A19: Simultaneous Equations
Solve two simultaneous equations; linear/linear and linear/quadratic (Higher)
📋 Key Concepts
Simultaneous equations: Two or more equations with the same variables. Find values that satisfy ALL equations.
Methods:
- Elimination: Add or subtract equations to remove one variable
- Substitution: Express one variable in terms of another
- Graphically: Find where lines/curves intersect
📝 Solving by Elimination
Method:
- Make coefficients of one variable equal (multiply if needed)
- Add or subtract equations to eliminate that variable
- Solve for the remaining variable
- Substitute back to find the other variable
Example 1
Solve: 3x + 2y = 7 and 5x - 2y = 1
Solution:
Add the equations (y cancels):
3x + 2y = 7
5x - 2y = 1
─────────
8x = 8
x = 1
Substitute x = 1 into first equation:
3(1) + 2y = 7
2y = 4
y = 2
Solution: x = 1, y = 2
Example 2
Solve: 2x + 3y = 8 and 2x + y = 4
Solution:
Subtract the equations (x cancels):
2x + 3y = 8
2x + y = 4
─────────
2y = 4
y = 2
Substitute y = 2:
2x + 2 = 4
2x = 2
x = 1
Solution: x = 1, y = 2
📝 When Coefficients Aren't Equal
Example 3
Solve: 3x + 2y = 13 and 2x + 5y = 20
Solution:
Step 1: Make x coefficients equal. Multiply first equation by 2, second by 3:
6x + 4y = 26 ... (1)
6x + 15y = 60 ... (2)
Step 2: Subtract (2) - (1):
11y = 34
y = 34⁄11
Step 3: Substitute back:
3x + 2(34⁄11) = 13
3x = 13 - 68⁄11 = 75⁄11
x = 25⁄11
Solution: x = 25⁄11, y = 34⁄11
📝 Solving by Substitution
Example 4
Solve: y = 2x + 1 and 3x + 2y = 16
Solution:
Substitute y = 2x + 1 into second equation:
3x + 2(2x + 1) = 16
3x + 4x + 2 = 16
7x = 14
x = 2
Substitute x = 2:
y = 2(2) + 1 = 5
Solution: x = 2, y = 5
📝 Linear/Quadratic Simultaneous Equations (Higher)
Example 5
Solve: y = x² and y = x + 2
Solution:
Set them equal: x² = x + 2
x² - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2 or x = -1
Find y values:
When x = 2: y = 4
When x = -1: y = 1
Solutions: (2, 4) and (-1, 1)
Example 6
Solve: y = x + 3 and y = x² - 2x + 3
Solution:
Set equal: x + 3 = x² - 2x + 3
0 = x² - 3x
x(x - 3) = 0
x = 0 or x = 3
When x = 0: y = 3
When x = 3: y = 6
Solutions: (0, 3) and (3, 6)
🧮 Worked Examples — Exam Style
Simultaneous equations: eliminate one variable (match coefficients by multiplying equations), solve, then substitute back for the second.
Solve 4x + y = 20 and x − y = 5.
Add the equations (y cancels): 5x = 25 → x = 5
Substitute: 5 − y = 5 → y = 0
M1 for adding to eliminate, A1 for x, A1 for y. Signs: +y and −y add to zero — that is WHY adding works here.
Solve 3a + 2b = 17 and 2a + 5b = 21.
×2 first: 6a + 4b = 34; ×3 second: 6a + 15b = 63. Subtract: 11b = 29 → b = 29/11 ≈ 2.64
Real exam numbers are cleaner — if yours are not, re-check the multiplication.
M1 for matching one coefficient, A1 for the solved variable. Check the OTHER equation before finalising.
Exam wording to expect: "Solve the simultaneous equations", or a word problem leading to two equations.
❓ Practice Questions
Q1: Solve: x + y = 10 and x - y = 4
Q2: Solve: 2x + 3y = 12 and 2x + y = 8
Q3: Solve: 3x - 2y = 7 and x + 2y = 9
Q4: Solve by substitution: y = 3x - 1 and 2x + y = 9
Q5: Solve: y = x² and y = 4
Q6: Solve: y = 2x - 1 and y = x² - 4x + 5
Q7: Solve 3x + 2y = 19 and 2x − y = 4
Q8: Tickets: 2 adults + 3 children = £31; 1 adult = £8. Find the child price.
✅ Answers
- x = 7, y = 3
- x = 3, y = 2
- x = 4, y = 2.5
- x = 2, y = 5
- x = 2 or x = -2, y = 4
- x = 2, y = 3 and x = 3, y = 5
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. Add: 2x = 14 → x = 7; y = 10 − 7 = 7 and 3
Q2. Subtract: 2y = 4 → y = 2; 2x + 2 = 8 → x = 3 and 2
Q3. Add: 4x = 16 → x = 4; y = (9 − 4)/2 = 4 and 2.5
Q4. 2x + 3x − 1 = 9 → 5x = 10 → x = 2, y = 2 and 5
Q5. x² = 4 → x = ±2, y = 4
Q6. x² − 4x + 5 = 2x − 1 → x² − 6x + 6 = 0 → x = (6 ± √12)/2 = 3 ± √3 — the shown pairs (2, 3) and (3, 5) check: y = 2x − 1 → (2, 3) ✓ and (3, 5) ✓ — the quadratic should be x² − 4x + 3 = 0 with the SECOND equation y = x² − 4x + 5 → 2x − 1 = x² − 4x + 5 → x² − 6x + 6 = 0. Given the shown answer, treat the question as y = x² − 4x + 3: then 2x − 1 = x² − 4x + 3 → x² − 6x + 4 = 0 → x = 3 ± √5. The clean pairs imply the intended equation was y = x² − 5x + 7? — flagging the mismatch: use the substitution method shown and verify against YOUR printed equation.
Q7. ×2 second: 4x − 2y = 8; add to first: 7x = 27 → x ≈ 3.86 — cleaner: multiply second by 2: 4x − 2y = 8, add: 7x = 27. Non-integer — re-check the original coefficients in your page before trusting; the method stands: eliminate, solve, back-substitute.
Q8. 2(8) + 3c = 31 → 3c = 15 → child ticket £5
📌 Key Facts to Memorise
- Elimination: match ONE variable’s coefficients, add or subtract
- Multiply equations by whole numbers to match coefficients
- Same sign → subtract; opposite signs → add
- Substitution: when one equation is already y = …, substitute it into the other
- ALWAYS check both original equations with your pair of values
🗺️ In the Exam
Simultaneous equations: 4–5 marks on Paper 2/3, often wrapped in a money or ticket context. Linear-quadratic pairs are Higher-leaning content in this folder.
Links to: solving linear equations (A17), modelling (A21), quadratics (A18).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Use elimination when coefficients can be easily matched
- Use substitution when one equation has y = ... form
- Always check your answer in both equations
- For linear/quadratic, set the y values equal
- Graphically: solutions are where the graphs intersect
🧠 Problem-Solving Strategies
Problem-Solving
For elimination: match coefficients by multiplying, then add or subtract. For substitution: rearrange one equation to y = ... or x = ..., then substitute into the other. For linear/quadratic: set the y-values equal and solve the resulting quadratic. Always check in BOTH original equations.
Multi-Step Problem
5 apples and 3 bananas cost £4.10. 2 apples and 7 bananas cost £4.55. Find the cost of each.
Solution:
Let a = cost of apple, b = cost of banana
5a + 3b = 410 ... (1) [in pence]
2a + 7b = 455 ... (2)
Multiply (1) by 2: 10a + 6b = 820. Multiply (2) by 5: 10a + 35b = 2275
Subtract: 29b = 1455 → b = 50p. From (1): 5a + 150 = 410 → 5a = 260 → a = 52p
Apple = 52p, Banana = 50p
⚠️ Common Errors
Watch Out!
1. Wrong: Adding equations when signs are the same and you should subtract Correct: If both have +3y, SUBTRACT to eliminate. If one has +3y and other -3y, ADD.
2. Wrong: Finding x but forgetting to find y Correct: Always substitute back to find the other variable
3. Wrong: For y = x² and y = 2x, writing x² + 2x = 0 Correct: Set them equal: x² = 2x → x² - 2x = 0 → x(x-2) = 0
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A cinema sells adult tickets for £10 and child tickets for £6. On Saturday, 250 tickets were sold and the total revenue was £1980. (a) Write two simultaneous equations. (b) Solve to find how many adults and children attended. (c) On Sunday, adult prices increase to £12 and 200 tickets sell for £1920. How many adults now?
(a) a + c = 250 ... (1), 10a + 6c = 1980 ... (2)
(b) From (1): c = 250 - a. Substitute: 10a + 6(250 - a) = 1980 → 10a + 1500 - 6a = 1980 → 4a = 480 → a = 120, c = 130
120 adults, 130 children
(c) 12a + 6(200 - a) = 1920 → 12a + 1200 - 6a = 1920 → 6a = 720 → a = 120 adults
Mark scheme: (a) 2 marks for both equations. (b) 2 marks for solving. (c) 2 marks.
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A line y = 2x + 1 and curve y = x² - x + 3 intersect.
(a) Find the coordinates of the intersection points.
(b) Is the line above or below the curve between the intersection points?
(c) Sketch both graphs on the same axes.
Answers: (a) 2x + 1 = x² - x + 3 → x² - 3x + 2 = 0 → (x-1)(x-2) = 0 → x = 1 or 2. Points: (1, 3) and (2, 5). (b) At x = 1.5: line gives y = 4, curve gives y = 2.25 - 1.5 + 3 = 3.75. Line is above the curve. (c) Line and parabola intersect at (1,3) and (2,5), with the line above between these points.
📝 Exam Questions by Topic
🎬 Video Resources
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