This key and the fresh checks are publicly accessible in the library, so they are formative rather than secure tests. Use a locally changed case if prior access matters; give the learner a clean prompt on Days 15/20. Record the learner's original working, access method and any content hint. A correct result obtained by a route you did not predict can be valid when the algebra, domain and interpretation withstand substitution. Do not score handwriting, pace, accent, device choice or family circumstances.
Daily models and exits
| Day | Model/guided/exit answers | If the exit is not yet secure |
|---|---|---|
| 11 | 3(x+4)=3x+12; at 2 both 18. Guided 4(y+2)=4y+8, at 3 both 20. Exit 2(z+6)=2z+12; at 1 original/corrected 14, false version 8. |
Build two groups and touch both bracket terms; then check a new value. |
| 12 | 4x+7+2x−3=6x+4, at 2 both 16. Guided 3a+9+a−5=4a+4, at 1 both 8. Exit 2n+5+3n−1=5n+4, at 2 both 14. |
Put sign and term on one card; regroup variables and constants separately. |
| 13 | 6x+12=6(x+2), at 3 both 30. Guided 4p+20=4(p+5), at 2 both 28. Exit blank 5; at t=1, both 18. |
Expand the proposed bracket back; use six equal rows for the model. |
| 14 | C=8+5n ⇒ n=(C−8)/5; C33→n5, C23→n3, C24→n16/5 outside whole-count domain. Exit C8→n0. |
Separate inverse operations from the real-world domain. Check original at candidate n. |
| 15 | Fresh key below; later practice keys in next table. | Diagnose expansion, collection, factoring and model interpretation separately. |
| 16 | 3x+7=25 ⇒ x6; guided 2y+5=12 ⇒ y7/2=3.5. Exit 4z−3=9 ⇒ z3. Each value checks in its original equation. |
Apply same inverse to both sides; keep exact fractions before rounding. |
| 17 | 6+4n=18+2n ⇒ n6, both30. At 5, A26/B28; at7, A34/B32. Exit B lower by2 tokens at7. |
Algebra gives exact crossing; a graph needs labelled axes and a domain note. |
| 18 | 9+5n≤34 ⇒ n≤5, allowed whole 0–5; 5 reaches34; 6 gives39 and fails. Exit ≤ includes 5, < excludes it. |
Test one below, at and beyond boundary; list the set, not only the maximum. |
| 19 | 15−3t>3 ⇒ −3t>−12 ⇒ t<4; domain 0–5 gives 0,1,2,3. At3 remaining6, at4 remaining3. Exit 8−2u≥0 ⇒ u≤4; domain must be named. |
Test a value that would violate the learner's unreversed sign; explain negative scaling. |
| 20 | Fresh key below; later practice keys in next table. | Diagnose equality, inequality, domain and interpretation separately. |
All 30 practice routes · answers and targeted feedback
| Day | A | B | C | Next move when reasoning breaks |
|---|---|---|---|---|
| 11 | 2(a+5)=2a+10; at a3 both16. |
4(b+3)=4b+12; at b2 both20. |
Correct 5(c+2)=5c+10; at c1 both15; false form7. |
Ask which term the outside multiplier reached. |
| 12 | 3t+8+4t−2=7t+6; at t2 both20. |
5r−4+2r+9=7r+5; at r1 both12. |
Correct 4p+10; at p2 both18, false 4p+4=12. |
Keep signed constants as separate marked tiles. |
| 13 | 8n+16=8(n+2). |
5p+20=5(p+4); at p2 both30. |
9q+18=9(q+2); proposed 9(q+18) expands 9q+162. |
Expand to verify before accepting a bracket. |
| 14 | n=(C−10)/4; C30→n5; check10+20=30. |
m=(L−6)/3; L21→m5; check6+15=21. |
h=(K−7)/2; K19→h6; check7+12=19, not12. |
Put original formula beside every inverse step. |
| 15 | After check 5(a+2)+a=6a+10; at a2 both22. |
8b+16=8(b+2); expand back. |
3(c+4)+c=4c+12; at c0 original12, false4. |
Do not reuse held-out numbers for reteach. |
| 16 | 2x+5=14 ⇒ x9/2=4.5; check9+5=14. |
5y−4=13 ⇒ y17/5=3.4; check17−4=13. |
4z+6=17 ⇒ z11/4=2.75; check11+6=17, not z11. |
Use inverse steps and substitution; fraction is exact. |
| 17 | 8+3n=20+n ⇒ n6; both26. |
5+5p=20+2p ⇒ p5; both30. |
4+6k=18+4k ⇒ k7; both46; k3 gives22 vs30. |
Compare both totals at the claimed crossing. |
| 18 | 4n+7≤27 ⇒ n≤5, whole0–5; 5→27, 6→31. |
3t+8<23 ⇒ t<5, whole0–4; 4→20, 5→23. |
2r+9≤21 ⇒ r≤6; 7→23 and fails; 6→21. |
Distinguish inclusive and exclusive boundary. |
| 19 | 14−2t≥4 ⇒ t≤5; at5 left4, at6 left2. |
21−3m>9 ⇒ m<4; at3 left12, at4 left9. |
10−4u≤2 ⇒ u≥2; at2 left2, at1 left6. |
Negative division reverses relation; test a value on either side. |
| 20 | After check 6+2p=18 ⇒ p6; 16−2p≥8 ⇒ p≤4, so p6 fails the second condition. |
5+3q=17+q ⇒ q6; 18−2q≥8 ⇒ q≤5, so q6 fails the second condition. |
3k+5≤20 ⇒ k≤5; 2k+4=14 ⇒ k5; at5 the inequality is 20≤20, so allowed. |
Keep later practice apart from the first check; solve both relations before combining. |
Day 15 new-case check · worked key and next moves
Item 1. Four trays each q+3 plus two additional q groups give E(q)=4(q+3)+2q=4q+12+2q=6q+12=6(q+2). Expanding 6(q+2) returns 6q+12. An alternative 2(3q+6) is equivalent but less reduced; accept with proof. No need to assume any real cards exist.
Item 2. 42=6q+12 ⇒ subtract 12: 30=6q ⇒ q=5. Original grouped form 4(5+3)+2×5=32+10=42; simplified 6×5+12=42. Both checks matter because an accidentally correct simplified form can hide a wrong original model.
Item 3. 39=6q+12 ⇒ 27=6q ⇒ q=27/6=9/2=4.5, not a non-negative whole number. So 39 is not a valid total under the stated identical whole-card grouping. This does not prove that 39 cards cannot exist in a real booth; the classroom model may omit arrangements.
Next moves: If expansion loses the 12, physically make four rows of three marker cards. If 4q+2q is combined incorrectly, group like terms and test q=1. If factorisation is wrong, expand the suggested form. If inverse arithmetic is wrong, keep 30÷6 separate from modelling. If a learner says 39 impossible in reality, ask what assumption made it impossible here. Record I independent, A neutral access, P content prompt or N not observed by criterion.
Day 20 new-case check · worked key and next moves
Item 1. Equality 7+4r=19+2r. Subtract 2r and 7: 2r=12, r=6; A 7+24=31, B 19+12=31. On a graph (6,31) means six whole batches and 31 model tokens for either layout. A line may pass through fractional x-values as a mathematical relation, but batch cases here are integers.
Item 2. At least 6 pins: 18−2r≥6. Subtract 18: −2r≥−12. Divide by −2, reversing relation: r≤6. With r≥0 whole and the nonnegative-pin model, allowed r∈{0,1,2,3,4,5,6}. At6 remaining6 (allowed); at7 remaining4 (excluded). The cap domain r≤9 from nonnegative pins is less restrictive here.
Item 3. At5, A 7+20=27, B 19+10=29, so A is 2 tokens lower. At7, A35, B33, so B is 2 tokens lower but seven batches violate the at-least-six-pins rule (only4 remain). A bounded answer: “Under the invented pin rule, r≤6; at five batches A uses fewer tokens, and at six both use31. The figures alone do not choose a real layout because actual supply, access or quality conditions are unknown.” Other accurate interpretations are valid; do not force a single preference.
Next moves: If equation setup is wrong, write a row for each layout before crossing. If algebra yields r6 but substitution disagrees, check constants. If inequality direction is wrong, test r7 to falsify it. If student gives only r6, test r5/r0 to reveal a solution set. If a learner recommends B at r7, ask them to apply the pin rule before comparing tokens. If the graph lacks units, label x/y with batches/tokens. Do not promote this to a real financial or accessibility decision.
These assessments are not validated scales. A teacher should use the observed steps to decide a next example, then check with a genuinely new item. Original SubjectNest staff key © NeuroForgeIO Pty Ltd 2026, CC BY 4.0.