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.
Printable Worksheets
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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?
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 fare | Distance (km) | Total cost ($) | Cost per km | Flag fall |
| Phone bill | Number of texts | Total cost ($) | Cost per text | Monthly fee |
| Distance travelled | Time (hours) | Distance (km) | Speed (km/h) | Starting distance |
| Temperature conversion | Celsius | Fahrenheit | Scale factor | Offset |
General strategy:
- Read the problem and identify the two changing quantities.
- Decide which one depends on the other, this is $y$.
- Look for a “per” rate, this is $m$.
- Look for a fixed starting amount, this is $c$.
- Write the equation: $y = mx + c$.
Cost models have the form:
$$\text{Total Cost} = (\text{rate} \times \text{quantity}) + \text{fixed cost}$$
Or simply: $$C = mx + c$$
-
Identify the fixed cost and the rate
Fixed cost (call-out fee) $= c = 80$
Rate (cost per hour) $= m = 50$
-
Write the equation
$$C = 50x + 80$$
-
Substitute $x = 3$ hours
$$C = 50 \times 3 + 80 = 150 + 80 = 230$$
-
State the answer with units
The total cost for 3 hours is $230.
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: $d = vt + d_0$
-
Identify the variables
Starting distance $= d_0 = 5$ km
Speed (gradient) $= v = 15$ km/h
-
Write the equation
$$d = 15t + 5$$
-
Substitute $t = 2$
$$d = 15 \times 2 + 5 = 30 + 5 = 35$$
-
State the answer in context
After 2 hours, the cyclist is 35 km from home.
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.
-
Find the gradient (exchange rate)
$$m = \frac{65 - 0}{100 - 0} = 0.65$$
This means 1 AUD = 0.65 USD.
-
Write the equation
Line passes through (0, 0), so $c = 0$: $y = 0.65x$
-
Substitute $x = 40$ AUD
$$y = 0.65 \times 40 = 26$$
-
State the answer
40 AUD = 26 USD.
The gradient $m$ in a linear model always tells you the rate of change. Its meaning depends entirely on what the variables represent.
| Context | What $m$ means | Example value |
|---|---|---|
| Taxi fare vs distance | Cost per kilometre | $m = 2.50$ means $2.50 per km |
| Distance vs time | Speed | $m = 60$ means 60 km/h |
| Water bill vs usage | Cost per kilolitre | $m = 1.50$ means $1.50 per kL |
| Wages vs hours worked | Hourly rate | $m = 25$ means $25 per hour |
| Temperature conversion | Scale factor | $m = \frac{9}{5}$ relates Celsius to Fahrenheit |
The y-intercept $c$ is the value of $y$ when $x = 0$. It represents a starting value, fixed cost, or base amount.
| Context | What $c$ means | Example |
|---|---|---|
| Taxi fare | Flag fall | $c = 3.60$ means $3.60 flag fall |
| Phone plan | Monthly base fee | $c = 30$ means $30/month regardless of usage |
| Gym membership | Joining fee | $c = 100$ means $100 one-time joining fee |
| Temperature conversion | Offset constant | $c = 32$ in $F = \frac{9}{5}C + 32$ |
Set a timer for 3 minutes. Solve as many as you can!
A taxi costs $5 flag fall + $3 per km. Write the equation for cost $C$.
Answer
$C = 3d + 5$
Using $C = 3d + 5$, what is the cost for 8 km?
Answer
$C = 3 \times 8 + 5 = 29$. Cost is $29.
A car travels at 60 km/h. Write the equation for distance $d$ after $t$ hours.
Answer
$d = 60t$
Using $d = 60t$, how far does the car travel in 2.5 hours?
Answer
$d = 60 \times 2.5 = 150$ km.
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.
What is the gradient of the line through $(0, 10)$ and $(5, 30)$?
Answer
$m = \frac{30-10}{5-0} = 4$
Write the equation of a line with $m = 2.5$ and $c = 12$.
Answer
$y = 2.5x + 12$
A gym charges $80 joining fee + $18/week. When does the total reach $260?
Answer
$260 = 18w + 80$ → $w = 10$ weeks.
Convert 30°C to Fahrenheit using $F = 1.8C + 32$.
Answer
$F = 1.8 \times 30 + 32 = 86$°F.
A water bill is $C = 1.2x + 10$ where $x$ is kL. What is the fixed charge?
Answer
$10 (the y-intercept).
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.
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.
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.
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, $C = R$:
$$1.2n + 200 = 3.5n$$
$$200 = 2.3n$$
$$n \approx 87 \text{ cupcakes}$$
You can now apply linear equations to real-world situations, identifying rates, fixed costs, and interpreting graphs in context.