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hscscience Maths Std · Y12
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Algebraic Relationships · L1 of 6 ~50 min MST-12-S2-01 ⚡ +90 XP available

Simultaneous Linear Equations

Two equations, one solution. When two linear equations share the same $x$ and $y$ values, their graphs cross at exactly one point — and that crossing point is the answer. Master the graphical method and two algebraic methods (substitution and elimination) to solve any pair of simultaneous linear equations.

Today's hook — A café charges $4 for a coffee and $6 for a sandwich. You spend exactly $38 and buy 7 items in total. How many of each did you buy? You have two unknowns and two conditions — that's a simultaneous equations problem. Can you solve it in your head?
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Recall — your gut answer first
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A café charges $4 for a coffee and $6 for a sandwich. A customer buys 7 items in total and spends exactly $38. How many coffees and how many sandwiches did they buy?

Without calculating — write your gut feeling. What two equations could you write to describe this situation? We'll solve it properly by the end of the lesson.

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The key idea you need to own
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Simultaneous equations describe two conditions that must both be satisfied at the same time. The solution is the single pair of values $(x, y)$ that makes both equations true.

Two methods exist: graphical (graph both lines and read off the intersection) and algebraic (substitution or elimination). The graphical method gives a visual interpretation; the algebraic methods give exact answers.

GRAPHICAL Graph both lines read intersection ALGEBRAIC Substitution: rearrange one, sub into other Elimination: add/subtract to remove a variable Gives a visual picture Useful in HSC contexts Both give exact answers Choose whichever is faster for the question
The solution is the only $(x, y)$ pair satisfying both equations simultaneously
Always check your answer
Substitute both values into both equations to verify. A value that works in only one equation is not the solution.
Graph gives approximate answers
Graphical solutions from hand-drawn graphs may not be exact. Use graphing technology (GeoGebra or Desmos) for accuracy, or use algebra to get exact values.
Two equations needed
One equation with two unknowns has infinitely many solutions. You need exactly two independent equations to find a unique pair $(x, y)$.
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What you'll master
Know

Key facts

  • A simultaneous equation system is two equations with two unknowns
  • The solution is the intersection point of the two graphs
  • Three methods: graphical, substitution, elimination
  • Parallel lines have no solution; identical lines have infinitely many
Understand

Concepts

  • Why two conditions are required to fix two unknowns
  • Why the intersection of two lines gives the simultaneous solution
  • When to use graphical vs algebraic methods
  • Why a graphical solution should be verified algebraically
Can do

Skills

  • Graph two linear equations and identify the intersection point
  • Solve by substitution: rearrange one equation and substitute into the other
  • Solve by elimination: make coefficients equal and add or subtract
  • Set up simultaneous equations from a worded problem
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Key terms
Simultaneous equationsA set of two (or more) equations that must all be satisfied by the same values of the unknowns.
SolutionThe pair of values $(x, y)$ that satisfies both equations at the same time — the intersection of the two lines.
Substitution methodRearrange one equation to isolate a variable, then substitute the expression into the other equation.
Elimination methodMultiply one or both equations so a variable has equal coefficients, then add or subtract to eliminate that variable.
Point of intersectionThe coordinates $(x, y)$ where two lines cross on a graph — this is the simultaneous solution.
No solutionWhen two lines are parallel (same gradient, different $y$-intercept) — they never intersect.
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The Graphical Method — Finding the Intersection
core concept

Every linear equation $y = mx + b$ is a straight line. If you graph two such lines on the same set of axes, the point where they cross is the only pair of values that satisfies both equations — that crossing point is the simultaneous solution.

To solve graphically:

  1. Rearrange both equations into gradient-intercept form $y = mx + b$ (or use a table of values)
  2. Graph both lines on the same axes — use graphing technology for accuracy
  3. Identify the coordinates of the intersection point
  4. Write the solution as $x = \ldots,\ y = \ldots$ and verify in both original equations
Graphing technology (NESA requirement): You must be able to solve graphically with and without graphing applications. In an exam you may use GeoGebra or Desmos. When graphing by hand, draw at least two clearly marked points on each line and use a ruler.
1 2 3 4 5 0 1 2 −1
1 2 3 4 5 x 0 1 2 3 4 y y = x + 1 y = −x + 5 (2, 3) x=2, y=3 is the solution
Verify the intersection: Substitute $(2, 3)$ into both equations. Line 1: $3 = 2 + 1$ ✓ Line 2: $3 = -2 + 5$ ✓ Both satisfied — $(2, 3)$ is correct.
What to write in your book
  • Graphical method: graph both lines (gradient-intercept form), identify intersection $(x, y)$.
  • Always verify: substitute both values back into both equations.
  • Use graphing technology for accuracy; hand-drawn graphs give approximations only.

Quick check: Two lines intersect at the point $(3, -1)$. What does this tell you?

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The Substitution Method
core concept

Substitution works by making one equation give you one variable in terms of the other, then replacing (substituting) that expression into the second equation — turning two equations into one.

The substitution method follows four steps:

  1. Choose a variable to isolate. Pick the equation and variable that will be easiest to rearrange (often where the coefficient is 1 or −1).
  2. Rearrange to get that variable alone. E.g., from $x + y = 7$ get $y = 7 - x$.
  3. Substitute into the other equation. Replace every occurrence of that variable with your expression, giving a one-variable equation.
  4. Solve, then back-substitute. Find the first unknown, substitute back to find the second, then verify in both original equations.
Strategy tip — pick the easiest variable: If one equation has a term like $y$ (coefficient 1) or $-y$ (coefficient −1), rearrange that equation first — the algebra stays clean. Avoid rearranging when both variables have coefficients that aren't 1, as this leads to fractions.
Common error: After substituting, students sometimes substitute back into the modified equation rather than one of the original equations. Always use an original equation for the back-substitution step.
What to write in your book
  • Substitution: isolate $y$ (or $x$) from one equation → substitute expression into the other → solve → back-substitute.
  • Choose the variable with coefficient 1 or −1 to avoid fractions.
  • Final check: sub both answers into both original equations.

True or false: After finding $x$ by substitution, you should back-substitute into one of the original equations to find $y$ (not into the rearranged expression).

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The Elimination Method
core concept

Elimination removes one variable by making its coefficients equal in both equations, then adding or subtracting the equations to cancel that variable entirely.

The elimination method follows these steps:

  1. Match coefficients. Multiply one or both equations by a constant so that one variable has the same (or equal and opposite) coefficient in both equations.
  2. Add or subtract the equations. If coefficients are equal and same sign, subtract. If equal and opposite sign, add. This eliminates one variable.
  3. Solve the resulting equation. You now have one equation in one unknown.
  4. Back-substitute and verify. Sub the found value back into either original equation; then verify the solution in both.
Add or subtract? The rule: If the matching coefficients have the same sign (e.g., both $+3y$), subtract the equations. If they have opposite signs (e.g., $+3y$ and $-3y$), add the equations. Adding opposite-sign terms gives zero — that's the point.
When to use elimination vs substitution: Elimination is often faster when neither equation has a variable with coefficient 1 (e.g., $3x + 4y = 10$ and $3x - 2y = 4$). Substitution is often faster when one equation already has $y$ or $x$ isolated.
What to write in your book
  • Elimination: match one variable's coefficients → add (opposite signs) or subtract (same signs) → solve → back-substitute.
  • Same sign → subtract. Opposite sign → add.
  • Multiply whole equations (both sides) to create matching coefficients.

Fill the blanks: To eliminate $y$ from the system $2x + y = 8$ and $3x + y = 11$, you should the equations (because the $y$ coefficients are the sign).

Worked examples · 3 problems, reveal step by step

PROBLEM 1 · GRAPHICAL METHOD

Solve the simultaneous equations $y = 2x - 1$ and $y = -x + 5$ graphically. Verify your answer algebraically.

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Identify key features of each line
Line 1: $y = 2x - 1$ — gradient 2, $y$-intercept $-1$
Line 2: $y = -x + 5$ — gradient $-1$, $y$-intercept 5
Both are in gradient-intercept form. Read off gradient ($m$) and $y$-intercept ($b$) directly.
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PROBLEM 2 · SUBSTITUTION METHOD

Solve: $4c + 6s = 38$ and $c + s = 7$ (the café problem from the hook — $c$ = coffees, $s$ = sandwiches).

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Isolate $c$ from the simpler equation
From $c + s = 7$: $\quad c = 7 - s$
The second equation has coefficient 1 on both variables — easy to rearrange. Isolate $c$ first.
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PROBLEM 3 · ELIMINATION METHOD

Solve: $3x + 2y = 16$ and $3x - y = 7$.

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Label the equations
(1) $3x + 2y = 16$
(2) $3x - y = \phantom{1}7$
Both equations have $3x$. The $x$-coefficients already match, so no multiplication is needed.
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Quick-fire practice · 4 problems

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Solve graphically (sketch and state the intersection): $y = x + 3$ and $y = -2x + 6$

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Solve by substitution: $y = 3x - 1$ and $2x + y = 9$

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Solve by elimination: $5x + 2y = 24$ and $5x - 3y = 4$

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A shop sells notebooks for $3 and pens for $1.50. Tyra buys 10 items and pays $21. How many of each item did she buy? (Set up and solve two simultaneous equations.)

Match each situation to the correct action:

Top 3 list: Name THREE steps you must complete after finding $x$ and $y$ to be sure you have the correct simultaneous solution.

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Revisit your thinking

Look back at the café problem from the Think First section. Using substitution: let $c + s = 7$, so $c = 7 - s$. Substitute into $4c + 6s = 38$: $4(7 - s) + 6s = 38 \Rightarrow 2s = 10 \Rightarrow s = 5$, $c = 2$. The answer is 2 coffees and 5 sandwiches.

What changed in your understanding? Which method — graphical, substitution or elimination — do you find most reliable?

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Short answer — exam-style questions
+5 XP per question · show full working

These questions are modelled on the HSC Maths Standard style. Show all working — method marks are available even if the final answer is wrong.

ApplyBand 33 marks

SA 1. Solve the simultaneous equations $y = 2x - 3$ and $y = -x + 6$ by any method. (3 marks)

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ApplyBand 43 marks

SA 2. Solve by elimination: $4x + 3y = 18$ and $4x - y = 6$. (3 marks)

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AnalyseBand 44 marks

SA 3. Two mobile plans cost the same for a certain number of minutes. Plan A costs $15/month plus $0.05 per minute. Plan B costs $5/month plus $0.10 per minute. (a) Write two equations representing the monthly cost $C$ in terms of minutes $m$. (b) Solve simultaneously to find the number of minutes at which the plans cost the same amount, and state that cost. (4 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: $y=x+3$ and $y=-2x+6$. Set equal: $x+3=-2x+6 \Rightarrow 3x=3 \Rightarrow x=1$, $y=4$. Intersection: $(1, 4)$.

Drill 2: Sub $y=3x-1$ into $2x+y=9$: $2x+3x-1=9 \Rightarrow 5x=10 \Rightarrow x=2$, $y=5$.

Drill 3: Subtract: $(5x+2y)-(5x-3y)=24-4 \Rightarrow 5y=20 \Rightarrow y=4$; sub back: $5x=16 \Rightarrow x=3.2$.

Drill 4: Let $n$ = notebooks, $p$ = pens. Equations: $n+p=10$ and $3n+1.5p=21$. Sub $n=10-p$: $3(10-p)+1.5p=21 \Rightarrow 30-1.5p=21 \Rightarrow p=6$, $n=4$. Answer: 4 notebooks, 6 pens.

SA 1 (3 marks): Set equal: $2x-3=-x+6$ [1] $\Rightarrow 3x=9 \Rightarrow x=3$ [1]; $y=2(3)-3=3$ [1]. Solution: $x=3$, $y=3$. Verify: $3=3(3)-3$ ✓ and $3=-3+6$ ✓.

SA 2 (3 marks): Label: (1) $4x+3y=18$; (2) $4x-y=6$. Subtract (2) from (1) [1]: $4y=12 \Rightarrow y=3$ [1]. Sub into (2): $4x=9 \Rightarrow x=2.25$ [1]. Verify: $4(2.25)+3(3)=9+9=18$ ✓.

SA 3 (4 marks): (a) $C = 0.05m + 15$ and $C = 0.10m + 5$ [1 for each equation]. (b) Set equal: $0.05m+15=0.10m+5 \Rightarrow 10=0.05m \Rightarrow m=200$ minutes [1]; $C = 0.05(200)+15 = \$25$ [1].

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Boss battle · The Simultaneous Solver
earn bronze · silver · gold

Five timed questions on simultaneous equations. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
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Science Jump · platform challenge

Climb platforms by answering questions on simultaneous equations.

Mark lesson as complete

Tick when you've finished the practice and review.

Algebraic Relationships · Maths Standard