Linear Modelling
Linear modelling represents real-world situations using straight-line relationships where one quantity changes at a constant rate relative to another. The general form y = mx + b captures scenarios from taxi fares to temperature changes, where m represents the rate of change and b the starting value. This mathematical tool transforms practical problems into predictable equations.
Why it matters
Linear modelling appears throughout GCSE mathematics and forms the foundation for understanding more complex mathematical relationships. Mobile phone contracts demonstrate linear models: a £25 monthly fee plus £0.10 per text creates the equation C = 25 + 0.1t. Energy companies use linear models for billing, combining standing charges of £120 annually with usage rates of £0.15 per kWh. In business, break-even analysis relies on linear models to determine when revenue equals costs. Weather forecasting uses linear models for short-term temperature predictions, whilst economists model inflation rates linearly over brief periods. These applications extend into A-level mathematics, economics, and physics, where linear relationships serve as building blocks for exponential, quadratic, and logarithmic models.
How to solve linear modelling
Linear Modelling
- Identify the variables: what is changing (x) and what depends on it (y)?
- Find the rate of change (slope) from the context.
- Find the starting value (y-intercept).
- Write the equation y = mx + b and use it to predict.
Example: Taxi: £2 base + £1.50/km → C = 1.5d + 2. Cost for 10 km = £17.
Worked examples
A taxi charges £50.00 base + £15.00 per km. What is the cost for 6 km?
Answer: £140.00
- Calculate the distance cost → 15 x 6 = £90.00 — Rate per km times distance.
- Add the base charge → 50 + 90 = £140.00 — Total = base + distance cost.
Write a formula: cost C for d km if base is £50.00 and rate is £10.00/km.
Answer: C = 50 + 10d
- Identify the fixed and variable parts → Fixed: £50.00, Variable: £10.00 per km — The base fee is fixed; the rate multiplied by distance is variable.
- Write the formula → C = 50 + 10d — Cost equals base plus rate times distance.
Temperature starts at 24 degrees C and drops 2 degrees C per hour. When is it 14 degrees C?
Answer: 5 hours
- Set up the equation → 24 - 2t = 14 — Temperature = start - rate x time.
- Solve for t → 2t = 24 - 14 = 10, t = 5 — Divide 10 by 2 to get 5 hours.
Common mistakes
- Confusing the order of variables in y = mx + b, writing cost = distance + rate × base instead of cost = base + rate × distance, giving £6 + £50 × 2 = £106 instead of £50 + £6 × 2 = £62
- Mixing up the rate per unit with the total change, calculating £15 per km for 4 km as £15 ÷ 4 = £3.75 instead of £15 × 4 = £60
- Forgetting to include the starting value when solving equations, writing 24 - 2t = 14 as t = 14 ÷ 2 = 7 instead of solving 2t = 10 to get t = 5