Mathematics • Year 8 • Unit 2 • Lesson 15

Linear Modelling

Build fluency turning real situations into the equation y = mx + c. Identify the rate (m), the fixed amount (c), write the equation, then substitute. One worked example, one guided fill-in, then eight independent problems.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. The same four-step recipe works for every linear model.

Problem. A plumber charges $80 call-out plus $50 per hour. Find the total cost C for 3 hours of work.

Step 1 — Identify the fixed amount (c) and the rate (m).

Fixed call-out fee: c = 80 (paid even for 0 hours).

Rate per hour: m = 50.

Reason: a "fee" or "starting fixed cost" is always c (y-intercept). A "per hour" / "per km" / "per text" is the gradient m.

Step 2 — Write the equation in y = mx + c form.

C = 50x + 80, where x is hours of work.

Step 3 — Substitute the value asked for.

C = 50 × 3 + 80 = 150 + 80 = 230.

Step 4 — State the answer in context with units.

The total cost for 3 hours is $230.

Answer: $230.

Stuck? Revisit lesson § Card 1 — "per" = gradient; "fixed/starting/call-out" = y-intercept.

2. We do — fill in the missing steps

A cyclist starts 5 km from home and rides at 15 km/h. Find the distance from home after 2 hours. Fill in each blank. 4 marks

Step 1 — Identify variables:

Fixed starting distance: c = ______ km (y-intercept).

Rate (speed): m = ______ km/h (gradient).

Step 2 — Write the equation:

d = ______ × t + ______

Step 3 — Substitute t = 2:

d = ______ × 2 + ______ = ______ + ______ = ______

Step 4 — Answer in context:

After 2 hours the cyclist is ______ km from home.

Stuck? Revisit lesson § Card 3 — for distance-speed-time, gradient = speed, y-intercept = starting distance.

3. You do — independent practice

For each, write the equation in y = mx + c form, then answer the question with units. Show your substitution.

Foundation — write the equation

3.1 A phone plan costs $30 per month plus $0.10 per text. Write C in terms of x texts.    1 mark

3.2 Water costs $1.50 per kilolitre plus a $15 service fee. Write C in terms of k kilolitres.    1 mark

3.3 A taxi has a $4 flag-fall and charges $2.50 per km. Write C in terms of d km.    1 mark

3.4 A car travels at 80 km/h starting from a point 20 km outside the city. Write d in terms of t hours.    1 mark

Standard — write AND evaluate

3.5 Water bill (from 3.2): find C when k = 20 kL. Show every step.    2 marks

3.6 Taxi (from 3.3): find C for a 12 km trip.    2 marks

Extension — solve for x

3.7 Using the phone plan from 3.1, how many texts will make the total bill exactly $42?    2 marks

3.8 A printing service charges $20 setup plus $0.40 per page. How many pages can you print for exactly $60?    2 marks

Stuck on 3.7 / 3.8? Set up the equation, replace C with the target value (e.g. 42 or 60), then solve algebraically for x.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (cyclist)

Step 1: c = 5, m = 15.
Step 2: d = 15 × t + 5.
Step 3: d = 15 × 2 + 5 = 30 + 5 = 35.
Step 4: after 2 hours the cyclist is 35 km from home.

3.1 — Phone plan

m = 0.10 (per text), c = 30 (monthly). C = 0.10x + 30.

3.2 — Water bill

m = 1.50 (per kL), c = 15 (service fee). C = 1.50k + 15.

3.3 — Taxi

m = 2.50 (per km), c = 4 (flag-fall). C = 2.50d + 4.

3.4 — Car

m = 80 (speed), c = 20 (starting distance). d = 80t + 20.

3.5 — Water bill at k = 20

C = 1.50 × 20 + 15 = 30 + 15 = $45.

3.6 — Taxi for 12 km

C = 2.50 × 12 + 4 = 30 + 4 = $34.

3.7 — Phone bill = $42

42 = 0.10x + 30 → 12 = 0.10x → x = 120. So 120 texts.

3.8 — Print bill = $60

60 = 0.40p + 20 → 40 = 0.40p → p = 100. So 100 pages.