Teaching unit · Year 9-10 (ages 14 to 16)

Simultaneous equations: substitution and elimination

Solving two linear equations in two variables at once, by substitution and by elimination, and reading the solution as the point where two lines cross

About three to four lessons of 45 to 60 minutes

Start here · hook

Two clues, one exact answer

A market stall sells apples and bananas. Grace buys 2 apples and 3 bananas for $7. Leo buys 4 apples and 1 banana for $9. Neither purchase alone tells you the price of a single apple or a single banana, but the two together do: only one pair of prices fits BOTH purchases at once. That is exactly what 'simultaneous' means, solving two equations at the same time so the answer satisfies every clue simultaneously, not just one.

Graphically, each equation draws a straight line, one price rule per line. The one point where both lines cross is the only (x, y) pair that makes both true together; everywhere else on either line satisfies only ONE of the two equations.

Two mixed purchases at a market stall2 apples + 3 bananas = $7, and 4 apples + 1 banana = $9; only one pair of prices fits both
Two trips at different speedsfind where a faster car catches up to one with a head start
Two phone plans crossing overthe break-even point where a cheaper-upfront plan starts costing more
Mixing two concentrationshow much of each solution to combine for an exact target strength

Learning objective

What students will be able to do

Students will solve a pair of simultaneous linear equations in two variables using substitution and using elimination, check a solution by substituting it into both original equations, and set up and solve simultaneous equations from a worded real-world problem.

Success criteria

I can check whether a given (x, y) pair is the solution to a pair of simultaneous equations by substituting it into both equations.
I can solve a pair of simultaneous equations by substitution, replacing one variable using an expression from the other equation.
I can solve a pair of simultaneous equations by elimination, adding or subtracting the equations to remove one variable.
I can set up a pair of simultaneous equations from a worded problem and solve it for both unknowns.
I can interpret the solution as the point where the two lines cross on a graph.

Curriculum anchor

Standards this unit teaches

AC9M10A02Australian Curriculum v9 (ACARA)
Simultaneous equations and inequalities
Solve linear inequalities and simultaneous linear equations in 2 variables; interpret solutions graphically and communicate solutions in terms of the situation.
8.EE.C.8Common Core (US)
Solve systems of two linear equations
Understand that the solution to a system of two linear equations corresponds to the point where their graphs intersect; solve systems of two linear equations algebraically and estimate solutions by graphing, and use systems to solve real-world problems.

Before you start

Prior knowledge

This unit builds on skills students should already have met. Revisit any that are shaky first.

Key vocabulary

Words to teach and display

Simultaneous equations: two or more equations that must all be true for the SAME values of the variables at once
Substitution: replacing a variable with an equivalent expression from another equation, to leave a single-variable equation
Elimination: adding or subtracting two equations to cancel out (eliminate) one variable
Solution (of a system): the (x, y) pair that makes every equation in the system true at the same time
Intersection point: the point where two lines cross on a graph; graphically, the solution to their simultaneous equations

Teaching sequence

Teach it: concrete, pictorial, abstract

The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.

1. What a simultaneous solution means

Concrete

Before solving two equations at once, it helps to know what you are looking for: a single (x, y) pair that makes BOTH equations true, not just one. Graphically, each linear equation draws a straight line, and the shared solution is the exact point where the two lines cross.

Take x + y = 7 and x - y = 1. The pair (4, 3) makes the first equation true (4 + 3 = 7) AND the second true (4 - 3 = 1), so (4, 3) is the simultaneous solution. The pair (5, 2) makes the first true (5 + 2 = 7) but the second false (5 - 2 = 3, not 1), so (5, 2) solves only one equation, not both together.

The line x + y = 7 passes through (4, 3), among many other points.

The line x - y = 1 ALSO passes through (4, 3). This shared point is the simultaneous solution, the only pair that lies on both lines.

Check for understanding, ask

Why isn't it enough for a pair of values to satisfy just one of the two equations?
How could you check whether a given (x, y) pair is a genuine simultaneous solution?

2. Solving by substitution

Pictorial

Substitution rearranges ONE equation to isolate a variable, then substitutes that expression into the OTHER equation, leaving a single-variable equation to solve exactly like a Year 7 one-variable equation.

For y = 2x + 1 and 3x + y = 16: the first equation already gives y in terms of x, so substitute (2x + 1) for y in the second equation: 3x + (2x + 1) = 16, giving 5x + 1 = 16, so x = 3. Then y = 2(3) + 1 = 7.

Worked example

Solve simultaneously: y = 2x + 1 and 3x + y = 16.

The first equation already isolates y: y = 2x + 1.
Substitute (2x + 1) for y in the second equation: 3x + (2x + 1) = 16.
Simplify and solve: 5x + 1 = 16, so 5x = 15, so x = 3.
Substitute x = 3 back into y = 2x + 1: y = 2(3) + 1 = 7.
Check in the second original equation: 3(3) + 7 = 9 + 7 = 16. Correct.

Answer: x = 3, y = 7.

Check for understanding, ask

Why does substitution need one equation already isolated for a single variable (or easy to rearrange into one)?
How do you check a simultaneous solution once you have found x and y?

3. Solving by elimination

Abstract

Elimination adds or subtracts the two equations directly to cancel one variable, useful when neither equation is already solved for a single variable.

For 2x + y = 12 and x - y = 3: the y-terms are +y and -y, opposites, so ADDING the two equations eliminates y directly. If instead both equations had matching signs on a variable (like +2y in both), you would SUBTRACT to eliminate it. Sometimes neither variable's coefficients already match: for 3x + 2y = 16 and x + y = 6, multiplying the second equation by 2 gives 2x + 2y = 12, matching the y-coefficient in the first; subtracting then gives (3x + 2y) - (2x + 2y) = 16 - 12, so x = 4, and y = 6 - 4 = 2.

Worked example

Solve simultaneously: 2x + y = 12 and x - y = 3.

The y-terms are +y and -y, opposites, so adding the two equations eliminates y.
Add: (2x + y) + (x - y) = 12 + 3, giving 3x = 15.
Solve: x = 5.
Substitute x = 5 into x - y = 3: 5 - y = 3, so y = 2.
Check in 2x + y = 12: 2(5) + 2 = 10 + 2 = 12. Correct.

Answer: x = 5, y = 2.

Check for understanding, ask

Why does adding (not subtracting) eliminate y in 2x + y = 12 and x - y = 3?
When neither variable's coefficients already match, what do you do before adding or subtracting?

4. Setting up simultaneous equations from a word problem

Abstract

Real problems rarely hand you two ready-made equations: the first skill is TRANSLATING a worded situation into a pair of equations, using one variable for each unknown quantity, before solving as before.

Back to the market stall: let a = the price of one apple and b = the price of one banana. Grace's purchase (2 apples, 3 bananas, $7) becomes 2a + 3b = 7. Leo's purchase (4 apples, 1 banana, $9) becomes 4a + b = 9. These two equations, solved simultaneously, reveal both prices at once.

Worked example

Grace buys 2 apples and 3 bananas for $7. Leo buys 4 apples and 1 banana for $9. Find the price of one apple and one banana.

Let a = price of one apple, b = price of one banana (in dollars).
Grace: 2a + 3b = 7. Leo: 4a + b = 9.
Rearrange Leo's equation to isolate b: b = 9 - 4a.
Substitute into Grace's equation: 2a + 3(9 - 4a) = 7, giving 2a + 27 - 12a = 7, so -10a = -20, so a = 2.
Substitute a = 2 into b = 9 - 4a: b = 9 - 8 = 1.
Check in Grace's equation: 2(2) + 3(1) = 4 + 3 = 7. Correct.

Answer: One apple costs $2 and one banana costs $1.

Check for understanding, ask

Why does each unknown quantity in the word problem become its own variable?
How would the two equations change if Leo had bought 4 apples and 2 bananas instead of 1 banana?

Watch for

Common misconceptions and how to address them

MisconceptionA solution only needs to make ONE of the two equations true.

Why it happens: Students stop as soon as one equation checks out.

How to address it: A genuine simultaneous solution must satisfy BOTH equations at once; always check the found values in the SECOND original equation too, not just the one used to solve for a variable.

MisconceptionElimination requires the same variable to already have identical or opposite coefficients.

Why it happens: Early examples often already match, so students assume they must and skip cases needing a multiply step.

How to address it: Multiply one or both equations by a suitable number first so the coefficients of one variable become equal or opposite, THEN add or subtract. The multiplication keeps both sides of an equation balanced, so the equation is still true.

MisconceptionYou always subtract when eliminating, regardless of the signs.

Why it happens: Students apply a fixed rule instead of checking whether the matching coefficients have the same or opposite signs.

How to address it: Add when the matching coefficients have OPPOSITE signs (they cancel); subtract when they have the SAME sign (the difference cancels). Check the signs before choosing.

MisconceptionTwo linear equations always have exactly one solution.

Why it happens: Students have only seen examples with a single crossing point.

How to address it: If the two lines are parallel (same gradient, different intercept), there is NO solution; if the two equations describe the SAME line, there are infinitely many solutions. Most school examples have exactly one crossing point, but it is not guaranteed.

Do it together

Guided practice (with answers)

1. Does (7, 3) solve the simultaneous equations x + y = 10 and x - y = 2?
Answer: No. It satisfies the first equation (7+3=10) but not the second (7-3=4, not 2), so it is not a solution to the system; only (6, 4) satisfies both equations at once.
2. Solve by substitution: y = 3x - 2 and x + y = 10.
Answer: x = 3, y = 7, because substituting y = 3x - 2 into x + y = 10 gives x + 3x - 2 = 10, so 4x = 12 and x = 3; then y = 3(3) - 2 = 7.
3. Solve by elimination: 3x + y = 14 and x + y = 6.
Answer: x = 4, y = 2, because subtracting the second equation from the first eliminates y: (3x+y)-(x+y)=14-6 gives 2x=8, so x=4, and y=6-4=2.
4. A gym charges a joining fee plus a fee per visit: 5 visits costs $70, and 12 visits costs $140. Let j = joining fee and v = cost per visit. Write the two equations for these amounts.
Answer: j + 5v = 70 and j + 12v = 140.
5. Using the gym equations j + 5v = 70 and j + 12v = 140, find the joining fee and the cost per visit.
Answer: Joining fee $20, cost per visit $10, because subtracting the equations eliminates j: (j+12v)-(j+5v)=140-70 gives 7v=70, so v=10; then j=70-5(10)=20.

On their own

Independent practice worksheets

Practise substitution, elimination and setting up systems from word problems, with computed, never-wrong answer keys.

Year 9-10 simultaneous equations & inequalities worksheetthe core skills from this unit, freshly generated each time Year 10 simultaneous equations & inequalities worksheetthe same skills at Year 10, for extra practice or a review lesson Year 9 quadratics: expand, factorise and solverevisit one-variable equation solving if isolating a variable felt shaky AC9M10A02 curriculum pagemore worksheets and the official curriculum wording

Reach every student

Differentiation

Support

Start with substitution only, using equations where one variable is already isolated (y = ...), before introducing elimination.
Provide coloured highlighting for matching coefficients before eliminating, so students see WHICH terms cancel before adding or subtracting.
Use a simple 'check both!' checklist after every solve: substitute into equation 1, then equation 2.
Keep early word problems to two clearly-labelled unknowns (e.g. 'a' and 'b') with the equations partly set up already.

Extension

Introduce systems needing a multiply-then-eliminate step, where neither variable's coefficients already match.
Explore what happens graphically (and algebraically) when a system has no solution (parallel lines) or infinite solutions (the same line).
Set up and solve a three-unknown word problem informally, discussing why a third equation would be needed.
Compare solving the same system by substitution AND elimination, and discuss which method was more efficient and why.

Check it stuck

Assessment: exit ticket

A three-question exit ticket sampling checking a solution, solving by elimination, and setting up a word problem.

1. Does (5, 1) solve the simultaneous equations x + y = 6 and 2x - y = 9?
Answer: Yes. It satisfies both: 5 + 1 = 6, and 2(5) - 1 = 10 - 1 = 9.
2. Solve by elimination: x + y = 9 and x - y = 3.
Answer: x = 6, y = 3, because adding the equations gives 2x = 12, so x = 6, and y = 9 - 6 = 3.
3. A movie ticket plus 2 snacks costs $22. A movie ticket plus 5 snacks costs $37. Let t = ticket price and s = snack price. Write the two equations, then find the price of one snack.
Answer: t + 2s = 22 and t + 5s = 37. Subtracting gives 3s = 15, so s = 5: one snack costs $5.

For the teacher

Teacher notes and timings

Rough timing across three to four lessons: Lesson 1 what a simultaneous solution means (section 1), Lesson 2 substitution (section 2), Lesson 3 elimination (section 3), Lesson 4 word problems plus the exit ticket (section 4 and assessment).
This unit assumes comfort with solving a one-variable linear equation (Year 7 linear equations unit). Revisit that first if isolating a variable feels shaky.
Honest coverage note: the Grade 8 slope/linear-equations/systems unit already teaches ONE narrow case (both equations already solved for y, set equal to each other) with a real worked example; this unit's own independent practice explicitly flags the exact-topic worksheet as 'the closest systems-of-equations practice available, one year level up' rather than claiming full coverage. This unit completes the gap: general substitution (not just the already-isolated case), true elimination (including the multiply-first case), and setting up word problems as systems.
Language to keep repeating: a simultaneous solution must satisfy EVERY equation at once; substitution replaces a variable with an equivalent expression; elimination adds or subtracts equations to cancel a variable.
Curriculum note: AC9M10A02 (Australian Curriculum v9) is the Year 10 descriptor for simultaneous equations and inequalities; this site's worksheet also spans Grade 8-9 as extra practice tiers, matching the US Common Core 8.EE.C.8 cluster which introduces systems of two linear equations at Grade 8.
The two coordinateLine figures in section 1 deliberately show the SAME intersection point (4, 3) on two separate line diagrams rather than one combined graph, since this site's coordinate figure component only draws a single line per diagram; the two figures side by side make the shared point equally clear.
Present and print both work: use the Print button for a clean handout, or work an elimination example live on the board, asking the class to predict add-or-subtract before you show the step.

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