Explore formulas with digital tools

01 / LEARN

The idea, step by step

Read the formula and its inputs

In the Formula lab, choose the cost formula $C = p n + f$ . Here $C$ is total cost in dollars, $p$ is price per visit in dollars, $n$ is the number of visits and $f$ is a fixed fee in dollars paid once. The letters $p n$ mean $p \times n$ .

Set $p = 3$ , $n = 2$ and $f = 5$ . The live result is $C = 3 \times 2 + 5 = 11$ , so the total is 11 dollars. Check one result by hand before investigating. These are invented activity prices for practising mathematics.

C = p n + f

Change one input, keep the others fixed

Keep $p = 3$ and $f = 5$ fixed in the digital tool. Change only $n$ through $2, 3, 4$ . Record all three inputs and the live output in a comparison table. This regular change is systematic variation.

The costs increase by 3 dollars each time: $14 - 11 = 3$ and $17 - 14 = 3$ . Each extra visit adds the price of one visit; the fixed fee is still paid only once. Predict the cost for $n = 5$ , then test it in the tool: $3 \times 5 + 5 = 20$ .

Cost investigation: only the number of visits changes
Price p (dollars)	Visits n	Fixed fee f (dollars)	Cost C (dollars)
3	2	5	11
3	3	5	14
3	4	5	17

Test a second formula

Switch the Formula lab to perimeter: $P = 2 l + 2 w$ , where $l$ is a rectangle's length, $w$ is its width and $P$ is its perimeter, all measured in centimetres. Keep $w = 4$ fixed and use the length control to test $l = 5, 6, 7$ . Record $P = 18, 20, 22$ in a table.

Increasing length by 1 cm increases perimeter by 2 cm because a rectangle has two sides of that length. Write a prediction before changing a control, record the settings and results, and compare differences afterwards. To investigate width instead, hold length fixed and change only width.

P = 2 l + 2 w

Formula lab

Change one input at a time. Keep the others fixed and record your results.

C = pn + f

p · Price per visit ($)n · Number of visitsf · Fixed fee ($)

3 × 2 + 5C = $11

0/6 rows

A common trap

If two inputs change together, the whole change in the output must be caused by the first input.

Instead: You cannot isolate one input's effect that way. Keep the other inputs fixed and compare results when only the chosen input changes. For $C = p n + f$ , changing visits and price together mixes two effects.

02 / WORKED EXAMPLES

See the thinking

Example 1

An extra visit

Use $C = p n + f$ with $p = 4$ and $f = 6$ . In the Formula lab, vary $n$ through $2, 3, 4$ . What is the effect of each extra visit?

Keep the price and fixed fee unchanged. Change only the number of visits.
Record $(n, C) = (2, 14), (3, 18), (4, 22)$ , since $4 \times 2 + 6 = 14$ , $4 \times 3 + 6 = 18$ and $4 \times 4 + 6 = 22$ .
Compare successive costs: $18 - 14 = 4$ and $22 - 18 = 4$ .

Each extra visit adds 4 dollars while

p = 4

and

f = 6

remain fixed.

Example 2

A changing fixed fee

Keep $p = 3$ and $n = 4$ . Use the digital cost formula to compare fixed fees $f = 2, 5, 8$ . What changes?

The visit cost stays $p n = 3 \times 4 = 12$ dollars.
Record $(f, C) = (2, 14), (5, 17), (8, 20)$ using $12 + 2 = 14$ , $12 + 5 = 17$ and $12 + 8 = 20$ .
Each fee rises by 3 dollars and each total also rises by 3 dollars: $17 - 14 = 3$ and $20 - 17 = 3$ .

Increasing the fixed fee by 3 dollars increases the total by 3 dollars when price and visits stay fixed.

Example 3

A longer rectangle

Use the digital formula $P = 2 l + 2 w$ . Keep $w = 3$ cm and change $l$ through $4, 6, 8$ cm. Predict and check the perimeter at $l = 10$ cm.

Record $(l, P) = (4, 14), (6, 18), (8, 22)$ : for example, $2 \times 4 + 2 \times 3 = 14$ cm.
Length rises by 2 cm each time. The perimeter rises by 4 cm: $18 - 14 = 4$ and $22 - 18 = 4$ .
Predict $P = 26$ cm for length 10 cm. Check in the tool and by substitution: $2 \times 10 + 2 \times 3 = 26$ .

With width fixed at 3 cm, each 2 cm increase in length adds 4 cm to the perimeter. The checked prediction is 26 cm.

Check your understanding

In $C = p n + f$ , keep $p = 2$ and $f = 6$ . Predict the change in cost when $n$ increases from 4 to 7, then check it in the Formula lab.

Show answer

The cost increases by 6 dollars, from 14 to 20 dollars.

There are 3 extra visits at 2 dollars each, so the increase is $3 \times 2 = 6$ . The checked totals are $2 \times 4 + 6 = 14$ and $2 \times 7 + 6 = 20$ .

03 / WORKSHEET

Your turn

8 questions · 28 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $C = p n + f$ , $p$ is the price per visit, $n$ is visits and $f$ is a fixed fee paid once. Set $p = 3$ , $n = 4$ , $f = 5$ in the Formula lab. Show a substitution that checks its output and state the total in dollars.
2 marks
In the digital cost formula, hold $p = 5$ and $f = 3$ fixed. Change $n$ through $2, 3, 4$ . Record a comparison table showing every input and output, then describe the effect of each extra visit.
4 marks
Use the Formula lab with $n = 3$ and $f = 4$ fixed. Vary $p$ through $2, 4, 6$ dollars per visit. Record each total and describe the cost change for each 2-dollar rise in price per visit.
4 marks
Keep $p = 4$ and $n = 3$ in the digital cost formula. Compare $f = 1, 4, 7$ . State the three totals, their successive differences, and why this investigation isolates the fixed fee's effect.
3 marks
Switch the Formula lab to $P = 2 l + 2 w$ . Hold $w = 4$ cm fixed and vary $l$ through $3, 5, 7$ cm. Record a table of lengths and perimeters, state the fixed width, and describe the change in perimeter for each 2 cm increase in length.
4 marks
A student predicts: 'Doubling a rectangle's length while keeping its width fixed always doubles its perimeter.' In the digital tool, keep $w = 3$ cm and compare $l = 4$ cm with $l = 8$ cm. Use the two outputs to decide whether the prediction is correct.
3 marks
For $C = p n + f$ , keep $p = 3$ and $f = 5$ . Predict whether changing $n$ from 2 to 4 doubles the total cost. Test both settings digitally, record the totals and explain the result using the fixed fee.
4 marks
A student records $(p, n, f) = (2, 3, 4)$ giving $C = 10$ , then $(3, 5, 4)$ giving $C = 19$ . They conclude: 'Raising price per visit by 1 dollar raises total cost by 9 dollars.' Explain the problem. Design and run a fair digital comparison using $n = 3$ and $f = 4$ , then state the price change's effect.
4 marks

Read the worked solutions

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

For $C = p n + f$ , $p$ is the price per visit, $n$ is visits and $f$ is a fixed fee paid once. Set $p = 3$ , $n = 4$ , $f = 5$ in the Formula lab. Show a substitution that checks its output and state the total in dollars.
2 marks
$C = 3 \times 4 + 5$ (1 mark for correct substitution).
$C = 12 + 5 = 17$ , so the total is 17 dollars (1 mark).
17 dollars.
In the digital cost formula, hold $p = 5$ and $f = 3$ fixed. Change $n$ through $2, 3, 4$ . Record a comparison table showing every input and output, then describe the effect of each extra visit.
4 marks
$5 \times 2 + 3 = 13$ ; record $(5, 2, 3, 13)$ (1 mark).
$5 \times 3 + 3 = 18$ ; record $(5, 3, 3, 18)$ (1 mark).
$5 \times 4 + 3 = 23$ ; record $(5, 4, 3, 23)$ (1 mark).
$18 - 13 = 5$ and $23 - 18 = 5$ : each extra visit adds 5 dollars while price and fee stay fixed (1 mark).
Rows $(p, n, f, C)$ : $(5, 2, 3, 13)$ , $(5, 3, 3, 18)$ , $(5, 4, 3, 23)$ . Each extra visit adds 5 dollars.
Use the Formula lab with $n = 3$ and $f = 4$ fixed. Vary $p$ through $2, 4, 6$ dollars per visit. Record each total and describe the cost change for each 2-dollar rise in price per visit.
4 marks
$C = 2 \times 3 + 4 = 10$ (1 mark).
$C = 4 \times 3 + 4 = 16$ (1 mark).
$C = 6 \times 3 + 4 = 22$ (1 mark).
$16 - 10 = 6$ and $22 - 16 = 6$ : the 2-dollar increase applies to each of the 3 visits, giving $2 \times 3 = 6$ dollars (1 mark).
The totals are 10, 16 and 22 dollars. Each 2-dollar rise in price per visit adds 6 dollars to the total.
Keep $p = 4$ and $n = 3$ in the digital cost formula. Compare $f = 1, 4, 7$ . State the three totals, their successive differences, and why this investigation isolates the fixed fee's effect.
3 marks
$4 \times 3 + 1 = 13$ , $4 \times 3 + 4 = 16$ , $4 \times 3 + 7 = 19$ (1 mark for all three totals).
$16 - 13 = 3$ and $19 - 16 = 3$ dollars (1 mark for both differences).
Price and visits stay fixed, so only the fixed fee can cause the change in this formula (1 mark).
The totals are 13, 16 and 19 dollars, with successive increases of 3 dollars. Only the fixed fee changes; price and visits are held fixed.
Switch the Formula lab to $P = 2 l + 2 w$ . Hold $w = 4$ cm fixed and vary $l$ through $3, 5, 7$ cm. Record a table of lengths and perimeters, state the fixed width, and describe the change in perimeter for each 2 cm increase in length.
4 marks
For $l = 3$ : $P = 2 \times 3 + 2 \times 4 = 14$ cm (1 mark).
For $l = 5$ : $P = 2 \times 5 + 2 \times 4 = 18$ cm (1 mark).
For $l = 7$ : $P = 2 \times 7 + 2 \times 4 = 22$ cm (1 mark).
Width stays 4 cm; $18 - 14 = 4$ and $22 - 18 = 4$ , so each 2 cm increase in length adds 4 cm to perimeter (1 mark).
With width 4 cm, $(l, P) = (3, 14), (5, 18), (7, 22)$ in centimetres. Each 2 cm increase in length adds 4 cm to perimeter.
A student predicts: 'Doubling a rectangle's length while keeping its width fixed always doubles its perimeter.' In the digital tool, keep $w = 3$ cm and compare $l = 4$ cm with $l = 8$ cm. Use the two outputs to decide whether the prediction is correct.
3 marks
At $l = 4$ , $P = 2 \times 4 + 2 \times 3 = 14$ cm (1 mark).
At $l = 8$ , $P = 2 \times 8 + 2 \times 3 = 22$ cm (1 mark).
$2 \times 14 = 28$ , not 22. The unchanged width contribution prevents the perimeter from doubling in this example (1 mark).
The prediction is incorrect. The perimeters are 14 cm and 22 cm; double 14 cm would be 28 cm.
For $C = p n + f$ , keep $p = 3$ and $f = 5$ . Predict whether changing $n$ from 2 to 4 doubles the total cost. Test both settings digitally, record the totals and explain the result using the fixed fee.
4 marks
Record a prediction with a reason before testing (1 process mark; an initial prediction may be revised after testing).
At $n = 2$ , $C = 3 \times 2 + 5 = 11$ dollars (1 mark).
At $n = 4$ , $C = 3 \times 4 + 5 = 17$ dollars (1 mark).
Double 11 is 22, not 17. The visit part doubles from 6 to 12 dollars, but the fee stays 5 dollars, so the total increases by 6 dollars (1 mark for the checked conclusion and explanation).
The total does not double: it changes from 11 dollars to 17 dollars. The 5-dollar fixed fee is paid once in each total.
A student records $(p, n, f) = (2, 3, 4)$ giving $C = 10$ , then $(3, 5, 4)$ giving $C = 19$ . They conclude: 'Raising price per visit by 1 dollar raises total cost by 9 dollars.' Explain the problem. Design and run a fair digital comparison using $n = 3$ and $f = 4$ , then state the price change's effect.
4 marks
The first comparison changes $p$ and $n$ together (1 mark).
Hold $n = 3$ and $f = 4$ fixed and vary only $p$ from 2 to 3 in the Formula lab (1 mark).
Record $C = 2 \times 3 + 4 = 10$ and $C = 3 \times 3 + 4 = 13$ dollars (1 mark for both totals).
$13 - 10 = 3$ : a 1-dollar rise per visit adds 3 dollars across 3 visits (1 mark).
Both price and visits changed, so the 9-dollar difference does not isolate price. With $n = 3$ and $f = 4$ , changing $p$ from 2 to 3 changes cost from 10 to 13 dollars: a 3-dollar increase.

Checking saved progress…

Ready for the topic test?

Review the six skills, then try the Algebra topic test.

Topic review

Curriculum reference & lesson scope

Australian Curriculum, verified against QCAA · 9.0

AC9M7A06: Digital formulas and systematic variation

Use the on-page Formula lab for controlled investigations of cost and perimeter formulas. Record inputs, outputs and changes, then explain the relationships. The digital activity and written record work together; the worksheet alone does not demonstrate use of a digital tool. Rearranging formulas is outside this lesson.

Official curriculum source ↗

Source checked 2026-11-02. Schools may sequence content differently.

Back to Algebra