Simultaneous equations and break-even situations | Year 11

01 / LEARN

The idea, step by step

Two rules must hold at once

A simultaneous solution satisfies both rules for the same input and output. Equate two output expressions, solve for the input, then substitute to find the output. A graph represents this shared solution as an intersection.

Parallel lines with different intercepts have no intersection. Equivalent equations describe the same line and have infinitely many shared points, subject to domain restrictions.

Break-even is zero profit

Revenue equals total cost at break-even. Positive profit requires revenue to be greater than cost. Include fixed costs in the cost rule rather than treating them as a per-item charge.

If quantities must be whole numbers, interpret the threshold. A non-integer solution may lie between two feasible sales quantities; the first profitable quantity is the next integer that makes profit positive.

profit = revenue - cost

A common trap

Reporting break-even as a positive profit.

Instead: At break-even profit is zero; the strict profit inequality needs a larger input.

02 / WORKED EXAMPLES

See the thinking

Example 1

Equal-cost plans

A costs $18 + 3 x$ dollars and B costs $6 + 5 x$ dollars. Find their equal-cost input.

$18 + 3 x = 6 + 5 x$ gives $12 = 2 x$ .
Thus $x = 6$ .
Both costs are $18 + 3 (6) = 36$ dollars.

6 units, at 36 dollars.

Example 2

First profitable sale

Revenue is 14 dollars per item; cost is $80 + 6 n$ dollars. Find break-even and the first profitable integer quantity.

Set $14 n = 80 + 6 n$ .
Then $8 n = 80$ , so break-even is 10 items.
Profit is $8 n - 80$ , positive for $n > 10$ .

Break-even 10 items; first profit at 11.

Check your understanding

Solve $5 x + 4 = 2 x + 19$ .

Show answer

$x = 5$ ; each side equals 29.

Subtract $2 x + 4$ to get $3 x = 15$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $y = 2 x + 1$ and $y = 5 x - 8$ .
2 marks
Plans cost $10 + 4 h$ and $22 + 2 h$ dollars. Find equal usage and cost.
2 marks
Explain the solutions of $y = 3 x + 2$ and $y = 3 x - 5$ .
2 marks
Compare $y = 2 x + 6$ and $2 y = 4 x + 12$ .
2 marks
Revenue is $9 n$ and cost is $50 + 4 n$ dollars. Find break-even.
2 marks
Revenue is $12 n$ and cost is $70 + 5 n$ . Find the smallest quantity giving at least 35 dollars profit.
3 marks

Read the worked solutions

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

Solve $y = 2 x + 1$ and $y = 5 x - 8$ .
2 marks
Equate: $2 x + 1 = 5 x - 8$ .
Then $x = 3$ , $y = 7$ .
$(3, 7)$ .
Plans cost $10 + 4 h$ and $22 + 2 h$ dollars. Find equal usage and cost.
2 marks
Equate to get $2 h = 12$ .
Substitute $h = 6$ .
6 h, at 34 dollars.
Explain the solutions of $y = 3 x + 2$ and $y = 3 x - 5$ .
2 marks
Same gradient but distinct intercepts give parallel lines.
No solution.
Compare $y = 2 x + 6$ and $2 y = 4 x + 12$ .
2 marks
Divide the second equation by 2.
It becomes the first equation.
Infinitely many common solutions.
Revenue is $9 n$ and cost is $50 + 4 n$ dollars. Find break-even.
2 marks
$9 n = 50 + 4 n$ gives $5 n = 50$ .
10 items.
Revenue is $12 n$ and cost is $70 + 5 n$ . Find the smallest quantity giving at least 35 dollars profit.
3 marks
Profit $= 7 n - 70$ .
Solve $7 n - 70 \geq 35$ , so $n \geq 15$ .
15 items.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: algebraic and graphical modelling of linear/exponential change, with interpretation and evaluation.

Official curriculum source ↗

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