Skip to content
mathlab
0
0
0 XP
Lvl 1
KJ
Lesson 15 ~35 min Unit 2 · Linear Relationships +100 XP

Linear Modelling

Apply linear equations to the real world — taxi fares, plumbers, cyclists and currency. Learn to identify rates and fixed costs in any context.

Today's hook: A taxi charges a $5 flag fall plus $2 per kilometre. A 3 km trip costs $11. A 10 km trip costs $25. Can you write the equation?
0/5QUESTS

Printable Worksheets

Print or save as PDF — or build a custom worksheet from any module's questions.

Build Apply Master Build custom
Think First
warm-up

A taxi charges a $5 flag fall plus $2 per kilometre travelled.

  • How much would a 3 km trip cost?
  • How much would a 10 km trip cost?
  • Can you write an equation for the cost $C$ of any trip of distance $d$ km?
3 km: $C = 2 \times 3 + 5 = \$11$. 10 km: $C = 2 \times 10 + 5 = \$25$. Equation: $C = 2d + 5$
1
What is Linear Modelling?
+5 XP

A linear model uses the equation $y = mx + c$ to describe a real-world relationship. The trick is to identify which quantity is $x$, which is $y$, what the rate $m$ is, and what the starting value $c$ is.

Real-world situation$x$ (input)$y$ (output)$m$ (rate)$c$ (fixed)
Taxi fareDistance (km)Total cost ($)Cost per kmFlag fall
Phone billNumber of textsTotal cost ($)Cost per textMonthly fee
Distance travelledTime (hours)Distance (km)Speed (km/h)Starting distance
Temperature conversionCelsiusFahrenheitScale factorOffset

General strategy:

  1. Read the problem and identify the two changing quantities.
  2. Decide which one depends on the other — this is $y$.
  3. Look for a “per” rate — this is $m$.
  4. Look for a fixed starting amount — this is $c$.
  5. Write the equation: $y = mx + c$.
Tip: “per”
Whenever you see “per”, think gradient.
Tip: starting fee
Whenever you see a starting fee or base charge, think y-intercept.
Always define
Always define what $x$ and $y$ represent in your model.
5 10 15 20 25 2 4 6 8 Distance (km) Cost ($) Fixed cost = $5 Rate = $2/km Taxi Cost: C = 2d + 5
2
Cost Models
+5 XP

Cost models have the form:

$$\text{Total Cost} = (\text{rate} \times \text{quantity}) + \text{fixed cost}$$

Or simply: $$C = mx + c$$

Worked Example — Plumber: $80 call-out + $50/hr. Total cost for 3 hours?
  1. Identify the fixed cost and the rate

    Fixed cost (call-out fee) $= c = 80$

    Rate (cost per hour) $= m = 50$

  2. Write the equation

    $$C = 50x + 80$$

  3. Substitute $x = 3$ hours

    $$C = 50 \times 3 + 80 = 150 + 80 = 230$$

  4. State the answer with units

    The total cost for 3 hours is $230.

Always include units in the answer and check the answer is reasonable.
0 100 200 300 400 0 1 2 3 4 5 Hours ($x$) Cost ($) $230 $80 fixed Plumber Cost: $C = 50x + 80$
3
Distance-Speed-Time
+5 XP

In distance-time graphs, the gradient represents speed. A steeper line means faster travel.

$$\text{Distance} = \text{speed} \times \text{time} + \text{starting distance}$$

Or: $d = vt + d_0$

Worked Example — Cyclist starts 5 km from home, rides at 15 km/h. Distance after 2 hours?
  1. Identify the variables

    Starting distance $= d_0 = 5$ km

    Speed (gradient) $= v = 15$ km/h

  2. Write the equation

    $$d = 15t + 5$$

  3. Substitute $t = 2$

    $$d = 15 \times 2 + 5 = 30 + 5 = 35$$

  4. State the answer in context

    After 2 hours, the cyclist is 35 km from home.

The gradient on a distance-time graph always has units of km/h or m/s — it is the speed.
0 10 20 30 40 50 0 1 2 3 4 Time (hours) 35 km Starts at 5 km 15 km/h Cyclist: $d = 15t + 5$
4
Conversion Graphs
+5 XP

A conversion graph is a straight line that converts one unit to another. You can read values directly from the graph or use the equation.

Worked Example — Currency graph through (0,0) and (100 AUD, 65 USD). Convert 40 AUD.
  1. Find the gradient (exchange rate)

    $$m = \frac{65 - 0}{100 - 0} = 0.65$$

    This means 1 AUD = 0.65 USD.

  2. Write the equation

    Line passes through (0, 0), so $c = 0$: $y = 0.65x$

  3. Substitute $x = 40$ AUD

    $$y = 0.65 \times 40 = 26$$

  4. State the answer

    40 AUD = 26 USD.

When the line passes through the origin, $c = 0$ — there is no fixed amount.
0 20 40 60 80 0 25 50 75 100 125 AUD ($x$) 26 USD Rate: 0.65 AUD to USD: $y = 0.65x$
5
Interpreting Gradient in Context
interpretation

The gradient $m$ in a linear model always tells you the rate of change. Its meaning depends entirely on what the variables represent.

ContextWhat $m$ meansExample value
Taxi fare vs distanceCost per kilometre$m = 2.50$ means $2.50 per km
Distance vs timeSpeed$m = 60$ means 60 km/h
Water bill vs usageCost per kilolitre$m = 1.50$ means $1.50 per kL
Wages vs hours workedHourly rate$m = 25$ means $25 per hour
Temperature conversionScale factor$m = \frac{9}{5}$ relates Celsius to Fahrenheit
Include units
The gradient always has units of $\frac{\text{y-units}}{\text{x-units}}$. Include units to show you understand the context.
6
Interpreting Intercept in Context
interpretation

The y-intercept $c$ is the value of $y$ when $x = 0$. It represents a starting value, fixed cost, or base amount.

ContextWhat $c$ meansExample
Taxi fareFlag fall$c = 3.60$ means $3.60 flag fall
Phone planMonthly base fee$c = 30$ means $30/month regardless of usage
Gym membershipJoining fee$c = 100$ means $100 one-time joining fee
Temperature conversionOffset constant$c = 32$ in $F = \frac{9}{5}C + 32$
Watch out: $c = 0$
Sometimes there is no fixed cost. Currency conversion has $c = 0$ — 0 AUD = 0 USD. The line passes through the origin.
Brain Trainer
speed drill

Set a timer for 3 minutes. Solve as many as you can!

1

A taxi costs $5 flag fall + $3 per km. Write the equation for cost $C$.

Answer

$C = 3d + 5$

2

Using $C = 3d + 5$, what is the cost for 8 km?

Answer

$C = 3 \times 8 + 5 = 29$. Cost is $29.

3

A car travels at 60 km/h. Write the equation for distance $d$ after $t$ hours.

Answer

$d = 60t$

4

Using $d = 60t$, how far does the car travel in 2.5 hours?

Answer

$d = 60 \times 2.5 = 150$ km.

5

A phone plan costs $25/month + $0.15 per text. Find the cost for 100 texts.

Answer

$C = 0.15 \times 100 + 25 = 40$. Cost is $40.

6

What is the gradient of the line through $(0, 10)$ and $(5, 30)$?

Answer

$m = \frac{30-10}{5-0} = 4$

7

Write the equation of a line with $m = 2.5$ and $c = 12$.

Answer

$y = 2.5x + 12$

8

A gym charges $80 joining fee + $18/week. When does the total reach $260?

Answer

$260 = 18w + 80$ → $w = 10$ weeks.

9

Convert 30°C to Fahrenheit using $F = 1.8C + 32$.

Answer

$F = 1.8 \times 30 + 32 = 86$°F.

10

A water bill is $C = 1.2x + 10$ where $x$ is kL. What is the fixed charge?

Answer

$10 (the y-intercept).

Q1
A plumber charges a $60 call-out fee plus $40 per hour. Which equation gives the total cost $C$ for $x$ hours?
10 XP
Q2
A car travels at 80 km/h. On a distance-time graph, which variable represents the gradient?
10 XP
Q3
A phone plan costs $30 per month plus $0.10 per text. In $C = 0.1x + 30$, what is the fixed cost?
10 XP
Q4
The graph shows printing cost against brochures. What is the cost per brochure?
10 XP
0 50 100 150 200 0 50 100 150 200 Number of brochures Printing Cost
Q5
Water is charged at $1.50 per kilolitre plus a $15 service fee. What is the cost for 20 kL?
10 XP
SAQ1
Short Answer
15 XP

A taxi charges a $4 flag fall plus $1.80 per kilometre.

(a) Write a linear equation for the cost $C$ in terms of distance $d$.

(b) Calculate the cost of a 15 km trip.

Sample solution

(a) The rate is $1.80 per km and the fixed cost is $4. $C = 1.8d + 4$

(b) Substitute $d = 15$: $C = 1.8 \times 15 + 4 = 27 + 4 = 31$. A 15 km trip costs $31.

SAQ2
Short Answer
15 XP

A temperature conversion graph passes through $(0, 32)$ and $(100, 212)$, where $x$ is Celsius and $y$ is Fahrenheit.

(a) Find the equation in the form $y = mx + c$.

(b) Convert 25°C to Fahrenheit.

Sample solution

(a) $m = \frac{212-32}{100-0} = \frac{180}{100} = \frac{9}{5}$. $c = 32$. Equation: $F = \frac{9}{5}C + 32$ (or $y = 1.8x + 32$).

(b) $y = \frac{9}{5} \times 25 + 32 = 45 + 32 = 77$. So 25°C = 77°F.

SAQ3
Short Answer
20 XP

A gym charges a $100 joining fee plus $20 per week.

(a) Write an equation for the total cost $C$ after $w$ weeks.

(b) After how many weeks has the total cost reached $500?

Sample solution

(a) $C = 20w + 100$

(b) $500 = 20w + 100$ → $400 = 20w$ → $w = 20$. The total reaches $500 after 20 weeks.

Stretch Challenge — Break-Even Analysis
extension

A bakery sells cupcakes. It costs $200 per day in rent and wages, plus $1.20 for ingredients per cupcake. Each cupcake sells for $3.50.

(a) Write a linear equation for the total cost $C$ of making $n$ cupcakes.

(b) Write a linear equation for the revenue $R$ from selling $n$ cupcakes.

(c) Find the break-even point — the number of cupcakes where cost equals revenue.

Hint

The break-even point is where $C = R$. Set the two equations equal and solve for $n$.

Full solution

(a) $C = 1.2n + 200$

(b) $R = 3.5n$

(c) At break-even, $C = R$:

$$1.2n + 200 = 3.5n$$

$$200 = 2.3n$$

$$n \approx 87 \text{ cupcakes}$$

0 200 400 600 800 0 50 100 150 200 Cupcakes sold ($n$) 87, $304 Cost Revenue Break-Even Analysis
Key Takeaways
copy these
Linear model form$y = mx + c$ — $m$ is the rate of change, $c$ is the starting/fixed value.
Gradient in contextAlways has units of $\frac{\text{y-units}}{\text{x-units}}$ e.g. $/km, km/h, $/hr.
y-intercept in contextThe value when $x = 0$ — flag fall, joining fee, base charge, offset.
StrategyFind “per” → $m$. Find fixed amount → $c$. Define variables. Write equation. Substitute. Check.
Lesson Complete!
+100 XP

You can now apply linear equations to real-world situations — identifying rates, fixed costs, and interpreting graphs in context.