Simultaneous linear equations and break-even analysis: HSC Maths Standard 1 Year 12

Syllabus dot point

“A3.1 Simultaneous linear equations: develop a pair of linear equations to model a practical situation, solve it graphically, and interpret the point of intersection, including break-even analysis (break-even point, profit and loss zones, and the meaning of the y-intercept)”

HSCMaths Standard 1Year 12: Algebra7 min readUpdated 23 Sep 2026

Quick answer

Model related quantities with linear equations $y = mx + c$ , graph them on the same axes, and interpret the point of intersection in context. In break-even analysis, the intersection of cost and income lines is where profit is zero; to the right is the profit zone and to the left the loss zone, and the y-intercept of the cost line is the fixed cost.

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What this dot point is asking
The answer
Practice questions

What this dot point is asking

You need to model two related quantities with linear equations, solve them graphically (and check algebraically), and interpret the point where the lines meet in context. Break-even problems are the most common application.

The answer

Linear models

A linear relationship has the form $y = mx + c$ (or with context letters, such as $C = 200 + 5n$ ):

$c$ , the y-intercept, is the starting value (for example fixed costs).
$m$ , the gradient, is the rate of change (for example cost per item).

Solving graphically

Draw both lines on the same axes (a table of values, or graphing software).
Find the point of intersection: its coordinates satisfy both equations.
Interpret it in words, with units.

You can check by substitution: set the two expressions equal and solve.

Break-even analysis

Break-even vocabulary

Cost: $C = \text{fixed costs} + \text{cost per item} \times n$ .
Income (revenue): $I = \text{price} \times n$ .
Break-even point: where $I = C$ (profit is zero).
Profit zone: $I > C$ (to the right of the break-even point).
Loss zone: $I < C$ (to the left).
Profit $= I - C$ .

Other applications: comparing plans or prices (the intersection shows where one becomes cheaper), supply and demand (the intersection is the equilibrium price and quantity), distance and time problems.

Worked example

A mobile coffee van has fixed costs of $360 per day and each coffee costs $1.20 to make. Coffees sell for $4.80.

$C = 360 + 1.2n$ , $I = 4.8n$ .
Break even: $4.8n = 360 + 1.2n$ , so $3.6n = 360$ and $n = 100$ coffees.
At 100 coffees, $C = I = \$480$ .
Selling 150 coffees: profit $= 4.8 \times 150 - (360 + 1.2 \times 150) = 720 - 540 = \$180$ .

Common traps

Rounding the break-even number down. You cannot sell part of an item; round up to reach at least break-even.

Reading the wrong axis. State both coordinates and what each means.

Forgetting fixed costs when calculating profit.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks

A candle maker has fixed costs of 300 dollars per month and it costs 4 dollars to make each candle. Candles sell for 10 dollars each. Write equations for cost C and income I for n candles, and find the break-even point.

Show worked solution →

$C = 300 + 4n$ and $I = 10n$ .

Break even when $I = C$ : $10n = 300 + 4n$ , so $6n = 300$ and $n = 50$ .

The candle maker breaks even at 50 candles, when income and costs are both $500.

Marking guide: 1 mark for each equation, 1 mark for the break-even point.

core4 marks

Phone plan A costs 20 dollars per month plus 0.10 dollars per minute of calls. Plan B costs 35 dollars per month with unlimited calls. Graph both plans for 0 to 300 minutes (describe the graph), find where they intersect and recommend a plan for someone who makes 200 minutes of calls a month.

Show worked solution →

Plan A: $C = 20 + 0.1m$ (a line starting at $20, rising $10 per 100 minutes). Plan B: $C = 35$ (a horizontal line).

Intersection: $20 + 0.1m = 35$ , so $0.1m = 15$ and $m = 150$ minutes.

For fewer than 150 minutes, Plan A is cheaper; for more than 150 minutes, Plan B is cheaper.

At 200 minutes, Plan A costs $20 + 0.1 \times 200 = \$40$ , so Plan B ($35) is better.

Marking guide: 1 mark for both equations, 1 mark for the intersection, 1 mark for interpreting the graph, 1 mark for the recommendation.

exam5 marks

A school fete stall has fixed costs of 150 dollars and each burger costs 2.50 dollars to make. Burgers sell for 6 dollars. (a) Find the break-even number of burgers. (b) Find the profit if 120 burgers are sold. (c) Explain what happens to the break-even point if the selling price is raised to 7 dollars, and whether this is a good idea.

Show worked solution →

(a) $6n = 150 + 2.5n$ , so $3.5n = 150$ and $n = 42.86$ . They must sell 43 burgers to break even (a whole number, rounded up).

(b) Profit $= 6 \times 120 - (150 + 2.5 \times 120) = 720 - 450 = \$270$ .

(c) At $7: $7n = 150 + 2.5n$ , so $4.5n = 150$ and $n = 33.3$ , so 34 burgers. The break-even point falls and each sale earns more, but a higher price may reduce how many people buy, so it is only better if sales do not drop much.

Marking guide: 2 marks for (a) including rounding up, 1 mark for (b), 2 marks for (c) with a sensible comment.

Practise this

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA