Codes: AC9M10A02, AC9M10A04. Goal: solve and interpret linear inequalities for whole-month commitments. Prepare: $1,000 practice cap, cost models.
- Ask · 2 min. “If the fictional cap is $1,000, does a solution of 9.5 months mean 9.5 payments are possible?” No: the model uses complete months.
- Model A · 5 min.
240+80m ≤ 1000→80m ≤ 760→m ≤ 9.5; with whole months 0–12, maximum 9. Check A(9)=$960 and A(10)=$1,040. - Guide B · 6 min.
480+50m ≤ 1000→50m ≤ 520→m ≤ 10.4; maximum 10. Check B(10)=$980 and B(11)=$1,030. - Apply · 7 min. A fictional planner needs 10 complete months. Students decide which plan fits the stated cap: B only, within this model. They must say what the calculation does not judge.
- Challenge · 3 min. Ask whether A at m=9 and B at m=10 is a fair price-only comparison for the same service duration. No; compare the same horizon.
- Exit · 2 min. Everyone writes A's maximum whole months and the pair of values that check the boundary.
Access: number-line inequality and table route; calculator as check after symbolic steps; oral explanation of whole-month boundary accepted.