VCE General Mathematics (Year 11) · Algebra and Number · 10m read

Simultaneous Equations and Graphs

Ever wondered how businesses figure out exactly when costs equal profits? Simultaneous equations let you solve two equations at once to find where two lines meet on a graph. You'll learn both the algebraic method (using algebra to find the answer) and the graphical method (plotting lines to see their intersection), skills that show up constantly in real-world scenarios like pricing strategies and physics problems.

Learning objective: Solve simultaneous linear equations algebraically and graphically.

By the end of this lesson you can:

• I use substitution and elimination.
• I find intersection points graphically.
• I apply simultaneous equations to problems.

Simultaneous equations find where two plans cost the same. Two phone plans cross at one usage level — the break-even point.

Solving simultaneous equations

SITUATION:
• Two variables, two equations.
• Find values satisfying BOTH.

METHOD 1 — SUBSTITUTION:
• Solve one equation for one variable.
• Substitute into the other.
• Example:
y = 2x + 1
3x + y = 11 → 3x + (2x + 1) = 11 → 5x = 10 → x = 2, y = 5.

METHOD 2 — ELIMINATION:
• Multiply equations to line up coefficients.
• Add/subtract to eliminate a variable.
• Example:
2x + y = 7
3x − y = 8 → add → 5x = 15 → x = 3, y = 1.

METHOD 3 — GRAPHICAL:
• Graph both lines.
• Intersection = solution.
• Less precise; good for visualisation.

METHOD 4 — CAS:
• Solve({eq1, eq2}, {x, y}) on most CAS calculators.

WORD PROBLEMS:
• Define 2 variables.
• Build 2 equations from two conditions.
• Solve.
• Interpret in context.

CASES:
• UNIQUE solution — lines intersect.
• NO solution — parallel lines (same gradient, different intercepts).
• INFINITE solutions — same line.

BREAK-EVEN:
• Revenue = Cost → break-even point.
• Classic simultaneous equation problem.

APPLICATIONS:
• Supply and demand.
• Mixture problems.
• Simple economics.

CHECKING:
• Substitute your solution back into BOTH equations.

COMMON ERRORS:
• Arithmetic slips.
• Forgetting to find BOTH variables.
• Misreading the question.

Pick your method

• One simple: Substitution.
• Balanced: Elimination.
• Visual: Graphical.

Simultaneous equations

• Substitution: y = m_1 x + c_1 = m_2 x + c_2
• Elimination: \text{add or subtract scaled equations}
• Break-even: R(x) = C(x)

Quick reference

| Method | Use |
|---|---|
| Graphical | Intersection point |
| Substitution | One variable isolated |
| Elimination | Add/subtract equations |
| CAS | Solve command |
| Context | Break-even, comparing options |

Worked example (Medium)

A baker makes small and large cakes. 2 small + 3 large = $43. 4 small + 1 large = $41. Price of each?

1. Variables. Let s = price of small, l = price of large.
2. Equations. 2s + 3l = 43 and 4s + l = 41.
3. Elimination. Multiply eq 2 by 3: 12s + 3l = 123. Subtract eq 1: 10s = 80 → s = 8.
4. Back-sub. 4(8) + l = 41 → l = 9.

Answer: Small cake $8, large cake $9. Check: 2(8)+3(9)=43 ✓; 4(8)+9=41 ✓.

Worked example (Hard)

A concert sells adult tickets for $35 and child tickets for $20. Total revenue is $5250 from 200 tickets. Find numbers of each.

1. Variables. Let a = adult tickets, c = child tickets.
2. Equations. a + c = 200 (total tickets). 35a + 20c = 5250 (revenue).
3. Solve. From eq 1: c = 200 − a. Substitute: 35a + 20(200 − a) = 5250 → 35a + 4000 − 20a = 5250 → 15a = 1250 → a = 83.33.
4. Interpret. Non-integer answer — check rounding. a ≈ 83, c ≈ 117 gives revenue $5245. Or a = 84, c = 116 gives $5260. Neither is exact $5250 — the problem data may need to be verified, or accept a = 83⅓ with the note that ticket sales must be whole numbers.

Answer: Approximately 83 adult and 117 child tickets. The non-integer result signals that real-world constraints (whole tickets) can conflict with algebraic solutions.

Common mistakes

• Only finding one variable and forgetting the other Both unknowns are required for a complete answer. Fix: After finding x, always substitute back to find y. State both values.
• Not checking the solution in BOTH original equations An arithmetic error in one step can produce a wrong answer that satisfies only one equation. Fix: Substitute (x, y) into both equations. Both must be satisfied.
• Setting up equations from wrong conditions Misreading the problem leads to wrong equations. Fix: Underline key information. Write what each equation represents (e.g., "total items", "total cost").

Break-even analysis — where simultaneous equations meet business

One of the most practical applications of simultaneous equations is break-even analysis, which every business uses to determine when revenue will cover costs.

A business has two cost components: fixed costs (rent, salaries, insurance — constant regardless of output) and variable costs (materials, shipping — proportional to output). Total cost: C = Fixed + Variable × quantity.

Revenue is typically: R = Price × quantity.

The break-even point is where R = C — the business starts making profit. Graphically, it is the intersection of the cost and revenue lines.

Example: A food truck has fixed costs of $800/week and variable costs of $5 per meal. Selling meals at $12 each: C = 800 + 5q, R = 12q. Break-even: 12q = 800 + 5q → 7q = 800 → q ≈ 114 meals per week.

Below 114 meals, the truck loses money. Above 114, it profits. This simple model helps entrepreneurs decide whether a business idea is viable before investing.

Sensitivity analysis extends this: what if rent increases? What if the selling price drops? By adjusting parameters and re-solving, businesses can stress-test their plans. This is exactly the kind of modelling VCE General Maths prepares you for.

Practice questions

1. Two lines with the same gradient but different intercepts are:

   1. Intersecting
   2. Parallel
   3. Perpendicular
   4. Identical

   (1 marks)

2. A café sells coffees for $4 and muffins for $6. On one day they sold 50 items total for $240. How many of each?

   (3 marks)

Summary

• Use substitution and elimination.
• Find intersection points graphically.
• Apply simultaneous equations to problems.