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Study GuideYear 10Level 2 · Standard

Free Year 10 Simultaneous Equations: Substitution and Elimination

Shows Year 10 students how to solve a pair of linear equations by substitution and by elimination, including when to multiply an equation first, how to handle subtraction with negatives, and how to set up and solve a word problem. A decision table helps students pick the quicker method, and every solution is checked in both equations.

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Year
Year 10
Subject
Mathematics
Topic
Simultaneous Equations
Difficulty
Level 2 · Standard
Estimated time
20 minutes
Curriculum
Australian Curriculum
Answers
Not applicable
Format
PDF (A4) + print

Students will practise

  • understanding that the solution to simultaneous equations satisfies both equations at once
  • solving by substitution when one variable is already the subject
  • solving by elimination, including multiplying an equation to match coefficients
  • setting up and solving simultaneous equations from a word problem

Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: Simultaneous Equations: Substitution and Elimination

  1. 1Year 10 Simultaneous Equations Worksheet — Level 2
  2. 2Year 10 Simultaneous Equations Worksheet — Level 3
Study guide

How to use this study guide

  1. Read it together first, pausing at each worked example to try the step before reading the answer.
  2. Attempt the "Check yourself" questions at the end without looking back.
  3. Then practise with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

Year 10 students (typically ages 15–16) working on simultaneous equations. It is pitched at level 2 · standard.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 20 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

Try Year 10 Simultaneous Equations Worksheet — Level 2.

Preview

Year 10 · Mathematics · Simultaneous Equations

Simultaneous Equations: Substitution and Elimination

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What you need to know

Simultaneous equations are two equations with two unknowns, usually x and y. The solution is the pair of values that makes both equations true at the same time. On a graph it is the point where the two lines cross. There are two algebraic methods, substitution and elimination, and whichever you use, always check your answer in both original equations.

Substitution. Use this when one equation already has a variable by itself, such as y = 2x + 1. Replace (substitute) that variable in the other equation with the expression it equals. You are left with one equation in one unknown, which you can solve.

Solve y = 2x + 1 and 3x + y = 11

  1. The first equation gives y in terms of x, so substitute 2x + 1 for y in the second: 3x + (2x + 1) = 11.
  2. Simplify: 5x + 1 = 11, so 5x = 10 and x = 2.
  3. Find y from the first equation: y = 2 × 2 + 1 = 5.
  4. Check in the second equation: 3 × 2 + 5 = 11. Correct. Solution: x = 2, y = 5.

Elimination. Line the equations up with x, y and the constants in columns. If one variable has the same coefficient in both equations, subtract one equation from the other; if the coefficients are opposite (such as +3y and −3y), add them. That variable disappears. If neither matches, multiply one or both equations first.

Solve 2x + 3y = 12 and 2x − y = 4

  1. Both equations have 2x, so subtract the second from the first: (2x + 3y) − (2x − y) = 12 − 4.
  2. Careful with the signs: 3y − (−y) = 4y. So 4y = 8 and y = 2.
  3. Substitute into the second equation: 2x − 2 = 4, so 2x = 6 and x = 3.
  4. Check in the first: 2 × 3 + 3 × 2 = 6 + 6 = 12. Correct. Solution: x = 3, y = 2.

Solve 3x + 2y = 16 and 5x − 4y = 1

  1. No coefficients match. Multiply the first equation by 2 to get 6x + 4y = 32; now the y terms are +4y and −4y.
  2. Add this to the second equation: (6x + 4y) + (5x − 4y) = 32 + 1, so 11x = 33 and x = 3.
  3. Substitute into the first equation: 3 × 3 + 2y = 16, so 2y = 7 and y = 3.5.
  4. Check in the second: 5 × 3 − 4 × 3.5 = 15 − 14 = 1. Correct. Solution: x = 3, y = 3.5.

Three pies and two drinks cost $23. One pie and two drinks cost $13. Find the price of each.

  1. Define the unknowns: let p be the price of a pie and d the price of a drink.
  2. Write the equations: 3p + 2d = 23 and p + 2d = 13.
  3. Both have 2d, so subtract: (3p + 2d) − (p + 2d) = 23 − 13, giving 2p = 10 and p = 5.
  4. Substitute: 5 + 2d = 13, so 2d = 8 and d = 4.
  5. Check: 3 × 5 + 2 × 4 = 15 + 8 = 23. A pie costs $5 and a drink costs $4.
SituationBest method
One equation is already y = … or x = …Substitution
A variable has the same coefficient in both equationsElimination: subtract
A variable has opposite coefficients (e.g. +4y and −4y)Elimination: add
No coefficients matchMultiply one or both equations, then eliminate
Common mistake: subtraction errors with negatives. When (2x − y) is subtracted from (2x + 3y), the y terms give 3y − (−y) = 4y, not 2y. Write the whole subtraction in brackets and distribute the minus sign to every term.
Common mistake: stopping after finding one variable. The solution is a pair of values. Substitute back to find the second unknown, then check both values in the equation you have not used yet.

Check yourself

  1. Solve y = x + 3 and x + y = 11.
  2. Solve y = 3x and 2x + y = 20.
  3. Solve x + y = 10 and x − y = 2.
  4. Solve 3x + y = 14 and x + y = 6.
  5. Solve 2x + 3y = 13 and x − y = −1.
  6. Two numbers add to 25 and their difference is 7. What are they?

Answers: 1. x = 4, y = 7 2. x = 4, y = 12 3. x = 6, y = 4 4. x = 4, y = 2 5. x = 2, y = 3 6. 16 and 9

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Last reviewed
1 October 2026
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