Solving Real Problems with Simultaneous Equations

Simultaneous equations might look like lines crossing on a graph, but they’re far more than that — they are tools for solving everyday problems. From comparing mobile phone tariffs to mixing chemical solutions, these equations allow students to find where two conditions balance perfectly.

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The Concept

Two equations with two unknowns can represent any situation where two rules or constraints overlap.
Graphically, each equation is a line, and the solution is the point of intersection — the one set of values that satisfies both conditions.

For example:

$$\begin{cases} 3x + 2y = 12 \\\\ x + y = 5 \end{cases}$$

Solving gives $x = 2$ and $y = 3$, the only pair that works in both equations.

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Real-World Applications

• Finance: Comparing two mobile tariffs or energy deals to find where costs are equal.

• Science: Determining the concentrations of two solutions when mixed.

• Engineering: Finding where stress or voltage levels balance between two systems.

• Business: Calculating production levels where cost equals revenue.

Students can plot the lines on graph paper or use algebraic substitution and elimination to find precise values.

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Skills Highlight

• Solving linear systems algebraically and graphically

• Modelling real-life problems with mathematical equations

• Interpreting points of intersection as meaningful, practical results

• Using technology to verify and visualise solutions

Worked Example: Comparing Two Linear Cost Models

Scenario:
Two phone plans charge a monthly fee plus a cost per GB of data.

• Plan A: £6 per month + £2 per GB
  $C_A = 6 + 2x$

• Plan B: £2 per month + £3 per GB
  $C_B = 2 + 3x$

Here $x$ is data used (GB), and $C$ is cost (£).

Question:
For what usage $x$ do the plans cost the same? Which plan is cheaper below and above that usage?

Algebraic solution

Set costs equal:

$$6 + 2x = 2 + 3x$$
$$6 - 2 = 3x - 2x \Rightarrow 4 = x$$

At $x = 4$ GB, both plans cost:

$$C = 6 + 2(4) = 14 \text{ (£)} \quad\text{and}\quad C = 2 + 3(4) = 14 \text{ (£)}$$

Conclusion:

• For $x < 4$ GB, Plan B is cheaper (higher per-GB but much lower fixed fee).

• For $x > 4$ GB, Plan A is cheaper (lower per-GB dominates as usage grows).

• At 4 GB, they are equal at £14.

How this looks on a graph

• Plot $C_A = 6 + 2x$ (y-intercept 6, gradient 2).

• Plot $C_B = 2 + 3x$ (y-intercept 2, gradient 3).

• The lines intersect at $(x, C) = (4, 14)$.

• To the left of $x=4$, the line for Plan B sits below Plan A (cheaper).

• To the right of $x=4$, Plan A sits below Plan B (cheaper).

Quick check with a table

Data $x$ (GB)	Plan A $C_A=6+2x$	Plan B $C_B=2+3x$	Cheaper
1	£8	£5	B
3	£12	£11	B
4	£14	£14	Tie
6	£18	£20	A

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Extension idea (optional)

Ask students to add a third plan (e.g., £10 flat for up to 3 GB, then £1.50 per extra GB) and find the break-even points against Plans A and B. This introduces piecewise linear models and multiple intersections.

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Why It Works in Teaching

Simultaneous equations provide a clear link between abstract maths and real decisions. Students see how equations model situations they recognise — and that solving them leads directly to useful, real-world answers.