Solve two linear equations by elimination or substitution
Eliminate a variable by equal operations · Use substitution as an alternative · Check both original equations and context
Exam scope and assessment notes
Core and Extended foundations. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Linear equations
- Expansion
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Eliminate a variable by equal operations
- Use substitution as an alternative
- Check both original equations and context
Two conditions, two unknowns
A simultaneous solution satisfies both equations at once. For substitution, isolate one variable and put that entire expression into the other equation. For elimination, scale equations if needed, then add or subtract to cancel one variable.
Keep all terms when scaling an equation. After finding one variable, substitute back for the other and check both original conditions.
Interpret special cases
Parallel inconsistent equations can reduce to a false statement such as 0 = 3: no solution. Equivalent equations can reduce to 0 = 0: infinitely many pairs satisfying the common line. In a count problem, algebraic solutions must also be non-negative integers.
Sources for this lesson
Pause and explain
What does a simultaneous solution satisfy?
Worked example
Solve 2x+y = 13 and x−y = 2.
Show the worked solution
- Add to get 3x = 15, so x = 5.
- 5−y = 2 gives y = 3.
- Check 10+3 = 13 and 5−3 = 2.
Answer x = 5, y = 3.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Solve x+y = 9 and x−y = 3.
Give me a hint
Add to eliminate y.
Compare my reasoning
- 2x = 12 gives x = 6.
- y = 3; check both sums.
x = 6, y = 3.
Look for these in your work
- Eliminate cleanly
- Check both conditions
Solve 2a+3b = 19 and a−b = 2.
Give me a hint
Substitute a = b+2.
Compare my reasoning
- 2(b+2)+3b = 19 gives 5b = 15.
- b = 3 and a = 5.
a = 5, b = 3.
Look for these in your work
- Substitute the full expression
- Recover the second unknown
Ten tickets cost 65 in total; adult tickets cost 8 and child tickets 5. How many of each were bought?
Give me a hint
Use one equation for count and one for cost.
Compare my reasoning
- a+c = 10; 8a+5c = 65.
- Subtract 5(a+c)=50 to get 3a = 15; a = 5, c = 5.
5 adult and 5 child tickets.
Look for these in your work
- Define both variables
- Check counts and revenue
Common mistakes
- A pair must satisfy both equations.
- Scale the constant as well as the variable terms.
What can you explain now?
What does a simultaneous solution satisfy?
Compare with the explanation
Both original equations at the same time.
Solving only one equation leaves many possible pairs.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
What does a simultaneous solution satisfy?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.