Forming linear equations from situations | Year 12

01 / LEARN

Understand the idea.

Choose the changing quantities

Define the input and output with units before choosing a rule. A linear model has a constant rate of change: each extra kilometre, minute or item changes the output by the same amount.

In $C = m d + b$ , the gradient $m$ measures the change in cost per unit of distance. The intercept $b$ is the predicted cost when $d = 0$ ; interpret it only if zero belongs to the practical domain.

Combine rate and fixed amount

A fixed charge is paid once, so it is added. A per-unit charge is multiplied by the input. Check that both terms have the output units: dollars per kilometre multiplied by kilometres gives dollars.

State where the model is useful. Distance can vary continuously, but tickets are whole numbers. A draining model must stop when the tank empties; extending the straight line beyond that point would predict an impossible negative volume.

C = m x + b

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A delivery fee

A courier charges AUD 18 plus AUD 2.40 per kilometre. Write its charge for $d$ kilometres and find the charge for 15 kilometres.

Worked solution

1
Define the quantities.
$d \geq 0, C in AUD, d in km$
Distance is non-negative, and the charge is measured in dollars.
2
Add the fixed and distance charges.
$C = 18 + 2.40 d$
The AUD 18 is paid once; each kilometre adds AUD 2.40.
3
Substitute the journey length.
$\begin{aligned} C (15) & = 18 + 2.40 (15) \\ = 18 + 36 \\ = 54 \end{aligned}$
Fifteen kilometres contributes AUD 36, giving a total charge of AUD 54.

Answer

$C = 18 + 2.40 d$ ; AUD 54.

EXAMPLE 02

A draining tank

A tank holds 950 litres and drains at 35 litres per minute. Write its volume after $t$ minutes and its useful domain.

Worked solution

1
Write the decreasing-volume rule.
$V = 950 - 35 t$
The initial volume is 950 L and each minute removes 35 L.
2
Require a non-negative volume.
$\begin{aligned} 950 - 35 t & \geq 0 \\ 35 t & \leq 950 \\ t & \leq \frac{190}{7} \end{aligned}$
The model remains physical only while water is left in the tank.
3
Include the starting time.
$0 \leq t \leq \frac{190}{7} min \approx 27.14 min$
Time starts at zero and the upper endpoint is the instant the tank empties.
4
Check the endpoint.
$V (\frac{190}{7}) = 950 - 35 (\frac{190}{7}) = 0$
At the upper boundary, the predicted volume is exactly zero litres.

Answer

$V = 950 - 35 t$ for $0 \leq t \leq 190 / 7$ minutes.

03 / YOUR TURN

Try it yourself.

Check your understanding

A room costs AUD 90 plus AUD 7 per participant. Write its total cost.

Show answer & reasoning

$C = 90 + 7 n$ for whole-number $n \geq 0$ .

The AUD 90 is paid once and each of $n$ participants adds AUD 7, so $C = 90 + 7 n$ . Participant counts require whole-number $n \geq 0$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

A printer charges AUD 12 setup and AUD 0.08 per page. Write $C$ for $p$ pages.
2 marks
A 24 cm candle burns down at 1.5 cm per hour. Write its height and useful time domain.
2 marks
Temperature rises from 8 degrees Celsius at 6 am to 20 degrees at noon at a constant rate. Let $t$ be hours after 6 am. Find $T (t)$ .
2 marks
An account starts with AUD 250 and receives AUD 40 each week, with no interest. Find its balance after 9 weeks.
2 marks
A linear cost is AUD 46 for 3 units and AUD 70 for 7 units. Find its rule.
2 marks
Interpret $M = 180 - 4 t$ , where mass is in grams and time in minutes. When does it become physically impossible?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A printer charges AUD 12 setup and AUD 0.08 per page. Write $C$ for $p$ pages.
2 marks
The setup is constant; multiply the page rate by the number of pages.
$C = 12 + 0.08 p$ .
A 24 cm candle burns down at 1.5 cm per hour. Write its height and useful time domain.
2 marks
Subtract the amount burnt.
Set height to zero for the 16-hour endpoint.
$h = 24 - 1.5 t$ , $0 \leq t \leq 16$ .
Temperature rises from 8 degrees Celsius at 6 am to 20 degrees at noon at a constant rate. Let $t$ be hours after 6 am. Find $T (t)$ .
2 marks
The rate is $\frac{20 - 8}{6} = 2$ degrees per hour.
$T = 8 + 2 t$ for $0 \leq t \leq 6$ .
An account starts with AUD 250 and receives AUD 40 each week, with no interest. Find its balance after 9 weeks.
2 marks
$B = 250 + 40 w$ .
$B (9) = 250 + 360 = 610$ .
AUD 610.
A linear cost is AUD 46 for 3 units and AUD 70 for 7 units. Find its rule.
2 marks
$m = \frac{70 - 46}{7 - 3} = 6$ .
$46 = 6 (3) + b$ gives $b = 28$ .
$C = 6 n + 28$ .
Interpret $M = 180 - 4 t$ , where mass is in grams and time in minutes. When does it become physically impossible?
2 marks
Solve $180 - 4 t = 0$ .
For $t > 45$ the rule predicts negative mass.
Initial mass 180 g, loss 4 g/min; stop at 45 minutes.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics: Modelling with Linear Relationships, a compulsory school-assessed topic. The external examination scope differs from the full course.

Official curriculum source ↗

In this lesson

Understand the idea.

Choose the changing quantities

Combine rate and fixed amount

See the method in action.

A delivery fee

Define the quantities.

Add the fixed and distance charges.

Substitute the journey length.

A draining tank

Write the decreasing-volume rule.

Require a non-negative volume.

Include the starting time.

Check the endpoint.

Try it yourself.

Check your understanding

Keep practising