Modelling with Simultaneous Equations & Break-Even Analysis
Simultaneous equations are the engine behind every break-even analysis in business. You set up cost and revenue as two linear equations, then find where they intersect, the break-even point where profit switches from negative to positive.
A food stall has fixed costs of $400 per day plus $2 per item in ingredients. Items sell for $8 each. How many items must the stall sell to cover costs exactly?
Without calculatingwrite a guess and how you'd think about it. What two quantities would you need to make equal?
A worded problem with two unknowns can always be modelled as two simultaneous equations. The key step is converting the words into algebra, then the methods from Lesson 1 take over.
For a break-even problem: write a cost equation ($C = \ldots$) and a revenue equation ($R = \ldots$), both in terms of units sold. At break-even, $C = R$. Solving gives the break-even quantity, and substituting gives the break-even dollar value.
Key facts
- A practical problem with two unknowns yields two simultaneous equations
- Break-even occurs where cost equals revenue: $C = R$
- Profit $= R - C$; profit is positive above break-even, negative below
- Spreadsheets can be used to identify or verify the break-even point
Concepts
- How to read a worded problem and extract two linear equations
- Why the break-even point is the intersection of cost and revenue lines
- The difference between fixed costs, variable costs, and revenue
- How profit and loss relate to the position relative to break-even
Skills
- Define variables and write two simultaneous equations from a worded problem
- Solve the system to find the answer in context
- Identify break-even quantity and dollar value for a linear cost/revenue model
- Use a spreadsheet to find or verify break-even
Every worded problem that can be solved with simultaneous equations contains two pieces of information, two relationships, each of which becomes one equation. The art is identifying which quantities are related and how.
A reliable strategy:
- Read carefully and underline the two unknown quantities. These become your variables (e.g. $x$ = number of adult tickets, $y$ = number of child tickets).
- Find the two conditions. Look for two sentences that describe a total or a constraint. Each becomes one equation.
- Translate into algebra. Write an equation for each condition using your variables.
- Solve and interpret. Solve the system, then write the answer in the context of the problem, include units and check it makes practical sense.
To set up simultaneous equations from a worded problem: define two variables clearly, write one equation per condition, then solve. Always state what each variable represents and verify the answer in context.
Pause, copy the three-step worded-problem setup: define each variable with its units, write one equation per stated condition, then solve, and always verify the answer makes sense in context.
Quick check: A problem states "there are 30 animals, some chickens and some cows, with 86 legs in total." Which pair of equations correctly models this?
Setting up equations from words, define two variables, write one equation per condition, is the foundation skill. The most common real-world application is break-even analysis: write Revenue = Price × quantity and Cost = Fixed cost + (Variable cost × quantity), then solve for the quantity where those two expressions are equal and profit is exactly zero.
In any business, the break-even point is where you stop losing money and start making it. Mathematically, it is the value of $n$ (units sold) where the revenue line crosses the cost line, the intersection of two linear equations.
Setting up a break-even model:
- Total cost: $C = F + vn$, where $F$ is the fixed cost (e.g. setup, rent) and $v$ is the variable cost per unit (e.g. cost of materials per item).
- Revenue: $R = pn$, where $p$ is the selling price per unit.
- Break-even: Set $C = R$ and solve for $n$. The result is the break-even quantity.
- Dollar value at break-even: Substitute the break-even $n$ into either equation to find the corresponding cost/revenue dollar amount.
Cost: $C = 400 + 2n$ Revenue: $R = 8n$
Break-even: $8n = 400 + 2n \Rightarrow 6n = 400 \Rightarrow n = 66.\overline{6}$. Since items are whole, the stall must sell 67 items to first exceed costs. Dollar value: $R = 8 \times 67 = \$536$.
Break-even occurs where total revenue equals total cost: Revenue = Price × quantity; Cost = Fixed cost + Variable cost × quantity. At break-even, profit is zero; above it the business is profitable.
Pause, copy the two equations: Revenue = Price × n and Cost = Fixed cost + Variable cost × n, and note break-even is where Revenue = Cost (profit = 0) into your book.
True or false: If the break-even quantity works out to 83.4 items, the business should sell 83 items to first make a profit (round down).
Solving Revenue = Fixed + Variable × q algebraically gives the exact break-even quantity. A spreadsheet confirms this numerically: enter quantity in one column, the revenue formula in the next, and the cost formula in another, then scan for the first row where Revenue ≥ Cost, useful when the algebra produces non-integer answers.
NESA explicitly requires you to find and verify the break-even point using a spreadsheet. This is not just a checking tool, it is a required method you may be assessed on.
Setting up a spreadsheet for break-even:
- Column A: Units produced/sold ($n$ = 0, 10, 20, … or 1, 2, 3, … depending on the scale)
- Column B: Total cost, formula
=F + v*A2(replacing $F$ and $v$ with numbers) - Column C: Revenue, formula
=p*A2 - Column D (optional): Profit, formula
=C2-B2 - Find break-even: Look for the row where cost and revenue are equal, or where profit changes from negative to positive (zero is the break-even).
A spreadsheet finds break-even by entering cost and revenue formulas in columns, then scanning for the row where Revenue ≥ Cost. This numerical approach confirms the algebraic solution and handles non-linear models.
Pause, copy the four spreadsheet columns (quantity, revenue, cost, profit) and the break-even scan rule: first row where Revenue ≥ Cost is the break-even point into your book.
Match each spreadsheet column to its formula/meaning:
Worked examples · 3 problems
A theatre sells adult tickets for $22 and child tickets for $12. One night, 350 tickets were sold for total revenue of $5,900. How many adult and child tickets were sold?
Let $a$ = number of adult tickets, $c$ = number of child tickets
(1) $a + c = 350$ (total tickets)
(2) $22a + 12c = 5900$ (total revenue)
From (1): $a = 350 - c$
Substitute: $22(350 - c) + 12c = 5900$
$7700 - 22c + 12c = 5900$
$-10c = -1800 \quad \Rightarrow \quad c = 180$
$a = 350 - 180 = 170$
Verify: $22(170) + 12(180) = 3740 + 2160 = 5900$ ✓
A market stall has fixed costs of $180/day and spends $4 per item on materials. Items are sold for $10 each. Find (a) the break-even quantity and (b) the profit if 50 items are sold.
Cost: $C = 180 + 4n$
Revenue: $R = 10n$
$180 + 4n = 10n$
$180 = 6n \quad \Rightarrow \quad n = 30$ items
$R = 10 \times 50 = \$500$
$C = 180 + 4 \times 50 = \$380$
Profit $= \$500 - \$380 = \$120$
A small business sets up a spreadsheet to model cost and revenue. The table shows: at $n = 80$, cost = $1,040, revenue = $960; at $n = 90$, cost = $1,080, revenue = $1,080. Identify the break-even point and verify algebraically if the cost equation is $C = 800 + 3n$ and revenue is $R = 12n$.
At $n = 80$: $C = 1040 > R = 960$ (loss)
At $n = 90$: $C = R = 1080$ (break-even)
$800 + 3n = 12n$
$800 = 9n \quad \Rightarrow \quad n = 88.\overline{8}$
At $n = 90$: $C = 800 + 270 = \$1{,}070$ and $R = 12 \times 90 = \$1{,}080$
These differ from the table (likely a different model was used in the table).
Top 3 list: List THREE things you must state in your answer when a question asks you to "find the break-even point."
Quick-fire practice · 3 problems
A festival stall has fixed costs of $250/day. Each item costs $3 to produce and is sold for $8. Find the break-even quantity.
A parking station has $500 in fixed costs per day. Each car pays $6 to park. If the variable operating cost per car is $0.50, how many cars are needed to break even?
A store sells two types of juice: mango ($3.50) and orange ($2.00). On Monday 80 bottles were sold for $214. How many of each type were sold?
Fill the blanks: In a break-even model, the cost equation is $C = F + vn$ where $F$ is the cost and $v$ is the cost per unit. At break-even, profit equals .
Food stall: $C = 400 + 2n$, $R = 8n$. Break-even: $8n = 400 + 2n \Rightarrow 6n = 400 \Rightarrow n = 66.\overline{6}$. The stall must sell at least 67 items to turn a profit. Was your initial estimate close?
The key insight: break-even is driven by the gap between selling price and variable cost ($8 - \$2 = \$6$ per item). The fixed cost is recovered at a rate of $6 per item, $400 \div \$6 = 66.\overline{6}$ items.
SA 1. A school canteen sells pies for $4.50 and sandwiches for $3.00. On Friday, 120 items were sold for a total of $459. How many of each item were sold? (3 marks)
SA 2. A craft business has weekly fixed costs of $600. Each item costs $12 in materials and is sold for $30. (a) Write equations for weekly cost $C$ and revenue $R$ in terms of items sold $n$. (b) Find the break-even quantity. (c) Find the profit if 45 items are sold in a week. (4 marks)
SA 3. A spreadsheet for a food van shows: at $n = 40$ items, cost = $560 and revenue = $440 (loss of $120); at $n = 60$, cost = $680 and revenue = $660 (loss of $20); at $n = 70$, cost = $740 and revenue = $770 (profit of $30). (a) Between which two values of $n$ does break-even occur? (b) If the cost equation is $C = 400 + 5n$ and revenue is $R = 11n$, find the exact break-even quantity and verify it is consistent with the spreadsheet. (4 marks)
📖 Comprehensive answers (click to reveal)
Drill 1: $C = 250 + 3n$, $R = 8n$. $8n = 250 + 3n \Rightarrow 5n = 250 \Rightarrow n = 50$ items.
Drill 2: $C = 500 + 0.5n$, $R = 6n$. $6n = 500 + 0.5n \Rightarrow 5.5n = 500 \Rightarrow n = 90.\overline{9} \approx 91$ cars.
Drill 3: Let $m$ = mango, $o$ = orange. $m+o=80$ and $3.5m+2o=214$. Sub $m=80-o$: $3.5(80-o)+2o=214 \Rightarrow 280-1.5o=214 \Rightarrow o=44$, $m=36$.
SA 1 (3 marks): Let $p$ = pies, $s$ = sandwiches. $p+s=120$ [1] and $4.5p+3s=459$ [1]. Sub $p=120-s$: $4.5(120-s)+3s=459 \Rightarrow 540-1.5s=459 \Rightarrow s=54$, $p=66$ [1]. Verify: $4.5(66)+3(54)=297+162=459$ ✓.
SA 2 (4 marks): (a) $C=600+12n$, $R=30n$ [1]. (b) $30n=600+12n \Rightarrow 18n=600 \Rightarrow n=33.\overline{3}$, so must sell 34 items [1]. (c) At $n=45$: $R=1350$, $C=600+540=1140$, Profit $=\$210$ [1 for each correct value, or 1 mark for correct method + 1 for answer].
SA 3 (4 marks): (a) Between $n=60$ and $n=70$ (profit changes from negative to positive) [1]. (b) Set $C=R$: $400+5n=11n \Rightarrow 400=6n \Rightarrow n=66.\overline{6}$ [1]. Must sell 67 items. Consistency: at $n=67$, $C=400+335=735$, $R=737$, profit=$+\$2$ (positive for first time), consistent with break-even between 60 and 70 ✓ [1]. State break-even quantity = 67 items [1].
Five timed break-even and modelling questions. Gold tier requires 90% accuracy and speed.
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