NSWMaths Standard 2Syllabus dot point

How are simultaneous linear equations solved algebraically and graphically, and how are they used to model practical situations?

Solve a pair of simultaneous linear equations graphically and algebraically, and use simultaneous equations to model practical situations

A focused answer to the HSC Maths Standard 2 dot point on simultaneous linear equations. Algebraic solution by substitution and elimination, graphical solution by intersection of lines, and modelling break-even and comparison problems with worked Australian examples.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to solve a pair of linear equations in two unknowns using either substitution, elimination, or a graph, and to set up the equations yourself from a worded scenario. Break-even and comparison-of-plans problems are the most common practical context.

The answer

What "simultaneous" means

Two linear equations in $x$ and $y$ each describe a straight line. The simultaneous solution is the $(x, y)$ pair where both equations are satisfied at the same time. Graphically, that is the point where the two lines cross.

Three cases:

One solution. Lines cross at a single point. Most common.
No solution. Lines are parallel (same gradient, different intercept).
Infinitely many solutions. Lines are identical.

Algebraic method 1: substitution

Use when one equation is already solved for one variable, or can be quickly rearranged.

Rearrange one equation for $y$ (or $x$ ).
Substitute into the other equation.
Solve for the remaining variable.
Substitute back to find the other variable.

Algebraic method 2: elimination

Use when the coefficients line up neatly, or can be matched by multiplying through.

Multiply one or both equations so the coefficients of one variable match (or are opposite).
Add or subtract the equations to eliminate that variable.
Solve for the remaining variable.
Substitute back.

Graphical method

Plot both lines on the same axes. Read off the intersection point. The HSC will sometimes give you a pre-drawn graph and ask you to read the solution from it. Always state coordinates as a pair $(x, y)$ .

Practical modelling

Worded problems usually compare two situations: two phone plans, two taxi fares, two energy providers, two job offers. The strategy is:

Define your variables explicitly (let $x$ be the number of months, let $C$ be the cost in dollars).
Write one equation per scenario.
Solve simultaneously to find the crossover point.
State your answer in context, often with a comparison ("plan A is cheaper for usage above X, plan B is cheaper below").

Worked example

Substitution

Solve $y = 3x - 5$ and $2x + y = 10$ .

Substitute $y = 3x - 5$ into the second: $2x + (3x - 5) = 10$ .

$5x - 5 = 10$ , so $5x = 15$ , giving $x = 3$ .

Back into $y = 3x - 5$ : $y = 9 - 5 = 4$ .

Solution: $(3, 4)$ .

Elimination

Solve $3x + 2y = 16$ and $5x - 2y = 8$ .

The $y$ coefficients are already opposite. Add: $8x = 24$ , so $x = 3$ .

Substitute: $3(3) + 2y = 16$ , so $2y = 7$ , giving $y = 3.5$ .

Solution: $(3, 3.5)$ .

Phone plan comparison (Australian example, 2025 retail rates)

Plan A: $\$ 25 $per month plus$ \ $0.10$ per call. Plan B: $\$ 15 $per month plus$ \ $0.20$ per call.

Let $n$ be number of calls, $C$ be monthly cost.

Plan A: $C = 25 + 0.10 n$ . Plan B: $C = 15 + 0.20 n$ .

Set equal: $25 + 0.10 n = 15 + 0.20 n$ , so $10 = 0.10 n$ , giving $n = 100$ calls.

At $100$ calls, both cost $\$ 35 $per month. Below$ 100 $calls, plan B is cheaper. Above$ 100$ calls, plan A is cheaper.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q243 marksSolve the simultaneous equations

2x + 3y = 12

and

x - y = 1

Show worked answer →

Use elimination. Multiply the second equation by $3$ to match the $y$ coefficients.

$3(x - y) = 3(1) \implies 3x - 3y = 3$ .

Add to the first equation: $(2x + 3y) + (3x - 3y) = 12 + 3$ , giving $5x = 15$ , so $x = 3$ .

Substitute back into $x - y = 1$ : $3 - y = 1$ , so $y = 2$ .

Check in the first equation: $2(3) + 3(2) = 6 + 6 = 12$ . Correct.

Markers reward a clear elimination or substitution step, both values, and a check or substitution back. Half marks for the right method but an arithmetic slip.

2021 HSC Q274 marksA taxi company charges a flag fall of

\

plus

2.20

per kilometre. A rival charges no flag fall but

\

2.80$ per kilometre. After how many kilometres do the two fares cost the same, and what is that fare?

Show worked answer →

Let $C$ be cost in dollars and $x$ be distance in kilometres.

Company A: $C = 4 + 2.20x$ . Company B: $C = 2.80x$ .

Set equal: $4 + 2.20x = 2.80x$ .

$4 = 2.80x - 2.20x = 0.60x$ , so $x = \frac{4}{0.60} \approx 6.67$ km.

Fare at the crossover: $C = 2.80 \times 6.67 \approx \$ 18.67$.

Markers reward both equations defined with variables stated, the algebraic step to solve, and a numerical answer to cents. Mention that company B is cheaper for trips under $6.67$ km and company A is cheaper for longer trips.

What this dot point is asking

The answer

What "simultaneous" means

Algebraic method 1: substitution

Algebraic method 2: elimination

Graphical method

Practical modelling

Past exam questions, worked

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