student writes solution steps here. Auto-grader compares final answer (
2) and rewards intermediate steps from the canonical solution above.#3intermediateanswer:
7Solve the equation: 5x + 7 = 3x + 21.
a=5b=7c=3d=21x=7variant=axb-cxd
A.
4
→ Added all terms instead of moving variables to one side
B.
7 ✓ correct
C.
-7
→ Subtracted x-coefficients in wrong order (denominator sign error)
D.
14
→ Collected constants correctly but forgot to divide by (a−c)
1.
Bring x-terms to LHS, constants to RHS.
5x − 3x = 21 − 7
2.
Simplify.
2x = 14
3.
Divide both sides by 2.
x = 14 ÷ 2 = 7
5(7) + 7 = 42 = 3(7) + 21. ✓
Fill-in / Short Answer
Solve the equation: 5x + 7 = 3x + 21.
Answer: ________
Expected:
7Long answer / Show working
Solve the equation: 5x + 7 = 3x + 21.
Student writes solution steps here. Auto-grader compares final answer (
7) and rewards intermediate steps from the canonical solution above.#4intermediateanswer:
10Solve the equation: 7x + 20 = 5x + 40.
a=7b=20c=5d=40x=10variant=axb-cxd
A.
10 ✓ correct
B.
20
→ Collected constants correctly but forgot to divide by (a−c)
C.
5
→ Added all terms instead of moving variables to one side
D.
-10
→ Subtracted x-coefficients in wrong order (denominator sign error)
1.
Bring x-terms to LHS, constants to RHS.
7x − 5x = 40 − 20
2.
Simplify.
2x = 20
3.
Divide both sides by 2.
x = 20 ÷ 2 = 10
7(10) + 20 = 90 = 5(10) + 40. ✓
Fill-in / Short Answer
Solve the equation: 7x + 20 = 5x + 40.
Answer: ________
Expected:
10Long answer / Show working
Solve the equation: 7x + 20 = 5x + 40.
Student writes solution steps here. Auto-grader compares final answer (
10) and rewards intermediate steps from the canonical solution above.#5intermediateanswer:
12Solve the equation: 10x + 7 = 7x + 43.
a=10b=7c=7d=43x=12variant=axb-cxd
A.
3
→ Added all terms instead of moving variables to one side
B.
36
→ Collected constants correctly but forgot to divide by (a−c)
C.
-12
→ Subtracted x-coefficients in wrong order (denominator sign error)
D.
12 ✓ correct
1.
Bring x-terms to LHS, constants to RHS.
10x − 7x = 43 − 7
2.
Simplify.
3x = 36
3.
Divide both sides by 3.
x = 36 ÷ 3 = 12
10(12) + 7 = 127 = 7(12) + 43. ✓
Fill-in / Short Answer
Solve the equation: 10x + 7 = 7x + 43.
Answer: ________
Expected:
12Long answer / Show working
Solve the equation: 10x + 7 = 7x + 43.
Student writes solution steps here. Auto-grader compares final answer (
12) and rewards intermediate steps from the canonical solution above.#6intermediateanswer:
2Solve the equation: 13x + 2 = 8x + 12.
a=13b=2c=8d=12x=2variant=axb-cxd
A.
-2
→ Subtracted x-coefficients in wrong order (denominator sign error)
B.
10
→ Collected constants correctly but forgot to divide by (a−c)
C.
2 ✓ correct
D.
1
→ Added all terms instead of moving variables to one side
1.
Bring x-terms to LHS, constants to RHS.
13x − 8x = 12 − 2
2.
Simplify.
5x = 10
3.
Divide both sides by 5.
x = 10 ÷ 5 = 2
13(2) + 2 = 28 = 8(2) + 12. ✓
Fill-in / Short Answer
Solve the equation: 13x + 2 = 8x + 12.
Answer: ________
Expected:
2Long answer / Show working
Solve the equation: 13x + 2 = 8x + 12.
Student writes solution steps here. Auto-grader compares final answer (
2) and rewards intermediate steps from the canonical solution above.#7intermediateanswer:
6Solve the equation: 8x + 13 = 5x + 31.
a=8b=13c=5d=31x=6variant=axb-cxd
A.
18
→ Collected constants correctly but forgot to divide by (a−c)
B.
6 ✓ correct
C.
3
→ Added all terms instead of moving variables to one side
D.
-6
→ Subtracted x-coefficients in wrong order (denominator sign error)
1.
Bring x-terms to LHS, constants to RHS.
8x − 5x = 31 − 13
2.
Simplify.
3x = 18
3.
Divide both sides by 3.
x = 18 ÷ 3 = 6
8(6) + 13 = 61 = 5(6) + 31. ✓
Fill-in / Short Answer
Solve the equation: 8x + 13 = 5x + 31.
Answer: ________
Expected:
6Long answer / Show working
Solve the equation: 8x + 13 = 5x + 31.
Student writes solution steps here. Auto-grader compares final answer (
6) and rewards intermediate steps from the canonical solution above.#8intermediateanswer:
2Solve the equation: 6x + 6 = 2x + 14.
a=6b=6c=2d=14x=2variant=axb-cxd
A.
-2
→ Subtracted x-coefficients in wrong order (denominator sign error)
B.
2 ✓ correct
C.
3
→ Added all terms instead of moving variables to one side
D.
8
→ Collected constants correctly but forgot to divide by (a−c)
1.
Bring x-terms to LHS, constants to RHS.
6x − 2x = 14 − 6
2.
Simplify.
4x = 8
3.
Divide both sides by 4.
x = 8 ÷ 4 = 2
6(2) + 6 = 18 = 2(2) + 14. ✓
Fill-in / Short Answer
Solve the equation: 6x + 6 = 2x + 14.
Answer: ________
Expected:
2Long answer / Show working
Solve the equation: 6x + 6 = 2x + 14.
Student writes solution steps here. Auto-grader compares final answer (
2) and rewards intermediate steps from the canonical solution above.s1s3-eq-ax-b-cx-dUnit 8 · Solve linear equations
Variables both sides (−): ax − b = cx + d
Same as above but with a negative constant on the LHS — common slip pattern.
Variety profile: 1 internal structural variant · Search loop + fixed fallback (same exam shape).
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1intermediateanswer:
5Solve the equation: 3x − 3 = x + 7.
a=3b=3c=1d=7x=5variant=ax-b-cxd
A.
5 ✓ correct
B.
-5
→ Both numerator and denominator signs wrong simultaneously
C.
2
→ When transposing −b, treated it as −b on RHS instead of +b (sign error)
D.
10
→ Correct numerator but forgot to divide by (a−c)
1.
Move x-terms to LHS and constants to RHS (transpose −3 as +3).
3x − 1x = 7 + 3
2.
Simplify.
2x = 10
3.
Divide both sides by 2.
x = 10 ÷ 2 = 5
3(5) − 3 = 12 = 1(5) + 7. ✓
Fill-in / Short Answer
Solve the equation: 3x − 3 = x + 7.
Answer: ________
Expected:
5Long answer / Show working
Solve the equation: 3x − 3 = x + 7.
Student writes solution steps here. Auto-grader compares final answer (
5) and rewards intermediate steps from the canonical solution above.#2intermediateanswer:
5Solve the equation: 6x − 7 = 4x + 3.
a=6b=7c=4d=3x=5variant=ax-b-cxd
A.
-5
→ Both numerator and denominator signs wrong simultaneously
B.
-2
→ When transposing −b, treated it as −b on RHS instead of +b (sign error)
C.
10
→ Correct numerator but forgot to divide by (a−c)
D.
5 ✓ correct
1.
Move x-terms to LHS and constants to RHS (transpose −7 as +7).
6x − 4x = 3 + 7
2.
Simplify.
2x = 10
3.
Divide both sides by 2.
x = 10 ÷ 2 = 5
6(5) − 7 = 23 = 4(5) + 3. ✓
Fill-in / Short Answer
Solve the equation: 6x − 7 = 4x + 3.
Answer: ________
Expected:
5Long answer / Show working
Solve the equation: 6x − 7 = 4x + 3.
Student writes solution steps here. Auto-grader compares final answer (
5) and rewards intermediate steps from the canonical solution above.#3intermediateanswer:
2Solve the equation: 6x − 3 = 4x + 1.
a=6b=3c=4d=1x=2variant=ax-b-cxd
A.
-2
→ Both numerator and denominator signs wrong simultaneously
B.
2 ✓ correct
C.
4
→ Correct numerator but forgot to divide by (a−c)
D.
-1
→ When transposing −b, treated it as −b on RHS instead of +b (sign error)
1.
Move x-terms to LHS and constants to RHS (transpose −3 as +3).
6x − 4x = 1 + 3
2.
Simplify.
2x = 4
3.
Divide both sides by 2.
x = 4 ÷ 2 = 2
6(2) − 3 = 9 = 4(2) + 1. ✓
Fill-in / Short Answer
Solve the equation: 6x − 3 = 4x + 1.
Answer: ________
Expected:
2Long answer / Show working
Solve the equation: 6x − 3 = 4x + 1.
Student writes solution steps here. Auto-grader compares final answer (
2) and rewards intermediate steps from the canonical solution above.#4intermediateanswer:
2Solve the equation: 6x − 1 = 5x + 1.
a=6b=1c=5d=1x=2variant=ax-b-cxd
A.
-2
→ Both numerator and denominator signs wrong simultaneously
B.
2 ✓ correct
C.
3
→ Off-by-one in root counting
D.
0
→ When transposing −b, treated it as −b on RHS instead of +b (sign error)
1.
Move x-terms to LHS and constants to RHS (transpose −1 as +1).
6x − 5x = 1 + 1
2.
Simplify.
1x = 2
3.
Divide both sides by 1.
x = 2 ÷ 1 = 2
6(2) − 1 = 11 = 5(2) + 1. ✓
Fill-in / Short Answer
Solve the equation: 6x − 1 = 5x + 1.
Answer: ________
Expected:
2Long answer / Show working
Solve the equation: 6x − 1 = 5x + 1.
Student writes solution steps here. Auto-grader compares final answer (
2) and rewards intermediate steps from the canonical solution above.#5intermediateanswer:
9Solve the equation: 5x − 1 = 2x + 26.
a=5b=1c=2d=26x=9variant=ax-b-cxd
A.
-9
→ Both numerator and denominator signs wrong simultaneously
B.
9 ✓ correct
C.
8
→ When transposing −b, treated it as −b on RHS instead of +b (sign error)
D.
27
→ Correct numerator but forgot to divide by (a−c)
1.
Move x-terms to LHS and constants to RHS (transpose −1 as +1).
5x − 2x = 26 + 1
2.
Simplify.
3x = 27
3.
Divide both sides by 3.
x = 27 ÷ 3 = 9
5(9) − 1 = 44 = 2(9) + 26. ✓
Fill-in / Short Answer
Solve the equation: 5x − 1 = 2x + 26.
Answer: ________
Expected:
9Long answer / Show working
Solve the equation: 5x − 1 = 2x + 26.
Student writes solution steps here. Auto-grader compares final answer (
9) and rewards intermediate steps from the canonical solution above.#6intermediateanswer:
4Solve the equation: 6x − 3 = 5x + 1.
a=6b=3c=5d=1x=4variant=ax-b-cxd
A.
-2
→ When transposing −b, treated it as −b on RHS instead of +b (sign error)
B.
-4
→ Both numerator and denominator signs wrong simultaneously
C.
5
→ Off-by-one in root counting
D.
4 ✓ correct
1.
Move x-terms to LHS and constants to RHS (transpose −3 as +3).
6x − 5x = 1 + 3
2.
Simplify.
1x = 4
3.
Divide both sides by 1.
x = 4 ÷ 1 = 4
6(4) − 3 = 21 = 5(4) + 1. ✓
Fill-in / Short Answer
Solve the equation: 6x − 3 = 5x + 1.
Answer: ________
Expected:
4Long answer / Show working
Solve the equation: 6x − 3 = 5x + 1.
Student writes solution steps here. Auto-grader compares final answer (
4) and rewards intermediate steps from the canonical solution above.#7intermediateanswer:
8Solve the equation: 7x − 2 = 5x + 14.
a=7b=2c=5d=14x=8variant=ax-b-cxd
A.
6
→ When transposing −b, treated it as −b on RHS instead of +b (sign error)
B.
8 ✓ correct
C.
-8
→ Both numerator and denominator signs wrong simultaneously
D.
16
→ Correct numerator but forgot to divide by (a−c)
1.
Move x-terms to LHS and constants to RHS (transpose −2 as +2).
7x − 5x = 14 + 2
2.
Simplify.
2x = 16
3.
Divide both sides by 2.
x = 16 ÷ 2 = 8
7(8) − 2 = 54 = 5(8) + 14. ✓
Fill-in / Short Answer
Solve the equation: 7x − 2 = 5x + 14.
Answer: ________
Expected:
8Long answer / Show working
Solve the equation: 7x − 2 = 5x + 14.
Student writes solution steps here. Auto-grader compares final answer (
8) and rewards intermediate steps from the canonical solution above.#8intermediateanswer:
10Solve the equation: 5x − 7 = 4x + 3.
a=5b=7c=4d=3x=10variant=ax-b-cxd
A.
-4
→ When transposing −b, treated it as −b on RHS instead of +b (sign error)
B.
11
→ Off-by-one in root counting
C.
10 ✓ correct
D.
-10
→ Both numerator and denominator signs wrong simultaneously
1.
Move x-terms to LHS and constants to RHS (transpose −7 as +7).
5x − 4x = 3 + 7
2.
Simplify.
1x = 10
3.
Divide both sides by 1.
x = 10 ÷ 1 = 10
5(10) − 7 = 43 = 4(10) + 3. ✓
Fill-in / Short Answer
Solve the equation: 5x − 7 = 4x + 3.
Answer: ________
Expected:
10Long answer / Show working
Solve the equation: 5x − 7 = 4x + 3.
Student writes solution steps here. Auto-grader compares final answer (
10) and rewards intermediate steps from the canonical solution above.s1s3-eq-fractionUnit 8 · Fractional linear equations
Fractional form: x/a ± b = c
Three difficulty tiers. Multiply through by denominator first.
Variety profile: 3 internal structural variants · 3 structural branches (basic / intermediate / advanced).
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1intermediateanswer:
14Solve the equation: x/2 − 2 = 5.
a=2b=2c=5x=14diff=5variant=fraction-minus
A.
2
→ Divided by a instead of multiplying (inverted operation)
B.
15
→ Off-by-one in root counting
C.
5
→ Forgot to multiply by a after isolating x/a term
D.
14 ✓ correct
1.
Subtract 2 from both sides.
x/2 = 5 − 2 = 5
2.
Multiply both sides by 2.
x = 5 × 2 = 14
14/2 + 2 = 5 + 2 = 5. ✓
Fill-in / Short Answer
Solve the equation: x/2 − 2 = 5.
Answer: ________
Expected:
14Long answer / Show working
Solve the equation: x/2 − 2 = 5.
Student writes solution steps here. Auto-grader compares final answer (
14) and rewards intermediate steps from the canonical solution above.#2basicanswer:
60Solve the equation: x/4 + 8 = 23.
a=4b=8c=23x=60diff=15variant=fraction
A.
124
→ Added b instead of subtracting — transposition sign error
B.
4
→ Divided by a instead of multiplying (inverted operation)
C.
60 ✓ correct
D.
15
→ Forgot to multiply by a after isolating x/a term
1.
Subtract 8 from both sides.
x/4 = 23 − 8 = 15
2.
Multiply both sides by 4.
x = 15 × 4 = 60
60/4 + 8 = 15 + 8 = 23. ✓
Fill-in / Short Answer
Solve the equation: x/4 + 8 = 23.
Answer: ________
Expected:
60Long answer / Show working
Solve the equation: x/4 + 8 = 23.
Student writes solution steps here. Auto-grader compares final answer (
60) and rewards intermediate steps from the canonical solution above.#3advancedanswer:
72Solve the equation: 2x/12 + 20 = 32.
a=12b=20c=32x=72diff=6variant=fraction-2x
A.
1
→ Divided by a instead of multiplying (inverted operation)
B.
72 ✓ correct
C.
624
→ Added b instead of subtracting — transposition sign error
D.
6
→ Forgot to multiply by a after isolating x/a term
1.
Subtract 20 from both sides.
x/12 = 32 − 20 = 6
2.
Multiply both sides by 12.
x = 6 × 12 = 72
72/12 + 20 = 6 + 20 = 32. ✓
Fill-in / Short Answer
Solve the equation: 2x/12 + 20 = 32.
Answer: ________
Expected:
72Long answer / Show working
Solve the equation: 2x/12 + 20 = 32.
Student writes solution steps here. Auto-grader compares final answer (
72) and rewards intermediate steps from the canonical solution above.#4basicanswer:
33Solve the equation: x/3 + 16 = 27.
a=3b=16c=27x=33diff=11variant=fraction
A.
4
→ Divided by a instead of multiplying (inverted operation)
B.
129
→ Added b instead of subtracting — transposition sign error
C.
11
→ Forgot to multiply by a after isolating x/a term
D.
33 ✓ correct
1.
Subtract 16 from both sides.
x/3 = 27 − 16 = 11
2.
Multiply both sides by 3.
x = 11 × 3 = 33
33/3 + 16 = 11 + 16 = 27. ✓
Fill-in / Short Answer
Solve the equation: x/3 + 16 = 27.
Answer: ________
Expected:
33Long answer / Show working
Solve the equation: x/3 + 16 = 27.
Student writes solution steps here. Auto-grader compares final answer (
33) and rewards intermediate steps from the canonical solution above.#5intermediateanswer:
108Solve the equation: x/12 − 1 = 8.
a=12b=1c=8x=108diff=8variant=fraction-minus
A.
1
→ Divided by a instead of multiplying (inverted operation)
B.
8
→ Forgot to multiply by a after isolating x/a term
C.
109
→ Off-by-one in root counting
D.
108 ✓ correct
1.
Subtract 1 from both sides.
x/12 = 8 − 1 = 8
2.
Multiply both sides by 12.
x = 8 × 12 = 108
108/12 + 1 = 8 + 1 = 8. ✓
Fill-in / Short Answer
Solve the equation: x/12 − 1 = 8.
Answer: ________
Expected:
108Long answer / Show working
Solve the equation: x/12 − 1 = 8.
Student writes solution steps here. Auto-grader compares final answer (
108) and rewards intermediate steps from the canonical solution above.#6intermediateanswer:
176Solve the equation: x/8 − 11 = 11.
a=8b=11c=11x=176diff=11variant=fraction-minus
A.
0
→ Divided by a instead of multiplying (inverted operation)
B.
11
→ Forgot to multiply by a after isolating x/a term
C.
176 ✓ correct
D.
177
→ Off-by-one in root counting
1.
Subtract 11 from both sides.
x/8 = 11 − 11 = 11
2.
Multiply both sides by 8.
x = 11 × 8 = 176
176/8 + 11 = 11 + 11 = 11. ✓
Fill-in / Short Answer
Solve the equation: x/8 − 11 = 11.
Answer: ________
Expected:
176Long answer / Show working
Solve the equation: x/8 − 11 = 11.
Student writes solution steps here. Auto-grader compares final answer (
176) and rewards intermediate steps from the canonical solution above.#7basicanswer:
99Solve the equation: x/11 + 7 = 16.
a=11b=7c=16x=99diff=9variant=fraction
A.
9
→ Forgot to multiply by a after isolating x/a term
B.
1
→ Divided by a instead of multiplying (inverted operation)
C.
253
→ Added b instead of subtracting — transposition sign error
D.
99 ✓ correct
1.
Subtract 7 from both sides.
x/11 = 16 − 7 = 9
2.
Multiply both sides by 11.
x = 9 × 11 = 99
99/11 + 7 = 9 + 7 = 16. ✓
Fill-in / Short Answer
Solve the equation: x/11 + 7 = 16.
Answer: ________
Expected:
99Long answer / Show working
Solve the equation: x/11 + 7 = 16.
Student writes solution steps here. Auto-grader compares final answer (
99) and rewards intermediate steps from the canonical solution above.#8basicanswer:
18Solve the equation: x/6 + 9 = 12.
a=6b=9c=12x=18diff=3variant=fraction
A.
3
→ Forgot to multiply by a after isolating x/a term
B.
1
→ Divided by a instead of multiplying (inverted operation)
C.
126
→ Added b instead of subtracting — transposition sign error
D.
18 ✓ correct
1.
Subtract 9 from both sides.
x/6 = 12 − 9 = 3
2.
Multiply both sides by 6.
x = 3 × 6 = 18
18/6 + 9 = 3 + 9 = 12. ✓
Fill-in / Short Answer
Solve the equation: x/6 + 9 = 12.
Answer: ________
Expected:
18Long answer / Show working
Solve the equation: x/6 + 9 = 12.
Student writes solution steps here. Auto-grader compares final answer (
18) and rewards intermediate steps from the canonical solution above.s1s3-eq-bracket-bothUnit 8 · Solve linear equations
Brackets both sides: a(x+b) = c(x+d)
Expand brackets, then solve. Tests distributive law + sign tracking.
Variety profile: 1 internal structural variant · Retry + fallback. Single algebraic shape.
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1advancedanswer:
1Solve the equation: 9(x + 1) = 6(x + 2).
a=9b=1c=6d=2x=1variant=bracket-both
A.
-1
→ Sign error in numerator — subtracted in wrong order
B.
0
→ Added a and c in denominator instead of subtracting
C.
2
→ Factor of 2 error
D.
1 ✓ correct
1.
Expand both brackets.
9x + 9 = 6x + 12
2.
Collect x-terms on LHS, constants on RHS.
3x = 3
3.
Divide both sides by 3.
x = 3 ÷ 3 = 1
9(1+1) = 18 = 6(1+2). ✓
Fill-in / Short Answer
Solve the equation: 9(x + 1) = 6(x + 2).
Answer: ________
Expected:
1Long answer / Show working
Solve the equation: 9(x + 1) = 6(x + 2).
Student writes solution steps here. Auto-grader compares final answer (
1) and rewards intermediate steps from the canonical solution above.#2advancedanswer:
2Solve the equation: 10(x + 2) = 8(x + 3).
a=10b=2c=8d=3x=2variant=bracket-both
A.
2 ✓ correct
B.
0
→ Added a and c in denominator instead of subtracting
C.
1
→ Equated bracket contents without expanding (forgot distributive law)
D.
-2
→ Sign error in numerator — subtracted in wrong order
1.
Expand both brackets.
10x + 20 = 8x + 24
2.
Collect x-terms on LHS, constants on RHS.
2x = 4
3.
Divide both sides by 2.
x = 4 ÷ 2 = 2
10(2+2) = 40 = 8(2+3). ✓
Fill-in / Short Answer
Solve the equation: 10(x + 2) = 8(x + 3).
Answer: ________
Expected:
2Long answer / Show working
Solve the equation: 10(x + 2) = 8(x + 3).
Student writes solution steps here. Auto-grader compares final answer (
2) and rewards intermediate steps from the canonical solution above.#3advancedanswer:
8Solve the equation: 6(x + 12) = 5(x + 16).
a=6b=12c=5d=16x=8variant=bracket-both
A.
8 ✓ correct
B.
4
→ Equated bracket contents without expanding (forgot distributive law)
C.
-8
→ Sign error in numerator — subtracted in wrong order
D.
1
→ Added a and c in denominator instead of subtracting
1.
Expand both brackets.
6x + 72 = 5x + 80
2.
Collect x-terms on LHS, constants on RHS.
1x = 8
3.
Divide both sides by 1.
x = 8 ÷ 1 = 8
6(8+12) = 120 = 5(8+16). ✓
Fill-in / Short Answer
Solve the equation: 6(x + 12) = 5(x + 16).
Answer: ________
Expected:
8Long answer / Show working
Solve the equation: 6(x + 12) = 5(x + 16).
Student writes solution steps here. Auto-grader compares final answer (
8) and rewards intermediate steps from the canonical solution above.#4advancedanswer:
3Solve the equation: 12(x + 4) = 6(x + 11).
a=12b=4c=6d=11x=3variant=bracket-both
A.
7
→ Equated bracket contents without expanding (forgot distributive law)
B.
-3
→ Sign error in numerator — subtracted in wrong order
C.
3 ✓ correct
D.
1
→ Added a and c in denominator instead of subtracting
1.
Expand both brackets.
12x + 48 = 6x + 66
2.
Collect x-terms on LHS, constants on RHS.
6x = 18
3.
Divide both sides by 6.
x = 18 ÷ 6 = 3
12(3+4) = 84 = 6(3+11). ✓
Fill-in / Short Answer
Solve the equation: 12(x + 4) = 6(x + 11).
Answer: ________
Expected:
3Long answer / Show working
Solve the equation: 12(x + 4) = 6(x + 11).
Student writes solution steps here. Auto-grader compares final answer (
3) and rewards intermediate steps from the canonical solution above.#5advancedanswer:
3Solve the equation: 6(x + 1) = 2(x + 9).
a=6b=1c=2d=9x=3variant=bracket-both
A.
8
→ Equated bracket contents without expanding (forgot distributive law)
B.
2
→ Added a and c in denominator instead of subtracting
C.
-3
→ Sign error in numerator — subtracted in wrong order
D.
3 ✓ correct
1.
Expand both brackets.
6x + 6 = 2x + 18
2.
Collect x-terms on LHS, constants on RHS.
4x = 12
3.
Divide both sides by 4.
x = 12 ÷ 4 = 3
6(3+1) = 24 = 2(3+9). ✓
Fill-in / Short Answer
Solve the equation: 6(x + 1) = 2(x + 9).
Answer: ________
Expected:
3Long answer / Show working
Solve the equation: 6(x + 1) = 2(x + 9).
Student writes solution steps here. Auto-grader compares final answer (
3) and rewards intermediate steps from the canonical solution above.#6advancedanswer:
15Solve the equation: 4(x + 1) = 2(x + 17).
a=4b=1c=2d=17x=15variant=bracket-both
A.
16
→ Equated bracket contents without expanding (forgot distributive law)
B.
5
→ Added a and c in denominator instead of subtracting
C.
15 ✓ correct
D.
-15
→ Sign error in numerator — subtracted in wrong order
1.
Expand both brackets.
4x + 4 = 2x + 34
2.
Collect x-terms on LHS, constants on RHS.
2x = 30
3.
Divide both sides by 2.
x = 30 ÷ 2 = 15
4(15+1) = 64 = 2(15+17). ✓
Fill-in / Short Answer
Solve the equation: 4(x + 1) = 2(x + 17).
Answer: ________
Expected:
15Long answer / Show working
Solve the equation: 4(x + 1) = 2(x + 17).
Student writes solution steps here. Auto-grader compares final answer (
15) and rewards intermediate steps from the canonical solution above.#7advancedanswer:
6Solve the equation: 12(x + 8) = 7(x + 18).
a=12b=8c=7d=18x=6variant=bracket-both
A.
10
→ Equated bracket contents without expanding (forgot distributive law)
B.
6 ✓ correct
C.
-6
→ Sign error in numerator — subtracted in wrong order
D.
2
→ Added a and c in denominator instead of subtracting
1.
Expand both brackets.
12x + 96 = 7x + 126
2.
Collect x-terms on LHS, constants on RHS.
5x = 30
3.
Divide both sides by 5.
x = 30 ÷ 5 = 6
12(6+8) = 168 = 7(6+18). ✓
Fill-in / Short Answer
Solve the equation: 12(x + 8) = 7(x + 18).
Answer: ________
Expected:
6Long answer / Show working
Solve the equation: 12(x + 8) = 7(x + 18).
Student writes solution steps here. Auto-grader compares final answer (
6) and rewards intermediate steps from the canonical solution above.#8advancedanswer:
1Solve the equation: 6(x + 6) = 3(x + 13).
a=6b=6c=3d=13x=1variant=bracket-both
A.
7
→ Equated bracket contents without expanding (forgot distributive law)
B.
0
→ Added a and c in denominator instead of subtracting
C.
1 ✓ correct
D.
-1
→ Sign error in numerator — subtracted in wrong order
1.
Expand both brackets.
6x + 36 = 3x + 39
2.
Collect x-terms on LHS, constants on RHS.
3x = 3
3.
Divide both sides by 3.
x = 3 ÷ 3 = 1
6(1+6) = 42 = 3(1+13). ✓
Fill-in / Short Answer
Solve the equation: 6(x + 6) = 3(x + 13).
Answer: ________
Expected:
1Long answer / Show working
Solve the equation: 6(x + 6) = 3(x + 13).
Student writes solution steps here. Auto-grader compares final answer (
1) and rewards intermediate steps from the canonical solution above.Unit 8
Solve word problems
Curriculum objective in Linear equations in one unknown. 2 generators currently serve this objective.
⚠ Reference shapes seen on disk but NOT yet served by these 2 generators (3 candidates):
- Consecutive integers / consecutive even numbers
- "k times age t years ago" structure (e.g. "Ceci is 3× her age 10 years ago")
- Fraction-of-population bus problems (passengers got off, then got on)
Variety dashboard · all 2 generators side-by-side
Compare structural variant counts, observed difficulty mix, and a representative stem from each. Click a row's button to jump to the full sample set below.
| Generator | Variants | Difficulty mix (8 samples) | First sample stem | |
|---|---|---|---|---|
Word: Ages1s3-eq-word-age | 3v 3 phrase variants. Reference bank also has "k times age t years ago" structure n… | intermediate: 8 | The total of Ben's age and Amy's age is 78. The difference between their ages is 20, with … | |
Word: Prices1s3-eq-word-price | 8v 8 item/context tuples (same structure). Variety from rotating items. | intermediate: 8 | A mango costs $6 more than a banana. 3 mangos and 4 bananas cost $137 altogether. Find the… |
Drill into one generator · 8 samples each
s1s3-eq-word-ageUnit 8 · Word problems
Word problem: Age
Two-person age problem. Translate "now / in N years" into a linear equation.
Variety profile: 3 internal structural variants · 3 phrase variants. Reference bank also has "k times age t years ago" structure not yet built.
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1intermediateanswer:
49The total of Ben's age and Amy's age is 78. The difference between their ages is 20, with Ben being older. Find the age of Ben.
K=20ageA=49ageB=29S=78nameA=BennameB=Amyvariant=word-age
A.
29
→ Found younger person's age instead of older person's (solved for wrong variable)
B.
39
→ Ignored the age gap — just halved the sum
C.
58
→ Subtracted age difference from sum directly without setting up equation
D.
49 ✓ correct
1.
Let Amy’s age = x. Then Ben’s age = x + 20.
2.
Sum equation: x + (x + 20) = 78.
2x + 20 = 78
3.
Solve for x (Amy’s age).
x = (78 − 20) ÷ 2 = 29
4.
Ben’s age = Amy + 20.
Ben = 29 + 20 = 49
49 + 29 = 78. Ben is 20 years older. ✓
Fill-in / Short Answer
The total of Ben's age and Amy's age is 78. The difference between their ages is 20, with Ben being older. Find the age of Ben.
Answer: ________
Expected:
49Long answer / Show working
The total of Ben's age and Amy's age is 78. The difference between their ages is 20, with Ben being older. Find the age of Ben.
Student writes solution steps here. Auto-grader compares final answer (
49) and rewards intermediate steps from the canonical solution above.#2intermediateanswer:
31The sum of Ray's age and Sue's age is 50. Ray is 12 years older than Sue. Find Ray's age.
K=12ageA=31ageB=19S=50nameA=RaynameB=Suevariant=word-age
A.
31 ✓ correct
B.
25
→ Ignored the age gap — just halved the sum
C.
19
→ Found younger person's age instead of older person's (solved for wrong variable)
D.
38
→ Subtracted age difference from sum directly without setting up equation
1.
Let Sue’s age = x. Then Ray’s age = x + 12.
2.
Sum equation: x + (x + 12) = 50.
2x + 12 = 50
3.
Solve for x (Sue’s age).
x = (50 − 12) ÷ 2 = 19
4.
Ray’s age = Sue + 12.
Ray = 19 + 12 = 31
31 + 19 = 50. Ray is 12 years older. ✓
Fill-in / Short Answer
The sum of Ray's age and Sue's age is 50. Ray is 12 years older than Sue. Find Ray's age.
Answer: ________
Expected:
31Long answer / Show working
The sum of Ray's age and Sue's age is 50. Ray is 12 years older than Sue. Find Ray's age.
Student writes solution steps here. Auto-grader compares final answer (
31) and rewards intermediate steps from the canonical solution above.#3intermediateanswer:
49Ben and Amy have a total age of 68. Amy is 30 years younger than Ben. How old is Ben?
K=30ageA=49ageB=19S=68nameA=BennameB=Amyvariant=word-age
A.
38
→ Subtracted age difference from sum directly without setting up equation
B.
19
→ Found younger person's age instead of older person's (solved for wrong variable)
C.
34
→ Ignored the age gap — just halved the sum
D.
49 ✓ correct
1.
Let Amy’s age = x. Then Ben’s age = x + 30.
2.
Sum equation: x + (x + 30) = 68.
2x + 30 = 68
3.
Solve for x (Amy’s age).
x = (68 − 30) ÷ 2 = 19
4.
Ben’s age = Amy + 30.
Ben = 19 + 30 = 49
49 + 19 = 68. Ben is 30 years older. ✓
Fill-in / Short Answer
Ben and Amy have a total age of 68. Amy is 30 years younger than Ben. How old is Ben?
Answer: ________
Expected:
49Long answer / Show working
Ben and Amy have a total age of 68. Amy is 30 years younger than Ben. How old is Ben?
Student writes solution steps here. Auto-grader compares final answer (
49) and rewards intermediate steps from the canonical solution above.#4intermediateanswer:
46Peter and Mary have a total age of 82. Mary is 10 years younger than Peter. How old is Peter?
K=10ageA=46ageB=36S=82nameA=PeternameB=Maryvariant=word-age
A.
41
→ Ignored the age gap — just halved the sum
B.
36
→ Found younger person's age instead of older person's (solved for wrong variable)
C.
72
→ Subtracted age difference from sum directly without setting up equation
D.
46 ✓ correct
1.
Let Mary’s age = x. Then Peter’s age = x + 10.
2.
Sum equation: x + (x + 10) = 82.
2x + 10 = 82
3.
Solve for x (Mary’s age).
x = (82 − 10) ÷ 2 = 36
4.
Peter’s age = Mary + 10.
Peter = 36 + 10 = 46
46 + 36 = 82. Peter is 10 years older. ✓
Fill-in / Short Answer
Peter and Mary have a total age of 82. Mary is 10 years younger than Peter. How old is Peter?
Answer: ________
Expected:
46Long answer / Show working
Peter and Mary have a total age of 82. Mary is 10 years younger than Peter. How old is Peter?
Student writes solution steps here. Auto-grader compares final answer (
46) and rewards intermediate steps from the canonical solution above.#5intermediateanswer:
36The total of Ken's age and Lily's age is 52. The difference between their ages is 20, with Ken being older. Find the age of Ken.
K=20ageA=36ageB=16S=52nameA=KennameB=Lilyvariant=word-age
A.
32
→ Subtracted age difference from sum directly without setting up equation
B.
26
→ Ignored the age gap — just halved the sum
C.
16
→ Found younger person's age instead of older person's (solved for wrong variable)
D.
36 ✓ correct
1.
Let Lily’s age = x. Then Ken’s age = x + 20.
2.
Sum equation: x + (x + 20) = 52.
2x + 20 = 52
3.
Solve for x (Lily’s age).
x = (52 − 20) ÷ 2 = 16
4.
Ken’s age = Lily + 20.
Ken = 16 + 20 = 36
36 + 16 = 52. Ken is 20 years older. ✓
Fill-in / Short Answer
The total of Ken's age and Lily's age is 52. The difference between their ages is 20, with Ken being older. Find the age of Ken.
Answer: ________
Expected:
36Long answer / Show working
The total of Ken's age and Lily's age is 52. The difference between their ages is 20, with Ken being older. Find the age of Ken.
Student writes solution steps here. Auto-grader compares final answer (
36) and rewards intermediate steps from the canonical solution above.#6intermediateanswer:
23The total of Mike's age and Anna's age is 36. The difference between their ages is 10, with Mike being older. Find the age of Mike.
K=10ageA=23ageB=13S=36nameA=MikenameB=Annavariant=word-age
A.
18
→ Ignored the age gap — just halved the sum
B.
23 ✓ correct
C.
13
→ Found younger person's age instead of older person's (solved for wrong variable)
D.
26
→ Subtracted age difference from sum directly without setting up equation
1.
Let Anna’s age = x. Then Mike’s age = x + 10.
2.
Sum equation: x + (x + 10) = 36.
2x + 10 = 36
3.
Solve for x (Anna’s age).
x = (36 − 10) ÷ 2 = 13
4.
Mike’s age = Anna + 10.
Mike = 13 + 10 = 23
23 + 13 = 36. Mike is 10 years older. ✓
Fill-in / Short Answer
The total of Mike's age and Anna's age is 36. The difference between their ages is 10, with Mike being older. Find the age of Mike.
Answer: ________
Expected:
23Long answer / Show working
The total of Mike's age and Anna's age is 36. The difference between their ages is 10, with Mike being older. Find the age of Mike.
Student writes solution steps here. Auto-grader compares final answer (
23) and rewards intermediate steps from the canonical solution above.#7intermediateanswer:
35The total of Peter's age and Mary's age is 66. The difference between their ages is 4, with Peter being older. Find the age of Peter.
K=4ageA=35ageB=31S=66nameA=PeternameB=Maryvariant=word-age
A.
62
→ Subtracted age difference from sum directly without setting up equation
B.
35 ✓ correct
C.
33
→ Ignored the age gap — just halved the sum
D.
31
→ Found younger person's age instead of older person's (solved for wrong variable)
1.
Let Mary’s age = x. Then Peter’s age = x + 4.
2.
Sum equation: x + (x + 4) = 66.
2x + 4 = 66
3.
Solve for x (Mary’s age).
x = (66 − 4) ÷ 2 = 31
4.
Peter’s age = Mary + 4.
Peter = 31 + 4 = 35
35 + 31 = 66. Peter is 4 years older. ✓
Fill-in / Short Answer
The total of Peter's age and Mary's age is 66. The difference between their ages is 4, with Peter being older. Find the age of Peter.
Answer: ________
Expected:
35Long answer / Show working
The total of Peter's age and Mary's age is 66. The difference between their ages is 4, with Peter being older. Find the age of Peter.
Student writes solution steps here. Auto-grader compares final answer (
35) and rewards intermediate steps from the canonical solution above.#8intermediateanswer:
31The total of Mike's age and Anna's age is 38. The difference between their ages is 24, with Mike being older. Find the age of Mike.
K=24ageA=31ageB=7S=38nameA=MikenameB=Annavariant=word-age
A.
31 ✓ correct
B.
14
→ Subtracted age difference from sum directly without setting up equation
C.
19
→ Ignored the age gap — just halved the sum
D.
7
→ Found younger person's age instead of older person's (solved for wrong variable)
1.
Let Anna’s age = x. Then Mike’s age = x + 24.
2.
Sum equation: x + (x + 24) = 38.
2x + 24 = 38
3.
Solve for x (Anna’s age).
x = (38 − 24) ÷ 2 = 7
4.
Mike’s age = Anna + 24.
Mike = 7 + 24 = 31
31 + 7 = 38. Mike is 24 years older. ✓
Fill-in / Short Answer
The total of Mike's age and Anna's age is 38. The difference between their ages is 24, with Mike being older. Find the age of Mike.
Answer: ________
Expected:
31Long answer / Show working
The total of Mike's age and Anna's age is 38. The difference between their ages is 24, with Mike being older. Find the age of Mike.
Student writes solution steps here. Auto-grader compares final answer (
31) and rewards intermediate steps from the canonical solution above.s1s3-eq-word-priceUnit 8 · Word problems
Word problem: Price difference
Two items with a price difference and a total. Set up and solve a linear equation.
Variety profile: 8 internal structural variants · 8 item/context tuples (same structure). Variety from rotating items.
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1intermediateanswer:
17A mango costs $6 more than a banana. 3 mangos and 4 bananas cost $137 altogether. Find the cost of a banana in dollars.
costA=23costB=17diff=6total=137m=3n=4itemA=mangoitemB=bananavariant=word-price
A.
20
→ Ignored the price difference entirely — just divided total by item count
B.
17 ✓ correct
C.
19
→ Subtracted diff only once instead of multiplying by item count
D.
18
→ Off-by-one in root counting
1.
Let cost of banana = $x. Cost of mango = $(x + 6).
2.
Set up equation.
3(x + 6) + 4x = 137
3.
Expand and simplify.
7x + 18 = 137 ⇒ 7x = 119
4.
Solve for x.
x = 119 ÷ 7 = 17
3(23) + 4(17) = 137 = 137. ✓
Fill-in / Short Answer
A mango costs $6 more than a banana. 3 mangos and 4 bananas cost $137 altogether. Find the cost of a banana in dollars.
Answer: ________
Expected:
17Long answer / Show working
A mango costs $6 more than a banana. 3 mangos and 4 bananas cost $137 altogether. Find the cost of a banana in dollars.
Student writes solution steps here. Auto-grader compares final answer (
17) and rewards intermediate steps from the canonical solution above.#2intermediateanswer:
18A ticket costs $5 more than a coupon. 4 tickets and 1 coupons cost $110 altogether. Find the cost of a coupon in dollars.
costA=23costB=18diff=5total=110m=4n=1itemA=ticketitemB=couponvariant=word-price
A.
18 ✓ correct
B.
21
→ Subtracted diff only once instead of multiplying by item count
C.
19
→ Off-by-one in root counting
D.
22
→ Ignored the price difference entirely — just divided total by item count
1.
Let cost of coupon = $x. Cost of ticket = $(x + 5).
2.
Set up equation.
4(x + 5) + 1x = 110
3.
Expand and simplify.
5x + 20 = 110 ⇒ 5x = 90
4.
Solve for x.
x = 90 ÷ 5 = 18
4(23) + 1(18) = 110 = 110. ✓
Fill-in / Short Answer
A ticket costs $5 more than a coupon. 4 tickets and 1 coupons cost $110 altogether. Find the cost of a coupon in dollars.
Answer: ________
Expected:
18Long answer / Show working
A ticket costs $5 more than a coupon. 4 tickets and 1 coupons cost $110 altogether. Find the cost of a coupon in dollars.
Student writes solution steps here. Auto-grader compares final answer (
18) and rewards intermediate steps from the canonical solution above.#3intermediateanswer:
10A apple costs $9 more than a orange. 4 apples and 3 oranges cost $106 altogether. Find the cost of a orange in dollars.
costA=19costB=10diff=9total=106m=4n=3itemA=appleitemB=orangevariant=word-price
A.
10 ✓ correct
B.
15
→ Ignored the price difference entirely — just divided total by item count
C.
11
→ Off-by-one in root counting
D.
14
→ Subtracted diff only once instead of multiplying by item count
1.
Let cost of orange = $x. Cost of apple = $(x + 9).
2.
Set up equation.
4(x + 9) + 3x = 106
3.
Expand and simplify.
7x + 36 = 106 ⇒ 7x = 70
4.
Solve for x.
x = 70 ÷ 7 = 10
4(19) + 3(10) = 106 = 106. ✓
Fill-in / Short Answer
A apple costs $9 more than a orange. 4 apples and 3 oranges cost $106 altogether. Find the cost of a orange in dollars.
Answer: ________
Expected:
10Long answer / Show working
A apple costs $9 more than a orange. 4 apples and 3 oranges cost $106 altogether. Find the cost of a orange in dollars.
Student writes solution steps here. Auto-grader compares final answer (
10) and rewards intermediate steps from the canonical solution above.#4intermediateanswer:
15A apple costs $6 more than a orange. 4 apples and 3 oranges cost $129 altogether. Find the cost of a orange in dollars.
costA=21costB=15diff=6total=129m=4n=3itemA=appleitemB=orangevariant=word-price
A.
18
→ Subtracted diff only once instead of multiplying by item count
B.
14
→ Off-by-one in root counting
C.
16
→ Off-by-one in root counting
D.
15 ✓ correct
1.
Let cost of orange = $x. Cost of apple = $(x + 6).
2.
Set up equation.
4(x + 6) + 3x = 129
3.
Expand and simplify.
7x + 24 = 129 ⇒ 7x = 105
4.
Solve for x.
x = 105 ÷ 7 = 15
4(21) + 3(15) = 129 = 129. ✓
Fill-in / Short Answer
A apple costs $6 more than a orange. 4 apples and 3 oranges cost $129 altogether. Find the cost of a orange in dollars.
Answer: ________
Expected:
15Long answer / Show working
A apple costs $6 more than a orange. 4 apples and 3 oranges cost $129 altogether. Find the cost of a orange in dollars.
Student writes solution steps here. Auto-grader compares final answer (
15) and rewards intermediate steps from the canonical solution above.#5intermediateanswer:
24A marker costs $11 more than a crayon. 2 markers and 3 crayons cost $142 altogether. Find the cost of a crayon in dollars.
costA=35costB=24diff=11total=142m=2n=3itemA=markeritemB=crayonvariant=word-price
A.
28
→ Ignored the price difference entirely — just divided total by item count
B.
26
→ Subtracted diff only once instead of multiplying by item count
C.
25
→ Off-by-one in root counting
D.
24 ✓ correct
1.
Let cost of crayon = $x. Cost of marker = $(x + 11).
2.
Set up equation.
2(x + 11) + 3x = 142
3.
Expand and simplify.
5x + 22 = 142 ⇒ 5x = 120
4.
Solve for x.
x = 120 ÷ 5 = 24
2(35) + 3(24) = 142 = 142. ✓
Fill-in / Short Answer
A marker costs $11 more than a crayon. 2 markers and 3 crayons cost $142 altogether. Find the cost of a crayon in dollars.
Answer: ________
Expected:
24Long answer / Show working
A marker costs $11 more than a crayon. 2 markers and 3 crayons cost $142 altogether. Find the cost of a crayon in dollars.
Student writes solution steps here. Auto-grader compares final answer (
24) and rewards intermediate steps from the canonical solution above.#6intermediateanswer:
19A cookie costs $2 more than a biscuit. 5 cookies and 2 biscuits cost $143 altogether. Find the cost of a biscuit in dollars.
costA=21costB=19diff=2total=143m=5n=2itemA=cookieitemB=biscuitvariant=word-price
A.
18
→ Off-by-one in root counting
B.
19 ✓ correct
C.
20
→ Subtracted diff only once instead of multiplying by item count
D.
21
→ Off-by-one in root counting
1.
Let cost of biscuit = $x. Cost of cookie = $(x + 2).
2.
Set up equation.
5(x + 2) + 2x = 143
3.
Expand and simplify.
7x + 10 = 143 ⇒ 7x = 133
4.
Solve for x.
x = 133 ÷ 7 = 19
5(21) + 2(19) = 143 = 143. ✓
Fill-in / Short Answer
A cookie costs $2 more than a biscuit. 5 cookies and 2 biscuits cost $143 altogether. Find the cost of a biscuit in dollars.
Answer: ________
Expected:
19Long answer / Show working
A cookie costs $2 more than a biscuit. 5 cookies and 2 biscuits cost $143 altogether. Find the cost of a biscuit in dollars.
Student writes solution steps here. Auto-grader compares final answer (
19) and rewards intermediate steps from the canonical solution above.#7intermediateanswer:
18A cookie costs $5 more than a biscuit. 5 cookies and 2 biscuits cost $151 altogether. Find the cost of a biscuit in dollars.
costA=23costB=18diff=5total=151m=5n=2itemA=cookieitemB=biscuitvariant=word-price
A.
22
→ Ignored the price difference entirely — just divided total by item count
B.
18 ✓ correct
C.
19
→ Off-by-one in root counting
D.
21
→ Subtracted diff only once instead of multiplying by item count
1.
Let cost of biscuit = $x. Cost of cookie = $(x + 5).
2.
Set up equation.
5(x + 5) + 2x = 151
3.
Expand and simplify.
7x + 25 = 151 ⇒ 7x = 126
4.
Solve for x.
x = 126 ÷ 7 = 18
5(23) + 2(18) = 151 = 151. ✓
Fill-in / Short Answer
A cookie costs $5 more than a biscuit. 5 cookies and 2 biscuits cost $151 altogether. Find the cost of a biscuit in dollars.
Answer: ________
Expected:
18Long answer / Show working
A cookie costs $5 more than a biscuit. 5 cookies and 2 biscuits cost $151 altogether. Find the cost of a biscuit in dollars.
Student writes solution steps here. Auto-grader compares final answer (
18) and rewards intermediate steps from the canonical solution above.#8intermediateanswer:
30A marker costs $11 more than a crayon. 2 markers and 3 crayons cost $172 altogether. Find the cost of a crayon in dollars.
costA=41costB=30diff=11total=172m=2n=3itemA=markeritemB=crayonvariant=word-price
A.
34
→ Ignored the price difference entirely — just divided total by item count
B.
30 ✓ correct
C.
31
→ Off-by-one in root counting
D.
32
→ Subtracted diff only once instead of multiplying by item count
1.
Let cost of crayon = $x. Cost of marker = $(x + 11).
2.
Set up equation.
2(x + 11) + 3x = 172
3.
Expand and simplify.
5x + 22 = 172 ⇒ 5x = 150
4.
Solve for x.
x = 150 ÷ 5 = 30
2(41) + 3(30) = 172 = 172. ✓
Fill-in / Short Answer
A marker costs $11 more than a crayon. 2 markers and 3 crayons cost $172 altogether. Find the cost of a crayon in dollars.
Answer: ________
Expected:
30Long answer / Show working
A marker costs $11 more than a crayon. 2 markers and 3 crayons cost $172 altogether. Find the cost of a crayon in dollars.
Student writes solution steps here. Auto-grader compares final answer (
30) and rewards intermediate steps from the canonical solution above.Unit 9
Solve simultaneous equations (substitution, elimination)
Curriculum objective in Linear equations in two unknowns. 4 generators currently serve this objective.
⚠ Reference shapes seen on disk but NOT yet served by these 4 generators (2 candidates):
- Method-step MC ("If we eliminate y first, which is the first step?")
- Pick which equation to substitute into (process-aware MC)
Variety dashboard · all 4 generators side-by-side
Compare structural variant counts, observed difficulty mix, and a representative stem from each. Click a row's button to jump to the full sample set below.
| Generator | Variants | Difficulty mix (8 samples) | First sample stem | |
|---|---|---|---|---|
Simul: substitutions1s3-simul-sub | 1v Single algebraic shape; variety from coefficient sampling. | intermediate: 8 | Solve the simultaneous equations: y = 4x − 36 6x + y = 64 | |
Simul: elim (add)s1s3-simul-elim-add | 1v Loop + fallback. Single shape. | intermediate: 8 | Solve the simultaneous equations: 8x + y = 54 8x − y = 42 | |
Simul: elim (subtract)s1s3-simul-elim-sub | 1v Single algebraic shape. | intermediate: 8 | Solve the simultaneous equations: 4x + 6y = 30 4x + 5y = 27 | |
Simul: fraction coeff.s1s3-simul-fraction-coeff | 4v 4 denominator pairs. | advanced: 8 | Solve the simultaneous equations: x/2 + y = 20 x/4 + y = 17 |
Drill into one generator · 8 samples each
s1s3-simul-subUnit 9 · Simultaneous (substitution)
Simultaneous: substitution method
One equation gives y in terms of x — substitute into the other and solve.
Variety profile: 1 internal structural variant · Single algebraic shape; variety from coefficient sampling.
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1intermediateanswer:
x = 10, y = 4Solve the simultaneous equations:
y = 4x − 36
6x + y = 64
a=4b=-36c=6d=1e=64x=10y=4variant=substitution
A.
x = 9, y = 4
→ Common computation error
B.
x = 11, y = 4
→ Common computation error
C.
x = 4, y = 10
→ Swapped x and y values in final answer
D.
x = 10, y = 4 ✓ correct
1.
Eq1 already gives y = 4x−36. Substitute into Eq2.
6x + 1(4x−36) = 64
2.
Expand and collect x-terms.
10x = 100
3.
Solve for x.
x = 100 ÷ 10 = 10
4.
Back-substitute into Eq1 for y.
y = 4(10)−36 = 4
6(10) + 1(4) = 64. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
y = 4x − 36
6x + y = 64
Answer: ________
Expected:
x = 10, y = 4Long answer / Show working
Solve the simultaneous equations:
y = 4x − 36
6x + y = 64
Student writes solution steps here. Auto-grader compares final answer (
x = 10, y = 4) and rewards intermediate steps from the canonical solution above.#2intermediateanswer:
x = 11, y = 6Solve the simultaneous equations:
y = 4x − 38
4x + 2y = 56
a=4b=-38c=4d=2e=56x=11y=6variant=substitution
A.
x = 11, y = 6 ✓ correct
B.
x = 12, y = 6
→ Common computation error
C.
x = 6, y = 11
→ Swapped x and y values in final answer
D.
x = 10, y = 6
→ Common computation error
1.
Eq1 already gives y = 4x−38. Substitute into Eq2.
4x + 2(4x−38) = 56
2.
Expand and collect x-terms.
12x = 132
3.
Solve for x.
x = 132 ÷ 12 = 11
4.
Back-substitute into Eq1 for y.
y = 4(11)−38 = 6
4(11) + 2(6) = 56. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
y = 4x − 38
4x + 2y = 56
Answer: ________
Expected:
x = 11, y = 6Long answer / Show working
Solve the simultaneous equations:
y = 4x − 38
4x + 2y = 56
Student writes solution steps here. Auto-grader compares final answer (
x = 11, y = 6) and rewards intermediate steps from the canonical solution above.#3intermediateanswer:
x = 3, y = 13Solve the simultaneous equations:
y = 3x + 4
4x + 5y = 77
a=3b=4c=4d=5e=77x=3y=13variant=substitution
A.
x = 3, y = 5
→ Sign error when computing y from the isolated equation (treated −b as +b)
B.
x = 13, y = 3
→ Swapped x and y values in final answer
C.
x = 3, y = 13 ✓ correct
D.
x = 4, y = 13
→ Common computation error
1.
Eq1 already gives y = 3x+4. Substitute into Eq2.
4x + 5(3x+4) = 77
2.
Expand and collect x-terms.
19x = 57
3.
Solve for x.
x = 57 ÷ 19 = 3
4.
Back-substitute into Eq1 for y.
y = 3(3)+4 = 13
4(3) + 5(13) = 77. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
y = 3x + 4
4x + 5y = 77
Answer: ________
Expected:
x = 3, y = 13Long answer / Show working
Solve the simultaneous equations:
y = 3x + 4
4x + 5y = 77
Student writes solution steps here. Auto-grader compares final answer (
x = 3, y = 13) and rewards intermediate steps from the canonical solution above.#4intermediateanswer:
x = 6, y = 12Solve the simultaneous equations:
y = 3x − 6
5x + 5y = 90
a=3b=-6c=5d=5e=90x=6y=12variant=substitution
A.
x = 6, y = 12 ✓ correct
B.
x = 12, y = 6
→ Swapped x and y values in final answer
C.
x = 7, y = 12
→ Common computation error
D.
x = 5, y = 12
→ Common computation error
1.
Eq1 already gives y = 3x−6. Substitute into Eq2.
5x + 5(3x−6) = 90
2.
Expand and collect x-terms.
20x = 120
3.
Solve for x.
x = 120 ÷ 20 = 6
4.
Back-substitute into Eq1 for y.
y = 3(6)−6 = 12
5(6) + 5(12) = 90. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
y = 3x − 6
5x + 5y = 90
Answer: ________
Expected:
x = 6, y = 12Long answer / Show working
Solve the simultaneous equations:
y = 3x − 6
5x + 5y = 90
Student writes solution steps here. Auto-grader compares final answer (
x = 6, y = 12) and rewards intermediate steps from the canonical solution above.#5intermediateanswer:
x = 1, y = 1Solve the simultaneous equations:
y = 4x − 3
4x + 2y = 6
a=4b=-3c=4d=2e=6x=1y=1variant=substitution
A.
x = 2, y = 1
→ Common computation error
B.
x = 3, y = 1
→ Common computation error
C.
x = 1, y = 1 ✓ correct
D.
x = 0, y = 1
→ Common computation error
1.
Eq1 already gives y = 4x−3. Substitute into Eq2.
4x + 2(4x−3) = 6
2.
Expand and collect x-terms.
12x = 12
3.
Solve for x.
x = 12 ÷ 12 = 1
4.
Back-substitute into Eq1 for y.
y = 4(1)−3 = 1
4(1) + 2(1) = 6. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
y = 4x − 3
4x + 2y = 6
Answer: ________
Expected:
x = 1, y = 1Long answer / Show working
Solve the simultaneous equations:
y = 4x − 3
4x + 2y = 6
Student writes solution steps here. Auto-grader compares final answer (
x = 1, y = 1) and rewards intermediate steps from the canonical solution above.#6intermediateanswer:
x = 9, y = 15Solve the simultaneous equations:
y = x + 6
x + 5y = 84
a=1b=6c=1d=5e=84x=9y=15variant=substitution
A.
x = 9, y = 3
→ Sign error when computing y from the isolated equation (treated −b as +b)
B.
x = 15, y = 9
→ Swapped x and y values in final answer
C.
x = 10, y = 15
→ Common computation error
D.
x = 9, y = 15 ✓ correct
1.
Eq1 already gives y = 1x+6. Substitute into Eq2.
1x + 5(1x+6) = 84
2.
Expand and collect x-terms.
6x = 54
3.
Solve for x.
x = 54 ÷ 6 = 9
4.
Back-substitute into Eq1 for y.
y = 1(9)+6 = 15
1(9) + 5(15) = 84. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
y = x + 6
x + 5y = 84
Answer: ________
Expected:
x = 9, y = 15Long answer / Show working
Solve the simultaneous equations:
y = x + 6
x + 5y = 84
Student writes solution steps here. Auto-grader compares final answer (
x = 9, y = 15) and rewards intermediate steps from the canonical solution above.#7intermediateanswer:
x = 2, y = 1Solve the simultaneous equations:
y = 4x − 7
4x + 2y = 10
a=4b=-7c=4d=2e=10x=2y=1variant=substitution
A.
x = 1, y = 1
→ Common computation error
B.
x = 3, y = 1
→ Common computation error
C.
x = 1, y = 2
→ Swapped x and y values in final answer
D.
x = 2, y = 1 ✓ correct
1.
Eq1 already gives y = 4x−7. Substitute into Eq2.
4x + 2(4x−7) = 10
2.
Expand and collect x-terms.
12x = 24
3.
Solve for x.
x = 24 ÷ 12 = 2
4.
Back-substitute into Eq1 for y.
y = 4(2)−7 = 1
4(2) + 2(1) = 10. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
y = 4x − 7
4x + 2y = 10
Answer: ________
Expected:
x = 2, y = 1Long answer / Show working
Solve the simultaneous equations:
y = 4x − 7
4x + 2y = 10
Student writes solution steps here. Auto-grader compares final answer (
x = 2, y = 1) and rewards intermediate steps from the canonical solution above.#8intermediateanswer:
x = 10, y = 15Solve the simultaneous equations:
y = x + 5
x + 3y = 55
a=1b=5c=1d=3e=55x=10y=15variant=substitution
A.
x = 15, y = 10
→ Swapped x and y values in final answer
B.
x = 10, y = 5
→ Sign error when computing y from the isolated equation (treated −b as +b)
C.
x = 10, y = 15 ✓ correct
D.
x = 11, y = 15
→ Common computation error
1.
Eq1 already gives y = 1x+5. Substitute into Eq2.
1x + 3(1x+5) = 55
2.
Expand and collect x-terms.
4x = 40
3.
Solve for x.
x = 40 ÷ 4 = 10
4.
Back-substitute into Eq1 for y.
y = 1(10)+5 = 15
1(10) + 3(15) = 55. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
y = x + 5
x + 3y = 55
Answer: ________
Expected:
x = 10, y = 15Long answer / Show working
Solve the simultaneous equations:
y = x + 5
x + 3y = 55
Student writes solution steps here. Auto-grader compares final answer (
x = 10, y = 15) and rewards intermediate steps from the canonical solution above.s1s3-simul-elim-addUnit 9 · Simultaneous (elimination)
Simultaneous: elimination by addition
Coefficients of one variable are opposite signs — add equations to eliminate.
Variety profile: 1 internal structural variant · Loop + fallback. Single shape.
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1intermediateanswer:
x = 6, y = 6Solve the simultaneous equations:
8x + y = 54
8x − y = 42
a=8b=1c=54d=8e=42x=6y=6variant=elim-add
A.
x = 7, y = 6
→ Solved first equation alone ignoring second (forgot elimination step)
B.
x = 6, y = 6 ✓ correct
C.
x = 6, y = 7
→ Arithmetic slip when back-substituting to find y
D.
x = Infinity, y = 6
→ Subtracted equations instead of adding — wrong sign for x-coefficient sum
1.
Add Eq1 and Eq2 to eliminate y (−1y and +1y cancel).
(8+8)x = 54+42
2.
Solve for x.
16x = 96 ⇒ x = 6
3.
Substitute x into Eq1 to find y.
8(6) + 1y = 54 ⇒ y = 6
8(6) − 1(6) = 42. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
8x + y = 54
8x − y = 42
Answer: ________
Expected:
x = 6, y = 6Long answer / Show working
Solve the simultaneous equations:
8x + y = 54
8x − y = 42
Student writes solution steps here. Auto-grader compares final answer (
x = 6, y = 6) and rewards intermediate steps from the canonical solution above.#2intermediateanswer:
x = 8, y = 10Solve the simultaneous equations:
2x + y = 26
2x − y = 6
a=2b=1c=26d=2e=6x=8y=10variant=elim-add
A.
x = 13, y = 10
→ Solved first equation alone ignoring second (forgot elimination step)
B.
x = 8, y = 11
→ Arithmetic slip when back-substituting to find y
C.
x = 8, y = 10 ✓ correct
D.
x = Infinity, y = 10
→ Subtracted equations instead of adding — wrong sign for x-coefficient sum
1.
Add Eq1 and Eq2 to eliminate y (−1y and +1y cancel).
(2+2)x = 26+6
2.
Solve for x.
4x = 32 ⇒ x = 8
3.
Substitute x into Eq1 to find y.
2(8) + 1y = 26 ⇒ y = 10
2(8) − 1(10) = 6. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
2x + y = 26
2x − y = 6
Answer: ________
Expected:
x = 8, y = 10Long answer / Show working
Solve the simultaneous equations:
2x + y = 26
2x − y = 6
Student writes solution steps here. Auto-grader compares final answer (
x = 8, y = 10) and rewards intermediate steps from the canonical solution above.#3intermediateanswer:
x = 5, y = 3Solve the simultaneous equations:
5x + 5y = 40
4x − 5y = 5
a=5b=5c=40d=4e=5x=5y=3variant=elim-add
A.
x = 45, y = 3
→ Subtracted equations instead of adding — wrong sign for x-coefficient sum
B.
x = 5, y = 4
→ Arithmetic slip when back-substituting to find y
C.
x = 5, y = 3 ✓ correct
D.
x = 8, y = 3
→ Solved first equation alone ignoring second (forgot elimination step)
1.
Add Eq1 and Eq2 to eliminate y (−5y and +5y cancel).
(5+4)x = 40+5
2.
Solve for x.
9x = 45 ⇒ x = 5
3.
Substitute x into Eq1 to find y.
5(5) + 5y = 40 ⇒ y = 3
4(5) − 5(3) = 5. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
5x + 5y = 40
4x − 5y = 5
Answer: ________
Expected:
x = 5, y = 3Long answer / Show working
Solve the simultaneous equations:
5x + 5y = 40
4x − 5y = 5
Student writes solution steps here. Auto-grader compares final answer (
x = 5, y = 3) and rewards intermediate steps from the canonical solution above.#4intermediateanswer:
x = 5, y = 9Solve the simultaneous equations:
7x + 2y = 53
7x − 2y = 17
a=7b=2c=53d=7e=17x=5y=9variant=elim-add
A.
x = 8, y = 9
→ Solved first equation alone ignoring second (forgot elimination step)
B.
x = 5, y = 9 ✓ correct
C.
x = Infinity, y = 9
→ Subtracted equations instead of adding — wrong sign for x-coefficient sum
D.
x = 5, y = 10
→ Arithmetic slip when back-substituting to find y
1.
Add Eq1 and Eq2 to eliminate y (−2y and +2y cancel).
(7+7)x = 53+17
2.
Solve for x.
14x = 70 ⇒ x = 5
3.
Substitute x into Eq1 to find y.
7(5) + 2y = 53 ⇒ y = 9
7(5) − 2(9) = 17. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
7x + 2y = 53
7x − 2y = 17
Answer: ________
Expected:
x = 5, y = 9Long answer / Show working
Solve the simultaneous equations:
7x + 2y = 53
7x − 2y = 17
Student writes solution steps here. Auto-grader compares final answer (
x = 5, y = 9) and rewards intermediate steps from the canonical solution above.#5intermediateanswer:
x = 5, y = 5Solve the simultaneous equations:
6x + 5y = 55
7x − 5y = 10
a=6b=5c=55d=7e=10x=5y=5variant=elim-add
A.
x = 5, y = 6
→ Arithmetic slip when back-substituting to find y
B.
x = 9, y = 5
→ Solved first equation alone ignoring second (forgot elimination step)
C.
x = 5, y = 5 ✓ correct
D.
x = -65, y = 5
→ Subtracted equations instead of adding — wrong sign for x-coefficient sum
1.
Add Eq1 and Eq2 to eliminate y (−5y and +5y cancel).
(6+7)x = 55+10
2.
Solve for x.
13x = 65 ⇒ x = 5
3.
Substitute x into Eq1 to find y.
6(5) + 5y = 55 ⇒ y = 5
7(5) − 5(5) = 10. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
6x + 5y = 55
7x − 5y = 10
Answer: ________
Expected:
x = 5, y = 5Long answer / Show working
Solve the simultaneous equations:
6x + 5y = 55
7x − 5y = 10
Student writes solution steps here. Auto-grader compares final answer (
x = 5, y = 5) and rewards intermediate steps from the canonical solution above.#6intermediateanswer:
x = 10, y = 7Solve the simultaneous equations:
4x + 2y = 54
6x − 2y = 46
a=4b=2c=54d=6e=46x=10y=7variant=elim-add
A.
x = 10, y = 7 ✓ correct
B.
x = -50, y = 7
→ Subtracted equations instead of adding — wrong sign for x-coefficient sum
C.
x = 10, y = 8
→ Arithmetic slip when back-substituting to find y
D.
x = 14, y = 7
→ Solved first equation alone ignoring second (forgot elimination step)
1.
Add Eq1 and Eq2 to eliminate y (−2y and +2y cancel).
(4+6)x = 54+46
2.
Solve for x.
10x = 100 ⇒ x = 10
3.
Substitute x into Eq1 to find y.
4(10) + 2y = 54 ⇒ y = 7
6(10) − 2(7) = 46. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
4x + 2y = 54
6x − 2y = 46
Answer: ________
Expected:
x = 10, y = 7Long answer / Show working
Solve the simultaneous equations:
4x + 2y = 54
6x − 2y = 46
Student writes solution steps here. Auto-grader compares final answer (
x = 10, y = 7) and rewards intermediate steps from the canonical solution above.#7intermediateanswer:
x = 15, y = 7Solve the simultaneous equations:
7x + 4y = 133
2x − 4y = 2
a=7b=4c=133d=2e=2x=15y=7variant=elim-add
A.
x = 27, y = 7
→ Subtracted equations instead of adding — wrong sign for x-coefficient sum
B.
x = 15, y = 7 ✓ correct
C.
x = 19, y = 7
→ Solved first equation alone ignoring second (forgot elimination step)
D.
x = 15, y = 8
→ Arithmetic slip when back-substituting to find y
1.
Add Eq1 and Eq2 to eliminate y (−4y and +4y cancel).
(7+2)x = 133+2
2.
Solve for x.
9x = 135 ⇒ x = 15
3.
Substitute x into Eq1 to find y.
7(15) + 4y = 133 ⇒ y = 7
2(15) − 4(7) = 2. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
7x + 4y = 133
2x − 4y = 2
Answer: ________
Expected:
x = 15, y = 7Long answer / Show working
Solve the simultaneous equations:
7x + 4y = 133
2x − 4y = 2
Student writes solution steps here. Auto-grader compares final answer (
x = 15, y = 7) and rewards intermediate steps from the canonical solution above.#8intermediateanswer:
x = 2, y = 1Solve the simultaneous equations:
5x + 2y = 12
4x − 2y = 6
a=5b=2c=12d=4e=6x=2y=1variant=elim-add
A.
x = 2, y = 2
→ Arithmetic slip when back-substituting to find y
B.
x = 3, y = 1
→ Common computation error
C.
x = 18, y = 1
→ Subtracted equations instead of adding — wrong sign for x-coefficient sum
D.
x = 2, y = 1 ✓ correct
1.
Add Eq1 and Eq2 to eliminate y (−2y and +2y cancel).
(5+4)x = 12+6
2.
Solve for x.
9x = 18 ⇒ x = 2
3.
Substitute x into Eq1 to find y.
5(2) + 2y = 12 ⇒ y = 1
4(2) − 2(1) = 6. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
5x + 2y = 12
4x − 2y = 6
Answer: ________
Expected:
x = 2, y = 1Long answer / Show working
Solve the simultaneous equations:
5x + 2y = 12
4x − 2y = 6
Student writes solution steps here. Auto-grader compares final answer (
x = 2, y = 1) and rewards intermediate steps from the canonical solution above.s1s3-simul-elim-subUnit 9 · Simultaneous (elimination)
Simultaneous: elimination by subtraction
Coefficients of one variable are equal — subtract one equation from the other.
Variety profile: 1 internal structural variant · Single algebraic shape.
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1intermediateanswer:
x = 3, y = 3Solve the simultaneous equations:
4x + 6y = 30
4x + 5y = 27
a=4b=6d=5c=30e=27x=3y=3variant=elim-sub
A.
x = 4, y = 3
→ Arithmetic slip when back-substituting y to find x
B.
x = 2, y = 3
→ Common computation error
C.
x = 3, y = 3 ✓ correct
D.
x = 3, y = 0
→ Added coefficients instead of subtracting when eliminating (sign error in y-coefficient difference)
1.
Subtract Eq2 from Eq1 to eliminate x (4x cancels).
(6−5)y = 30−27
2.
Solve for y.
1y = 3 ⇒ y = 3
3.
Substitute y into Eq1 to find x.
4x + 6(3) = 30 ⇒ x = 3
4(3) + 5(3) = 27. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
4x + 6y = 30
4x + 5y = 27
Answer: ________
Expected:
x = 3, y = 3Long answer / Show working
Solve the simultaneous equations:
4x + 6y = 30
4x + 5y = 27
Student writes solution steps here. Auto-grader compares final answer (
x = 3, y = 3) and rewards intermediate steps from the canonical solution above.#2intermediateanswer:
x = 8, y = 1Solve the simultaneous equations:
4x + 5y = 37
4x + y = 33
a=4b=5d=1c=37e=33x=8y=1variant=elim-sub
A.
x = 1, y = 8
→ Swapped x and y values in final answer
B.
x = 7, y = 1
→ Common computation error
C.
x = 9, y = 1
→ Arithmetic slip when back-substituting y to find x
D.
x = 8, y = 1 ✓ correct
1.
Subtract Eq2 from Eq1 to eliminate x (4x cancels).
(5−1)y = 37−33
2.
Solve for y.
4y = 4 ⇒ y = 1
3.
Substitute y into Eq1 to find x.
4x + 5(1) = 37 ⇒ x = 8
4(8) + 1(1) = 33. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
4x + 5y = 37
4x + y = 33
Answer: ________
Expected:
x = 8, y = 1Long answer / Show working
Solve the simultaneous equations:
4x + 5y = 37
4x + y = 33
Student writes solution steps here. Auto-grader compares final answer (
x = 8, y = 1) and rewards intermediate steps from the canonical solution above.#3intermediateanswer:
x = 3, y = 2Solve the simultaneous equations:
3x + 5y = 19
3x + 2y = 13
a=3b=5d=2c=19e=13x=3y=2variant=elim-sub
A.
x = 3, y = 2 ✓ correct
B.
x = 4, y = 2
→ Arithmetic slip when back-substituting y to find x
C.
x = 3, y = 1
→ Added coefficients instead of subtracting when eliminating (sign error in y-coefficient difference)
D.
x = 2, y = 3
→ Swapped x and y values in final answer
1.
Subtract Eq2 from Eq1 to eliminate x (3x cancels).
(5−2)y = 19−13
2.
Solve for y.
3y = 6 ⇒ y = 2
3.
Substitute y into Eq1 to find x.
3x + 5(2) = 19 ⇒ x = 3
3(3) + 2(2) = 13. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
3x + 5y = 19
3x + 2y = 13
Answer: ________
Expected:
x = 3, y = 2Long answer / Show working
Solve the simultaneous equations:
3x + 5y = 19
3x + 2y = 13
Student writes solution steps here. Auto-grader compares final answer (
x = 3, y = 2) and rewards intermediate steps from the canonical solution above.#4intermediateanswer:
x = 8, y = 6Solve the simultaneous equations:
2x + 4y = 40
2x + 3y = 34
a=2b=4d=3c=40e=34x=8y=6variant=elim-sub
A.
x = 8, y = 1
→ Added coefficients instead of subtracting when eliminating (sign error in y-coefficient difference)
B.
x = 6, y = 8
→ Swapped x and y values in final answer
C.
x = 8, y = 6 ✓ correct
D.
x = 9, y = 6
→ Arithmetic slip when back-substituting y to find x
1.
Subtract Eq2 from Eq1 to eliminate x (2x cancels).
(4−3)y = 40−34
2.
Solve for y.
1y = 6 ⇒ y = 6
3.
Substitute y into Eq1 to find x.
2x + 4(6) = 40 ⇒ x = 8
2(8) + 3(6) = 34. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
2x + 4y = 40
2x + 3y = 34
Answer: ________
Expected:
x = 8, y = 6Long answer / Show working
Solve the simultaneous equations:
2x + 4y = 40
2x + 3y = 34
Student writes solution steps here. Auto-grader compares final answer (
x = 8, y = 6) and rewards intermediate steps from the canonical solution above.#5intermediateanswer:
x = 1, y = 6Solve the simultaneous equations:
5x + 7y = 47
5x + y = 11
a=5b=7d=1c=47e=11x=1y=6variant=elim-sub
A.
x = 2, y = 6
→ Arithmetic slip when back-substituting y to find x
B.
x = 6, y = 1
→ Swapped x and y values in final answer
C.
x = 1, y = 5
→ Added coefficients instead of subtracting when eliminating (sign error in y-coefficient difference)
D.
x = 1, y = 6 ✓ correct
1.
Subtract Eq2 from Eq1 to eliminate x (5x cancels).
(7−1)y = 47−11
2.
Solve for y.
6y = 36 ⇒ y = 6
3.
Substitute y into Eq1 to find x.
5x + 7(6) = 47 ⇒ x = 1
5(1) + 1(6) = 11. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
5x + 7y = 47
5x + y = 11
Answer: ________
Expected:
x = 1, y = 6Long answer / Show working
Solve the simultaneous equations:
5x + 7y = 47
5x + y = 11
Student writes solution steps here. Auto-grader compares final answer (
x = 1, y = 6) and rewards intermediate steps from the canonical solution above.#6intermediateanswer:
x = 6, y = 5Solve the simultaneous equations:
6x + 5y = 61
6x + 3y = 51
a=6b=5d=3c=61e=51x=6y=5variant=elim-sub
A.
x = 6, y = 5 ✓ correct
B.
x = 5, y = 6
→ Swapped x and y values in final answer
C.
x = 6, y = 1
→ Added coefficients instead of subtracting when eliminating (sign error in y-coefficient difference)
D.
x = 7, y = 5
→ Arithmetic slip when back-substituting y to find x
1.
Subtract Eq2 from Eq1 to eliminate x (6x cancels).
(5−3)y = 61−51
2.
Solve for y.
2y = 10 ⇒ y = 5
3.
Substitute y into Eq1 to find x.
6x + 5(5) = 61 ⇒ x = 6
6(6) + 3(5) = 51. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
6x + 5y = 61
6x + 3y = 51
Answer: ________
Expected:
x = 6, y = 5Long answer / Show working
Solve the simultaneous equations:
6x + 5y = 61
6x + 3y = 51
Student writes solution steps here. Auto-grader compares final answer (
x = 6, y = 5) and rewards intermediate steps from the canonical solution above.#7intermediateanswer:
x = 3, y = 5Solve the simultaneous equations:
6x + 5y = 43
6x + y = 23
a=6b=5d=1c=43e=23x=3y=5variant=elim-sub
A.
x = 3, y = 3
→ Added coefficients instead of subtracting when eliminating (sign error in y-coefficient difference)
B.
x = 4, y = 5
→ Arithmetic slip when back-substituting y to find x
C.
x = 3, y = 5 ✓ correct
D.
x = 5, y = 3
→ Swapped x and y values in final answer
1.
Subtract Eq2 from Eq1 to eliminate x (6x cancels).
(5−1)y = 43−23
2.
Solve for y.
4y = 20 ⇒ y = 5
3.
Substitute y into Eq1 to find x.
6x + 5(5) = 43 ⇒ x = 3
6(3) + 1(5) = 23. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
6x + 5y = 43
6x + y = 23
Answer: ________
Expected:
x = 3, y = 5Long answer / Show working
Solve the simultaneous equations:
6x + 5y = 43
6x + y = 23
Student writes solution steps here. Auto-grader compares final answer (
x = 3, y = 5) and rewards intermediate steps from the canonical solution above.#8intermediateanswer:
x = 2, y = 8Solve the simultaneous equations:
3x + 6y = 54
3x + y = 14
a=3b=6d=1c=54e=14x=2y=8variant=elim-sub
A.
x = 3, y = 8
→ Arithmetic slip when back-substituting y to find x
B.
x = 2, y = 8 ✓ correct
C.
x = 8, y = 2
→ Swapped x and y values in final answer
D.
x = 2, y = 6
→ Added coefficients instead of subtracting when eliminating (sign error in y-coefficient difference)
1.
Subtract Eq2 from Eq1 to eliminate x (3x cancels).
(6−1)y = 54−14
2.
Solve for y.
5y = 40 ⇒ y = 8
3.
Substitute y into Eq1 to find x.
3x + 6(8) = 54 ⇒ x = 2
3(2) + 1(8) = 14. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
3x + 6y = 54
3x + y = 14
Answer: ________
Expected:
x = 2, y = 8Long answer / Show working
Solve the simultaneous equations:
3x + 6y = 54
3x + y = 14
Student writes solution steps here. Auto-grader compares final answer (
x = 2, y = 8) and rewards intermediate steps from the canonical solution above.s1s3-simul-fraction-coeffUnit 9 · Simultaneous (advanced)
Simultaneous: fractional coefficients
At least one equation has fractional coefficients — clear denominators first.
Variety profile: 4 internal structural variants · 4 denominator pairs.
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1advancedanswer:
x = 12, y = 14Solve the simultaneous equations:
x/2 + y = 20
x/4 + y = 17
p=2q=4lcm=4k=3x=12y=14c1=20c2=17variant=fraction-coeff
A.
x = 2, y = 14
→ Forgot to multiply by 6 after finding x/6 = c1−c2
B.
x = 12, y = 13
→ Back-substituted into wrong equation to find y
C.
x = 12, y = 14 ✓ correct
D.
x = -18, y = 29
→ Subtracted equations in wrong order — got negative x (sign error)
1.
Subtract Eq2 from Eq1 to eliminate y: x/2 − x/3 = 20−17.
x/6 = 3
2.
Multiply both sides by 6.
x = 6 × 3 = 12
3.
Substitute x into Eq1.
12/2 + y = 20 ⇒ y = 20 − 6 = 14
12/2 + 14 = 20; 12/3 + 14 = 17. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
x/2 + y = 20
x/4 + y = 17
Answer: ________
Expected:
x = 12, y = 14Long answer / Show working
Solve the simultaneous equations:
x/2 + y = 20
x/4 + y = 17
Student writes solution steps here. Auto-grader compares final answer (
x = 12, y = 14) and rewards intermediate steps from the canonical solution above.#2advancedanswer:
x = 24, y = 20Solve the simultaneous equations:
x/2 + y = 32
x/3 + y = 28
p=2q=3lcm=6k=4x=24y=20c1=32c2=28variant=fraction-coeff
A.
x = 24, y = 20 ✓ correct
B.
x = 4, y = 20
→ Forgot to multiply by 6 after finding x/6 = c1−c2
C.
x = 25, y = 20
→ Common computation error
D.
x = -24, y = 44
→ Subtracted equations in wrong order — got negative x (sign error)
1.
Subtract Eq2 from Eq1 to eliminate y: x/2 − x/3 = 32−28.
x/6 = 4
2.
Multiply both sides by 6.
x = 6 × 4 = 24
3.
Substitute x into Eq1.
24/2 + y = 32 ⇒ y = 32 − 12 = 20
24/2 + 20 = 32; 24/3 + 20 = 28. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
x/2 + y = 32
x/3 + y = 28
Answer: ________
Expected:
x = 24, y = 20Long answer / Show working
Solve the simultaneous equations:
x/2 + y = 32
x/3 + y = 28
Student writes solution steps here. Auto-grader compares final answer (
x = 24, y = 20) and rewards intermediate steps from the canonical solution above.#3advancedanswer:
x = 54, y = 20Solve the simultaneous equations:
x/2 + y = 47
x/6 + y = 29
p=2q=6lcm=6k=9x=54y=20c1=47c2=29variant=fraction-coeff
A.
x = 9, y = 20
→ Forgot to multiply by 6 after finding x/6 = c1−c2
B.
x = -108, y = 101
→ Subtracted equations in wrong order — got negative x (sign error)
C.
x = 54, y = 20 ✓ correct
D.
x = 54, y = 11
→ Back-substituted into wrong equation to find y
1.
Subtract Eq2 from Eq1 to eliminate y: x/2 − x/3 = 47−29.
x/6 = 18
2.
Multiply both sides by 6.
x = 6 × 18 = 54
3.
Substitute x into Eq1.
54/2 + y = 47 ⇒ y = 47 − 27 = 20
54/2 + 20 = 47; 54/3 + 20 = 29. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
x/2 + y = 47
x/6 + y = 29
Answer: ________
Expected:
x = 54, y = 20Long answer / Show working
Solve the simultaneous equations:
x/2 + y = 47
x/6 + y = 29
Student writes solution steps here. Auto-grader compares final answer (
x = 54, y = 20) and rewards intermediate steps from the canonical solution above.#4advancedanswer:
x = 60, y = 18Solve the simultaneous equations:
x/2 + y = 48
x/6 + y = 28
p=2q=6lcm=6k=10x=60y=18c1=48c2=28variant=fraction-coeff
A.
x = 10, y = 18
→ Forgot to multiply by 6 after finding x/6 = c1−c2
B.
x = -120, y = 108
→ Subtracted equations in wrong order — got negative x (sign error)
C.
x = 60, y = 8
→ Back-substituted into wrong equation to find y
D.
x = 60, y = 18 ✓ correct
1.
Subtract Eq2 from Eq1 to eliminate y: x/2 − x/3 = 48−28.
x/6 = 20
2.
Multiply both sides by 6.
x = 6 × 20 = 60
3.
Substitute x into Eq1.
60/2 + y = 48 ⇒ y = 48 − 30 = 18
60/2 + 18 = 48; 60/3 + 18 = 28. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
x/2 + y = 48
x/6 + y = 28
Answer: ________
Expected:
x = 60, y = 18Long answer / Show working
Solve the simultaneous equations:
x/2 + y = 48
x/6 + y = 28
Student writes solution steps here. Auto-grader compares final answer (
x = 60, y = 18) and rewards intermediate steps from the canonical solution above.#5advancedanswer:
x = 42, y = 20Solve the simultaneous equations:
x/2 + y = 41
x/6 + y = 27
p=2q=6lcm=6k=7x=42y=20c1=41c2=27variant=fraction-coeff
A.
x = -84, y = 83
→ Subtracted equations in wrong order — got negative x (sign error)
B.
x = 42, y = 13
→ Back-substituted into wrong equation to find y
C.
x = 7, y = 20
→ Forgot to multiply by 6 after finding x/6 = c1−c2
D.
x = 42, y = 20 ✓ correct
1.
Subtract Eq2 from Eq1 to eliminate y: x/2 − x/3 = 41−27.
x/6 = 14
2.
Multiply both sides by 6.
x = 6 × 14 = 42
3.
Substitute x into Eq1.
42/2 + y = 41 ⇒ y = 41 − 21 = 20
42/2 + 20 = 41; 42/3 + 20 = 27. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
x/2 + y = 41
x/6 + y = 27
Answer: ________
Expected:
x = 42, y = 20Long answer / Show working
Solve the simultaneous equations:
x/2 + y = 41
x/6 + y = 27
Student writes solution steps here. Auto-grader compares final answer (
x = 42, y = 20) and rewards intermediate steps from the canonical solution above.#6advancedanswer:
x = 48, y = 10Solve the simultaneous equations:
x/2 + y = 34
x/4 + y = 22
p=2q=4lcm=4k=12x=48y=10c1=34c2=22variant=fraction-coeff
A.
x = 48, y = 6
→ Back-substituted into wrong equation to find y
B.
x = 8, y = 10
→ Forgot to multiply by 6 after finding x/6 = c1−c2
C.
x = 48, y = 10 ✓ correct
D.
x = -72, y = 70
→ Subtracted equations in wrong order — got negative x (sign error)
1.
Subtract Eq2 from Eq1 to eliminate y: x/2 − x/3 = 34−22.
x/6 = 12
2.
Multiply both sides by 6.
x = 6 × 12 = 48
3.
Substitute x into Eq1.
48/2 + y = 34 ⇒ y = 34 − 24 = 10
48/2 + 10 = 34; 48/3 + 10 = 22. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
x/2 + y = 34
x/4 + y = 22
Answer: ________
Expected:
x = 48, y = 10Long answer / Show working
Solve the simultaneous equations:
x/2 + y = 34
x/4 + y = 22
Student writes solution steps here. Auto-grader compares final answer (
x = 48, y = 10) and rewards intermediate steps from the canonical solution above.#7advancedanswer:
x = 6, y = 17Solve the simultaneous equations:
x/2 + y = 20
x/6 + y = 18
p=2q=6lcm=6k=1x=6y=17c1=20c2=18variant=fraction-coeff
A.
x = 6, y = 17 ✓ correct
B.
x = 6, y = 16
→ Back-substituted into wrong equation to find y
C.
x = -12, y = 26
→ Subtracted equations in wrong order — got negative x (sign error)
D.
x = 1, y = 17
→ Forgot to multiply by 6 after finding x/6 = c1−c2
1.
Subtract Eq2 from Eq1 to eliminate y: x/2 − x/3 = 20−18.
x/6 = 2
2.
Multiply both sides by 6.
x = 6 × 2 = 6
3.
Substitute x into Eq1.
6/2 + y = 20 ⇒ y = 20 − 3 = 17
6/2 + 17 = 20; 6/3 + 17 = 18. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
x/2 + y = 20
x/6 + y = 18
Answer: ________
Expected:
x = 6, y = 17Long answer / Show working
Solve the simultaneous equations:
x/2 + y = 20
x/6 + y = 18
Student writes solution steps here. Auto-grader compares final answer (
x = 6, y = 17) and rewards intermediate steps from the canonical solution above.#8advancedanswer:
x = 32, y = 17Solve the simultaneous equations:
x/2 + y = 33
x/4 + y = 25
p=2q=4lcm=4k=8x=32y=17c1=33c2=25variant=fraction-coeff
A.
x = 5.333333333333333, y = 17
→ Forgot to multiply by 6 after finding x/6 = c1−c2
B.
x = -48, y = 57
→ Subtracted equations in wrong order — got negative x (sign error)
C.
x = 32, y = 14.333333333333334
→ Back-substituted into wrong equation to find y
D.
x = 32, y = 17 ✓ correct
1.
Subtract Eq2 from Eq1 to eliminate y: x/2 − x/3 = 33−25.
x/6 = 8
2.
Multiply both sides by 6.
x = 6 × 8 = 32
3.
Substitute x into Eq1.
32/2 + y = 33 ⇒ y = 33 − 16 = 17
32/2 + 17 = 33; 32/3 + 17 = 25. ✓
Fill-in / Short Answer
Solve the simultaneous equations:
x/2 + y = 33
x/4 + y = 25
Answer: ________
Expected:
x = 32, y = 17Long answer / Show working
Solve the simultaneous equations:
x/2 + y = 33
x/4 + y = 25
Student writes solution steps here. Auto-grader compares final answer (
x = 32, y = 17) and rewards intermediate steps from the canonical solution above.Unit 9
Formulate and solve word problems
Curriculum objective in Linear equations in two unknowns. 1 generator currently serves this objective.
⚠ Reference shapes seen on disk but NOT yet served by these 1 generators (4 candidates):
- Chickens-and-pigs (heads + feet) as equation-pair-selection
- Exam scoring (+/− marks) systems
- Digit / place-value two-unknown stories
- "Pick 2 equations from a list" formulation MC
Variety dashboard · all 1 generator side-by-side
Compare structural variant counts, observed difficulty mix, and a representative stem from each. Click a row's button to jump to the full sample set below.
| Generator | Variants | Difficulty mix (8 samples) | First sample stem | |
|---|---|---|---|---|
Simul: word problems1s3-simul-word-two-unknowns | 5v 5 scenario stems (fruit/stationery/tickets/bakery/cinema). Reference also has ch… | intermediate: 8 | A shop sells two types of fruit. Apples cost $5 each and oranges cost $2 each. A customer … |
Drill into one generator · 8 samples each
s1s3-simul-word-two-unknownsUnit 9 · Word problems (simultaneous)
Word problem: Two unknowns
Five rotating contexts: fruit, stationery, tickets, bakery, cinema.
Variety profile: 5 internal structural variants · 5 scenario stems (fruit/stationery/tickets/bakery/cinema). Reference also has chickens-pigs, exam-marks, digit problems not yet built.
8 live-generated samples
Each card is freshly drawn from the generator. Toggle between MC options (with trap analysis), canonical solution steps, or alternate formats (fill-in / long-answer).
#1intermediateanswer:
x = 5, y = 8A shop sells two types of fruit. Apples cost $5 each and oranges cost $2 each. A customer buys 13 fruits altogether and pays $41 in total. Find the number of apples and the number of oranges bought.
x=5y=8p1=5p2=2N=13T=41scenarioIdx=0aLabel=applesbLabel=orangesvariant=word-two
A.
x = 5, y = 8 ✓ correct
B.
x = 8, y = 5
→ Swapped apple and orange counts in the answer
C.
x = 6, y = 8
→ Common computation error
D.
x = 2, y = 11
→ Used (p1+p2) in denominator instead of (p1−p2) — added prices instead of taking difference
1.
Let x = apples, y = oranges. Form two equations.
x + y = 13 (Eq1); 5x + 2y = 41 (Eq2)
2.
From Eq1: y = 13 − x. Substitute into Eq2.
5x + 2(13−x) = 41
3.
Expand and solve.
3x = 15 ⇒ x = 5
4.
Find y.
y = 13 − 5 = 8
5+8=13; $5×5+$2×8=$41. ✓
Fill-in / Short Answer
A shop sells two types of fruit. Apples cost $5 each and oranges cost $2 each. A customer buys 13 fruits altogether and pays $41 in total. Find the number of apples and the number of oranges bought.
Answer: ________
Expected:
x = 5, y = 8Long answer / Show working
A shop sells two types of fruit. Apples cost $5 each and oranges cost $2 each. A customer buys 13 fruits altogether and pays $41 in total. Find the number of apples and the number of oranges bought.
Student writes solution steps here. Auto-grader compares final answer (
x = 5, y = 8) and rewards intermediate steps from the canonical solution above.#2intermediateanswer:
x = 10, y = 11Tickets for a school show cost $5 for adults and $2 for children. A family buys 21 tickets and pays $72 in total. Find the number of adult tickets and the number of children's tickets bought.
x=10y=11p1=5p2=2N=21T=72scenarioIdx=2aLabel=adult ticketsbLabel=children's ticketsvariant=word-two
A.
x = 10, y = 11 ✓ correct
B.
x = 14, y = 7
→ Ignored price of oranges — divided total by apple price only
C.
x = 11, y = 10
→ Swapped apple and orange counts in the answer
D.
x = 4, y = 17
→ Used (p1+p2) in denominator instead of (p1−p2) — added prices instead of taking difference
1.
Let x = apples, y = oranges. Form two equations.
x + y = 21 (Eq1); 5x + 2y = 72 (Eq2)
2.
From Eq1: y = 21 − x. Substitute into Eq2.
5x + 2(21−x) = 72
3.
Expand and solve.
3x = 30 ⇒ x = 10
4.
Find y.
y = 21 − 10 = 11
10+11=21; $5×10+$2×11=$72. ✓
Fill-in / Short Answer
Tickets for a school show cost $5 for adults and $2 for children. A family buys 21 tickets and pays $72 in total. Find the number of adult tickets and the number of children's tickets bought.
Answer: ________
Expected:
x = 10, y = 11Long answer / Show working
Tickets for a school show cost $5 for adults and $2 for children. A family buys 21 tickets and pays $72 in total. Find the number of adult tickets and the number of children's tickets bought.
Student writes solution steps here. Auto-grader compares final answer (
x = 10, y = 11) and rewards intermediate steps from the canonical solution above.#3intermediateanswer:
x = 7, y = 3A shop sells two types of fruit. Apples cost $10 each and oranges cost $6 each. A customer buys 10 fruits altogether and pays $88 in total. Find the number of apples and the number of oranges bought.
x=7y=3p1=10p2=6N=10T=88scenarioIdx=0aLabel=applesbLabel=orangesvariant=word-two
A.
x = 3, y = 7
→ Swapped apple and orange counts in the answer
B.
x = 7, y = 3 ✓ correct
C.
x = 2, y = 8
→ Used (p1+p2) in denominator instead of (p1−p2) — added prices instead of taking difference
D.
x = 9, y = 1
→ Ignored price of oranges — divided total by apple price only
1.
Let x = apples, y = oranges. Form two equations.
x + y = 10 (Eq1); 10x + 6y = 88 (Eq2)
2.
From Eq1: y = 10 − x. Substitute into Eq2.
10x + 6(10−x) = 88
3.
Expand and solve.
4x = 28 ⇒ x = 7
4.
Find y.
y = 10 − 7 = 3
7+3=10; $10×7+$6×3=$88. ✓
Fill-in / Short Answer
A shop sells two types of fruit. Apples cost $10 each and oranges cost $6 each. A customer buys 10 fruits altogether and pays $88 in total. Find the number of apples and the number of oranges bought.
Answer: ________
Expected:
x = 7, y = 3Long answer / Show working
A shop sells two types of fruit. Apples cost $10 each and oranges cost $6 each. A customer buys 10 fruits altogether and pays $88 in total. Find the number of apples and the number of oranges bought.
Student writes solution steps here. Auto-grader compares final answer (
x = 7, y = 3) and rewards intermediate steps from the canonical solution above.#4intermediateanswer:
x = 8, y = 6Tickets for a school show cost $7 for adults and $1 for children. A family buys 14 tickets and pays $62 in total. Find the number of adult tickets and the number of children's tickets bought.
x=8y=6p1=7p2=1N=14T=62scenarioIdx=2aLabel=adult ticketsbLabel=children's ticketsvariant=word-two
A.
x = 9, y = 6
→ Common computation error
B.
x = 8, y = 6 ✓ correct
C.
x = 6, y = 8
→ Used (p1+p2) in denominator instead of (p1−p2) — added prices instead of taking difference
D.
x = 9, y = 5
→ Ignored price of oranges — divided total by apple price only
1.
Let x = apples, y = oranges. Form two equations.
x + y = 14 (Eq1); 7x + 1y = 62 (Eq2)
2.
From Eq1: y = 14 − x. Substitute into Eq2.
7x + 1(14−x) = 62
3.
Expand and solve.
6x = 48 ⇒ x = 8
4.
Find y.
y = 14 − 8 = 6
8+6=14; $7×8+$1×6=$62. ✓
Fill-in / Short Answer
Tickets for a school show cost $7 for adults and $1 for children. A family buys 14 tickets and pays $62 in total. Find the number of adult tickets and the number of children's tickets bought.
Answer: ________
Expected:
x = 8, y = 6Long answer / Show working
Tickets for a school show cost $7 for adults and $1 for children. A family buys 14 tickets and pays $62 in total. Find the number of adult tickets and the number of children's tickets bought.
Student writes solution steps here. Auto-grader compares final answer (
x = 8, y = 6) and rewards intermediate steps from the canonical solution above.#5intermediateanswer:
x = 11, y = 5A shop sells two types of fruit. Apples cost $8 each and oranges cost $4 each. A customer buys 16 fruits altogether and pays $108 in total. Find the number of apples and the number of oranges bought.
x=11y=5p1=8p2=4N=16T=108scenarioIdx=0aLabel=applesbLabel=orangesvariant=word-two
A.
x = 5, y = 11
→ Swapped apple and orange counts in the answer
B.
x = 4, y = 12
→ Used (p1+p2) in denominator instead of (p1−p2) — added prices instead of taking difference
C.
x = 14, y = 2
→ Ignored price of oranges — divided total by apple price only
D.
x = 11, y = 5 ✓ correct
1.
Let x = apples, y = oranges. Form two equations.
x + y = 16 (Eq1); 8x + 4y = 108 (Eq2)
2.
From Eq1: y = 16 − x. Substitute into Eq2.
8x + 4(16−x) = 108
3.
Expand and solve.
4x = 44 ⇒ x = 11
4.
Find y.
y = 16 − 11 = 5
11+5=16; $8×11+$4×5=$108. ✓
Fill-in / Short Answer
A shop sells two types of fruit. Apples cost $8 each and oranges cost $4 each. A customer buys 16 fruits altogether and pays $108 in total. Find the number of apples and the number of oranges bought.
Answer: ________
Expected:
x = 11, y = 5Long answer / Show working
A shop sells two types of fruit. Apples cost $8 each and oranges cost $4 each. A customer buys 16 fruits altogether and pays $108 in total. Find the number of apples and the number of oranges bought.
Student writes solution steps here. Auto-grader compares final answer (
x = 11, y = 5) and rewards intermediate steps from the canonical solution above.#6intermediateanswer:
x = 9, y = 6A shop sells two types of fruit. Apples cost $12 each and oranges cost $8 each. A customer buys 15 fruits altogether and pays $156 in total. Find the number of apples and the number of oranges bought.
x=9y=6p1=12p2=8N=15T=156scenarioIdx=0aLabel=applesbLabel=orangesvariant=word-two
A.
x = 6, y = 9
→ Swapped apple and orange counts in the answer
B.
x = 9, y = 6 ✓ correct
C.
x = 2, y = 13
→ Used (p1+p2) in denominator instead of (p1−p2) — added prices instead of taking difference
D.
x = 13, y = 2
→ Ignored price of oranges — divided total by apple price only
1.
Let x = apples, y = oranges. Form two equations.
x + y = 15 (Eq1); 12x + 8y = 156 (Eq2)
2.
From Eq1: y = 15 − x. Substitute into Eq2.
12x + 8(15−x) = 156
3.
Expand and solve.
4x = 36 ⇒ x = 9
4.
Find y.
y = 15 − 9 = 6
9+6=15; $12×9+$8×6=$156. ✓
Fill-in / Short Answer
A shop sells two types of fruit. Apples cost $12 each and oranges cost $8 each. A customer buys 15 fruits altogether and pays $156 in total. Find the number of apples and the number of oranges bought.
Answer: ________
Expected:
x = 9, y = 6Long answer / Show working
A shop sells two types of fruit. Apples cost $12 each and oranges cost $8 each. A customer buys 15 fruits altogether and pays $156 in total. Find the number of apples and the number of oranges bought.
Student writes solution steps here. Auto-grader compares final answer (
x = 9, y = 6) and rewards intermediate steps from the canonical solution above.#7intermediateanswer:
x = 4, y = 4Seats in a cinema cost $4 in Section A and $1 in Section B. A group of 8 people pay $20 in total. Find the number of people sitting in Section A and Section B respectively.
x=4y=4p1=4p2=1N=8T=20scenarioIdx=4aLabel=Section A seatsbLabel=Section B seatsvariant=word-two
A.
x = 4, y = 4 ✓ correct
B.
x = 5, y = 3
→ Ignored price of oranges — divided total by apple price only
C.
x = 2, y = 6
→ Used (p1+p2) in denominator instead of (p1−p2) — added prices instead of taking difference
D.
x = 5, y = 4
→ Common computation error
1.
Let x = apples, y = oranges. Form two equations.
x + y = 8 (Eq1); 4x + 1y = 20 (Eq2)
2.
From Eq1: y = 8 − x. Substitute into Eq2.
4x + 1(8−x) = 20
3.
Expand and solve.
3x = 12 ⇒ x = 4
4.
Find y.
y = 8 − 4 = 4
4+4=8; $4×4+$1×4=$20. ✓
Fill-in / Short Answer
Seats in a cinema cost $4 in Section A and $1 in Section B. A group of 8 people pay $20 in total. Find the number of people sitting in Section A and Section B respectively.
Answer: ________
Expected:
x = 4, y = 4Long answer / Show working
Seats in a cinema cost $4 in Section A and $1 in Section B. A group of 8 people pay $20 in total. Find the number of people sitting in Section A and Section B respectively.
Student writes solution steps here. Auto-grader compares final answer (
x = 4, y = 4) and rewards intermediate steps from the canonical solution above.#8intermediateanswer:
x = 12, y = 8A stationery shop sells pens for $12 each and pencils for $8 each. A student buys 20 items altogether and pays $208 in total. Find the number of pens and the number of pencils bought.
x=12y=8p1=12p2=8N=20T=208scenarioIdx=1aLabel=pensbLabel=pencilsvariant=word-two
A.
x = 17, y = 3
→ Ignored price of oranges — divided total by apple price only
B.
x = 2, y = 18
→ Used (p1+p2) in denominator instead of (p1−p2) — added prices instead of taking difference
C.
x = 12, y = 8 ✓ correct
D.
x = 8, y = 12
→ Swapped apple and orange counts in the answer
1.
Let x = apples, y = oranges. Form two equations.
x + y = 20 (Eq1); 12x + 8y = 208 (Eq2)
2.
From Eq1: y = 20 − x. Substitute into Eq2.
12x + 8(20−x) = 208
3.
Expand and solve.
4x = 48 ⇒ x = 12
4.
Find y.
y = 20 − 12 = 8
12+8=20; $12×12+$8×8=$208. ✓
Fill-in / Short Answer
A stationery shop sells pens for $12 each and pencils for $8 each. A student buys 20 items altogether and pays $208 in total. Find the number of pens and the number of pencils bought.
Answer: ________
Expected:
x = 12, y = 8Long answer / Show working
A stationery shop sells pens for $12 each and pencils for $8 each. A student buys 20 items altogether and pays $208 in total. Find the number of pens and the number of pencils bought.
Student writes solution steps here. Auto-grader compares final answer (
x = 12, y = 8) and rewards intermediate steps from the canonical solution above.