Simultaneous equations and break-even situations | Year 11

01 / LEARN

Understand the idea.

Two rules must hold at once

A simultaneous solution uses the same input and output in both rules. When both formulas give cost, equate their expressions, solve for the input, then calculate the common cost. Graphically this is their intersection.

Always check both rules at the proposed input. Parallel lines with different intercepts never meet; identical lines share every point in their common domain. A practical restriction can also exclude a mathematical intersection.

Break-even is zero profit

At break-even, revenue covers total cost exactly, so profit is zero. Include the fixed setup cost as well as the cost per item. To find profitable quantities, solve the strict inequality revenue greater than cost.

Item counts are integers. Break-even itself is not profitable: if equality occurs at ten items, the next possible profitable quantity is eleven, provided the profit increases with each additional item.

profit = revenue - cost

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Equal-cost plans

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

Worked solution

1
Equate the two costs.
$18 + 3 x = 6 + 5 x$
An equal-cost input gives the same output under both plans.
2
Collect the variable terms.
$\begin{aligned} 12 & = 2 x \\ x & = 6 \end{aligned}$
Subtract six and three times the input from both sides.
3
Check the common cost.
$\begin{aligned} C_{A} & = 18 + 3 (6) = 36 \\ C_{B} & = 6 + 5 (6) = 36 \end{aligned}$
At six units both plans charge thirty-six dollars.

Answer

6 units, at 36 dollars.

EXAMPLE 02

First profitable sale

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

Worked solution

1
Write revenue and total cost.
$R = 14 n, C = 80 + 6 n$
Revenue is fourteen dollars per sale; cost includes the eighty-dollar fixed charge.
2
Solve the break-even equality.
$\begin{aligned} 14 n & = 80 + 6 n \\ 8 n & = 80 \\ n & = 10 \end{aligned}$
The eight-dollar contribution per item must cover the fixed cost.
3
Find the profitable inputs.
$\begin{aligned} 14 n - (80 + 6 n) & > 0 \\ 8 n - 80 & > 0 \\ n & > 10 \end{aligned}$
Profit requires a positive difference, rather than equality.
4
Choose the first feasible integer.
$n = 11, 8 (11) - 80 = 8$
Eleven is the smallest integer above ten and produces eight dollars profit.

Answer

Break-even 10 items; first profit at 11.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$x = 5$ ; each side equals 29.

Subtract $2 x + 4$ from both sides to get $3 x = 15$ , so $x = 5$ . Each original side then equals $29$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Two rules must hold at once

Break-even is zero profit

See the method in action.

Equal-cost plans

Equate the two costs.

Collect the variable terms.

Check the common cost.

First profitable sale

Write revenue and total cost.

Solve the break-even equality.

Find the profitable inputs.

Choose the first feasible integer.

Try it yourself.

Check your understanding

Keep practising