Simultaneous equations and break-even (SACE Stage 2 General Mathematics)

Syllabus dot point

“Solve pairs of simultaneous linear equations algebraically and graphically, and interpret the solution as a break-even point in cost and revenue models.”

8 min readUpdated 2 Jun 2026

Quick answer

Two linear equations are solved simultaneously by finding the point $(x, y)$ that satisfies both, using substitution, elimination, or a graph. In a cost-revenue model the break-even point is where total cost equals total revenue, so profit is zero; selling more than that quantity gives a profit.

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What this dot point is asking
Solving simultaneous equations
Break-even analysis
Profit, loss, and interpretation
Interpreting break-even in context
When the model has no break-even
Exam-style questions

What this dot point is asking

You must solve a pair of linear equations and interpret the intersection, especially as a break-even quantity in a business model.

Solving simultaneous equations

The solution is the point where the two lines cross. There are three methods.

Substitution: Rearrange one equation for a variable and substitute into the other.
Elimination: Add or subtract multiples of the equations to cancel one variable.
Graphical: Plot both lines; the intersection is the solution.

Each method has its place. Substitution is cleanest when one equation already has a variable by itself, such as $y = 2x + 1$ . Elimination is fastest when the same variable has matching or opposite coefficients, so adding or subtracting cancels it outright. The graphical method is the one to reach for when a question supplies a graph or asks you to read the intersection visually, though it gives only an approximate answer unless the crossing point lands on grid lines.

Solving by elimination

Solve the simultaneous equations $\;2x + 3y = 12\;$ and $\;2x - y = 4$ .

Step 1: Choose which variable to eliminate

Both equations contain $2x$ , so the coefficients of $x$ already match. Subtracting one equation from the other will cancel $x$ completely, leaving a single equation in $y$ that we can solve directly. This is why elimination is the natural choice here.

Step 2: Subtract the equations to eliminate $x$

Subtract the second equation from the first. Each term on the left is matched and subtracted, and the right-hand sides are subtracted too:

(2x + 3y) - (2x - y) = 12 - 4 \implies 4y = 8 \implies y = 2.

The $2x$ terms cancel, leaving $3y - (-y) = 4y$ , which gives $y = 2$ .

Step 3: Substitute back to find $x$

Now that we know $y = 2$ , substitute this value into either original equation to find $x$ . Using the simpler equation $2x - y = 4$ :

2x - 2 = 4 \implies 2x = 6 \implies x = 3.

Step 4: Verify both equations hold

Substituting $(x, y) = (3, 2)$ into both original equations confirms the answer is correct: $2(3) + 3(2) = 12$ and $2(3) - 2 = 4$ . Both check out.

Final answer: $x = 3$ , $y = 2$ , so the solution is the point $(3, 2)$ .

Break-even analysis

A business has a cost function and a revenue function:

C = (\text{variable cost per item})\,x + (\text{fixed costs}), \qquad R = (\text{selling price})\,x

Break-even point

The break-even point is the production level $x$ at which total cost equals total revenue, $C = R$ , so the profit $P = R - C$ is zero. Below it the business makes a loss; above it, a profit.

Worked example

Finding the break-even quantity

A stall has fixed costs of 200 dollars and each item costs 3 dollars to make. Items sell for 7 dollars. Find the break-even quantity.

C = 3x + 200, \qquad R = 7x.

Set $C = R$ :

7x = 3x + 200 \implies 4x = 200 \implies x = 50.

The stall breaks even at 50 items. The revenue there is $R = 7(50) = 350$ dollars, matching the cost. Selling a 51st item onwards yields profit.

Profit, loss, and interpretation

Profit is the vertical gap between the revenue and cost lines:

P = R - C

For the stall, $P = 7x - (3x + 200) = 4x - 200$ . Each extra item adds 4 dollars of profit (the gradient of $P$ ). The graph shows revenue and cost lines crossing at $x = 50$ ; to the left is the loss region, to the right the profit region.

Number of solutions

Two lines usually meet at one point (one solution). Parallel lines (equal gradients, different intercepts) never meet, so there is no solution. Identical lines have infinitely many solutions.

Interpreting break-even in context

The break-even quantity answers the practical question "how many must we sell before we stop losing money?" Below it, total cost exceeds revenue and the venture runs at a loss; above it, every additional sale contributes to profit at a rate equal to the difference between the selling price and the variable cost per item. This per-item margin is exactly the gradient of the profit line. A common follow-up question asks how many items are needed to reach a target profit: set the profit function equal to the target and solve, rather than re-deriving from cost and revenue separately.

When the model has no break-even

If the selling price per item is less than or equal to the variable cost per item, the revenue line is never steeper than the cost line, so the two never cross for positive output and the business can never break even. Recognising this from the gradients, before attempting to solve, is a useful check: a break-even point only exists when the price exceeds the variable cost so that the profit gradient is positive.

Check by substituting back

After solving, substitute your $(x, y)$ into both original equations. If both are satisfied, the solution is correct. This catches arithmetic slips quickly under exam pressure.

Common mistake

Do not forget the fixed costs when building the cost function. Writing $C = 3x$ instead of $C = 3x + 200$ makes the business appear to break even immediately and gives a meaningless answer. Fixed costs always appear as the intercept of the cost line.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

Calculator-assumed. A stall has fixed costs of 200 dollars and each item costs 3 dollars to make. Items sell for 7 dollars each. (a) Write the cost

C

and revenue

R

functions for

x

items. (b) Find the break-even quantity. (c) State the profit on 80 items.

Show worked answer →

(a) $C = 3x + 200$ and $R = 7x$ . (1 mark)

(b) Break-even where $C = R$ : $7x = 3x + 200$ , so $4x = 200$ and $x = 50$ items. (2 marks)

Marks: one for the two functions, two for solving the break-even, one for the profit on 80 items.

Practice3 marks

Calculator-free. Solve the simultaneous equations

2x + 3y = 12

and

2x - y = 4

Show worked answer →

Subtract the second from the first to eliminate $x$ : $(2x + 3y) - (2x - y) = 12 - 4$ , so $4y = 8$ and $y = 2$ . [1 mark]

Substitute into $2x - y = 4$ : $2x - 2 = 4$ , so $2x = 6$ and $x = 3$ . [1 mark]

The solution is $(3, 2)$ ; substituting confirms both equations hold. [1 mark]

Practise this

Sources & how we know this

Stage 2 General Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 General Mathematics: external assessment — SACE Board