Linear Modelling
Linear modelling transforms real-world scenarios into mathematical equations that students can solve systematically. When teaching CCSS 8.F and LK20 Grade 10 standards, students learn to identify variables, determine rates of change, and write equations like C = 50 + 15d for practical situations.
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Why it matters
Linear modelling appears everywhere in daily life, from calculating taxi fares to comparing phone plans. Students who master these skills can analyze situations like choosing between a gym membership at $40 monthly with $5 per class versus $15 per drop-in class. They can determine break-even points, predict costs, and make informed decisions. In business contexts, linear models help calculate profit margins, shipping costs, and pricing strategies. For example, a delivery service charging $25 base fee plus $3 per kilometer can predict that a 12-kilometer delivery costs $61. These mathematical tools prepare students for advanced topics in economics, engineering, and data analysis while building critical thinking skills for personal financial decisions.
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 $60.00 base + $10.00 per km. What is the cost for 5 km?
Answer: $110.00
- Calculate the distance cost → 10 x 5 = $50.00 — Rate per km times distance.
- Add the base charge → 60 + 50 = $110.00 — Total = base + distance cost.
Write a formula: cost C for d km if base is $50.00 and rate is $15.00/km.
Answer: C = 50 + 15d
- Identify the fixed and variable parts → Fixed: $50.00, Variable: $15.00 per km — The base fee is fixed; the rate multiplied by distance is variable.
- Write the formula → C = 50 + 15d — 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
- ✗Students confuse the base cost with the rate, writing C = 15 + 50d instead of C = 50 + 15d when the base is $50 and rate is $15 per unit.
- ✗When finding time in decay problems, students forget the negative sign and write 20 + 3t = 8 instead of 20 - 3t = 8 for temperature dropping 3 degrees per hour.
- ✗Students mix up variables in break-even problems, setting 100 + 5x = 200 + 3y instead of 100 + 5x = 200 + 3x when comparing two plans.
- ✗In distance-rate problems, students calculate 60 + 12 × 4 = 108 but forget units, writing $108 instead of 108 kilometers or 108 minutes depending on context.
Practice on your own
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