Mathematics • Year 8 • Unit 2 • Lesson 20
Real-World Linear Stories
Five real-world problems that need a full Unit 2 toolkit: build a linear model, compare two plans, solve a system from a word problem. Define your variables, write the equations, choose substitution or elimination, then check.
1. Word problems
For each: (i) define variables (with units), (ii) write the equation(s), (iii) solve — choose substitution or elimination if it's a system, (iv) verify and INTERPRET in context.
1.1 — Streaming plans. Plan A charges $15 per month plus $0.50 per movie rented. Plan B charges $5 per month plus $1 per movie rented. Let n = number of movies in a month, C = monthly cost in dollars.
(a) Write a linear equation C = mn + c for each plan.
(b) Find the value of n where the two plans cost the same.
(c) Below that value, which plan is cheaper? 3 marks
1.2 — Cinema tickets. 2 adults and 3 children pay $72. 1 adult and 4 children pay $58. Let a = adult price, c = child price (in dollars).
(a) Write the two equations.
(b) Solve by either substitution OR elimination — state your method.
(c) Predict the cost of 3 adults and 2 children at this cinema. 4 marks
1.3 — Mobile data. A SIM plan starts with 2 GB free, then $5 per extra GB used. Let G = total GB used in a month, C = monthly cost in dollars (above 2 GB only).
(a) Write a linear model for C in terms of G (valid when G ≥ 2).
(b) Use it to find the cost when G = 7 GB.
(c) If a user is charged $30 extra one month, how many GB did they use in total? 3 marks
1.4 — Two trains. Train P leaves a station heading east at a steady 60 km/h. Train Q leaves the SAME station 1 hour later, also east, but at 80 km/h. Let t = hours since Train P left.
(a) Write a linear model for the distance from the station for EACH train (note: Train Q's model is valid only when t ≥ 1).
(b) Find the value of t when Train Q catches up to Train P.
(c) How far from the station does this happen? 4 marks
1.5 — Fence and gate. Liam buys 12 fence panels and 1 gate for a total of $445. Each panel costs $30 more than the gate. Let p = price of a panel and g = price of the gate (in dollars).
(a) Write two equations relating p and g.
(b) Solve by substitution.
(c) State the prices and check both equations. 3 marks
2. Explain your thinking
This question is about interpretation. Use full sentences. 4 marks
2.1 A taxi company models its fare with C = 2k + 5, where k is the number of kilometres and C is the cost in dollars.
(i) State the gradient and y-intercept of the model. (ii) Explain in everyday language what each one MEANS in this context (don't just say "m = 2", say what the 2 represents). (iii) Why does the y-intercept of $5 make sense even when k = 0? (iv) Why is the gradient $2 (not, say, $2 per minute or $2 per ride)? Use the phrase "rate of change" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Streaming plans
(a) Plan A: C = 0.5n + 15. Plan B: C = n + 5.
(b) Set equal: 0.5n + 15 = n + 5 → 10 = 0.5n → n = 20. At 20 movies/month both plans cost $25.
(c) Below n = 20, Plan B is cheaper (lower fixed fee dominates when you rent few movies). Above n = 20, Plan A is cheaper (lower per-movie rate dominates).
1.2 — Cinema tickets
(a) Eq 1: 2a + 3c = 72. Eq 2: a + 4c = 58.
(b) Substitution. From Eq 2: a = 58 − 4c. Sub: 2(58 − 4c) + 3c = 72 → 116 − 8c + 3c = 72 → −5c = −44 → c = $8.80. Then a = 58 − 4(8.80) = $22.80.
(c) 3a + 2c = 3(22.80) + 2(8.80) = 68.40 + 17.60 = $86.
1.3 — Mobile data
(a) C = 5(G − 2) = 5G − 10 (for G ≥ 2).
(b) At G = 7: C = 5(7) − 10 = 35 − 10 = $25.
(c) 30 = 5G − 10 → 5G = 40 → G = 8 GB total (i.e., 6 GB above the free 2 GB).
1.4 — Two trains
(a) Train P: d = 60t. Train Q: d = 80(t − 1) for t ≥ 1.
(b) Set equal: 60t = 80(t − 1) → 60t = 80t − 80 → 80 = 20t → t = 4 hours (after Train P left).
(c) Distance = 60(4) = 240 km. Check: Train Q at t = 4: 80(4 − 1) = 80(3) = 240 ✓.
1.5 — Fence and gate
(a) Eq 1: 12p + g = 445. Eq 2: p = g + 30.
(b) Sub Eq 2 into Eq 1: 12(g + 30) + g = 445 → 12g + 360 + g = 445 → 13g = 85 → g = 85/13 ≈ $6.54. Then p = 85/13 + 30 = 85/13 + 390/13 = 475/13 ≈ $36.54.
(c) Check Eq 1: 12(475/13) + 85/13 = 5700/13 + 85/13 = 5785/13 = 445 ✓. (Note: numbers come out as awkward fractions — the problem could equally well be designed with nicer values like total $480, panel $30 more → g = $10, p = $40.)
2.1 — Explain your thinking (sample response)
The gradient is m = 2 and the y-intercept is c = 5. The gradient represents the rate of change of cost with kilometres — every extra kilometre driven adds $2 to the fare. The y-intercept of $5 makes sense even at k = 0 because it is the flagfall: a fixed charge the taxi company applies the moment the meter starts, before the car has moved at all. The gradient is $2 per kilometre (not per minute or per ride) because in this model x is kilometres travelled, so m tells you the cost-per-extra-unit of WHATEVER x measures — and here x measures distance, in km. So the units of m are dollars per kilometre.
Marking: 1 mark for correctly identifying m = 2 and c = 5; 1 mark for explaining m in context (cost added per extra km); 1 mark for explaining c in context (flagfall when k = 0); 1 mark for using "rate of change" correctly and noting the units of m match the units of x.