EDEXCEL GCSE MATHS · FOUNDATION & HIGHER
More simultaneous equations
Equations and inequalities · Lesson 6 of 7
Teacher copy - includes the notes for whoever is teaching from it.
Warm-up
Answer each one, then check.
1. Solve x + y = 5 and x − y = 1.
x = 3, y = 2
2. Multiply 2x + 3y = 5 by 3.
6x + 9y = 15
3. What is the lowest common multiple of 2 and 3?
6
4. Solve 4y = 12.
y = 3
5. What is −6 + 6?
0
Learning Objectives
1. Multiply one equation to match coefficients.
2. Multiply both equations to match coefficients.
3. Solve worded problems using two equations.
4. Check that answers satisfy both equations.
Matching the Coefficients
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Multiply so that one pair of coefficients matches, then add or subtract. |
The equations 2x plus 3y equals 13 and 5x minus 2y equals 4 multiplied by 2 and 3 so that the y terms match, then added to find x equals 2 and y equals 3.
Multiplying Before Eliminating
When no pair of coefficients matches.
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1 Choose the unknown to eliminate Look for the easiest pair of coefficients |
2 Find the lowest common multiple For 3 and 2 it is 6 |
3 Multiply each equation So the coefficients of that unknown match |
4 Add or subtract Same signs subtract, opposite signs add |
5 Solve, substitute, check Two values, checked in both equations |
Multiplying Both Equations
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Solve 2x + 3y = 13 and 5x − 2y = 4. |
1. Multiply (1) by 2 and (2) by 3 to match the y terms
4x + 6y = 26 and 15x − 6y = 12
2. The y terms have opposite signs, so add
19x = 38
3. Solve
x = 2
4. Substitute into (1)
4 + 3y = 13, so y = 3
5. Check in (2)
5 × 2 − 2 × 3 = 4
Answer: x = 2 and y = 3
Another Pair
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Solve 3x + 2y = 12 and 5x − 3y = 1. |
1. Multiply (1) by 3 and (2) by 2
9x + 6y = 36 and 10x − 6y = 2
2. Add
19x = 38, so x = 2
3. Substitute into (1)
6 + 2y = 12, so y = 3
4. Check in (2)
10 − 9 = 1
Answer: x = 2 and y = 3
A Number Problem
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The sum of two numbers is 25. Twice the first number plus three times the second number is 62. Find the numbers. |
1. Let the numbers be x and y
x + y = 25 and 2x + 3y = 62
2. Multiply the first by 2
2x + 2y = 50
3. Subtract from the second
y = 12
4. Find x
x = 25 − 12 = 13
Answer: The numbers are 13 and 12.
A Cost Problem
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3 apples and 2 bananas cost £1.80. 2 apples and 5 bananas cost £2.30. Find the cost of one apple and one banana. |
1. Write the equations in pence
3a + 2b = 180 and 2a + 5b = 230
2. Multiply (1) by 2 and (2) by 3
6a + 4b = 360 and 6a + 15b = 690
3. Subtract
11b = 330, so b = 30
4. Find a
3a + 60 = 180, so a = 40
Answer: An apple costs 40p and a banana costs 30p.
Common Slips
Check every step.
▸ Multiply every term. Both sides and every term of the equation.
▸ Signs when subtracting. Subtracting −6y adds 6y.
▸ Wrong equation for substitution. Substitute into either original equation, then check in the other.
▸ Units. In a money problem, keep everything in pence or everything in pounds.
Key Terms
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Simultaneous equations Equations that must both be true for the same unknown values. |
Elimination Removing an unknown by adding or subtracting. |
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Lowest common multiple The smallest number that is a multiple of two numbers. |
Coefficient The number multiplying an unknown. |
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Substitute Replace an unknown by its value. |
Check Put the answers back into both equations. |
Your Task: Spot the Method
12 minutes
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For each pair, say how you would eliminate an unknown, then solve. (a) x + 2y = 8 and 3x − y = 3 (b) 3x + 4y = 25 and 5x − 2y = 7 (c) 2x + 5y = 16 and 3x − 2y = 5. 1. Choose the unknown to eliminate. 2. Multiply to match. 3. Solve and check. |
A good answer shows: (a) Multiply the second by 2: 6x − 2y = 6, add: 7x = 14, so x = 2, y = 3. (b) Multiply the second by 2: 10x − 4y = 14, add: 13x = 39, so x = 3, y = 4. (c) Multiply the first by 2 and the second by 5: 4x + 10y = 32 and 15x − 10y = 25, add: 19x = 57, so x = 3, y = 2.
Note: Check that each answer satisfies both original equations.
Can I...?
☐ Multiply one equation to match coefficients.
☐ Multiply both equations.
☐ Choose add or subtract correctly.
☐ Find both unknowns.
☐ Check in both equations.
☐ Form equations from a worded problem.
☐ Work in consistent units.
☐ Explain each step.
Summary
✓ Multiply so that one pair of coefficients matches.
✓ Same signs: subtract. Opposite signs: add.
✓ Substitute back and check in both equations.
✓ In problems, define the letters and write two equations.
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EXAM FOCUS Solve the simultaneous equations 2x + 3y = 13 and 5x − 2y = 4. (4 marks) Multiply both equations so that the y-coefficients match (6 and 6), then add because the signs are opposite. |
Exam Practice: More Simultaneous Equations
Answer all questions. Show your working. · 35 minutes
▸ Question 1 · 4 marks · Non-calculator. Solve the simultaneous equations 2x + 3y = 13 and 5x − 2y = 4.
▸ Question 2 · 4 marks · Non-calculator. Solve the simultaneous equations 3x + 2y = 12 and 5x − 3y = 1.
▸ Question 3 · 4 marks · Non-calculator. The sum of two numbers is 25. Twice the first number plus three times the second number is 62. Work out the two numbers.
▸ Question 4 · 4 marks · Calculator. 3 apples and 2 bananas cost £1.80. 2 apples and 5 bananas cost £2.30. Work out the cost of one apple and the cost of one banana.
▸ Question 5 · 3 marks · Non-calculator. Find the coordinates of the point where the lines y = 3x − 2 and y = x + 4 cross.
▸ Question 6 · 4 marks · Non-calculator. Solve the simultaneous equations 4x + y = 14 and 2x − 3y = 0.
Question 1 · 4 marks · Non-calculator
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“Solve the simultaneous equations 2x + 3y = 13 and 5x − 2y = 4.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 4 marks, so plan before writing.
Question 1 · mark scheme
4 marks available. Award a mark for each point made.
▸ Multiplying to match coefficients. M1
▸ 19x = 38. M1
▸ x = 2. A1
▸ y = 3. A1
▸ Model answer. Multiplying gives 4x + 6y = 26 and 15x − 6y = 12. Adding gives 19x = 38, so x = 2. Then 4 + 3y = 13, so y = 3.
Question 2 · 4 marks · Non-calculator
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“Solve the simultaneous equations 3x + 2y = 12 and 5x − 3y = 1.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 4 marks, so plan before writing.
Question 2 · mark scheme
4 marks available. Award a mark for each point made.
▸ Multiplying to match coefficients. M1
▸ 19x = 38. M1
▸ x = 2. A1
▸ y = 3. A1
▸ Model answer. Multiplying gives 9x + 6y = 36 and 10x − 6y = 2. Adding gives 19x = 38, so x = 2 and y = 3.
Question 3 · 4 marks · Non-calculator
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“The sum of two numbers is 25. Twice the first number plus three times the second number is 62. Work out the two numbers.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 4 marks, so plan before writing.
Question 3 · mark scheme
4 marks available. Award a mark for each point made.
▸ Forming both equations. M1
▸ A correct elimination step. M1
▸ y = 12. A1
▸ x = 13. A1
▸ Model answer. x + y = 25 and 2x + 3y = 62. Subtracting twice the first from the second gives y = 12, so x = 13.
Question 4 · 4 marks · Calculator
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“3 apples and 2 bananas cost £1.80. 2 apples and 5 bananas cost £2.30. Work out the cost of one apple and the cost of one banana.” |
HOW TO ANSWER IT Command word: Calculator. Worth 4 marks, so plan before writing.
Question 4 · mark scheme
4 marks available. Award a mark for each point made.
▸ Forming both equations. M1
▸ Multiplying to match. M1
▸ b = 30 or a = 40. A1
▸ Both correct with units. A1
▸ Model answer. 3a + 2b = 180 and 2a + 5b = 230 (pence). Multiplying gives 6a + 4b = 360 and 6a + 15b = 690. Subtracting gives 11b = 330, b = 30, and a = 40. An apple costs 40p and a banana 30p.
Question 5 · 3 marks · Non-calculator
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“Find the coordinates of the point where the lines y = 3x − 2 and y = x + 4 cross.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 3 marks, so plan before writing.
Question 5 · mark scheme
3 marks available. Award a mark for each point made.
▸ Equating the two expressions. M1
▸ x = 3. A1
▸ (3, 7). A1
▸ Model answer. 3x − 2 = x + 4, so 2x = 6 and x = 3. Then y = 7. The point is (3, 7).
Question 6 · 4 marks · Non-calculator
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“Solve the simultaneous equations 4x + y = 14 and 2x − 3y = 0.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 4 marks, so plan before writing.
Question 6 · mark scheme
4 marks available. Award a mark for each point made.
▸ A method to eliminate or substitute. M1
▸ 7y = 14 or 14x = 42. M1
▸ y = 2. A1
▸ x = 3. A1
▸ Model answer. From the second, x = 1.5y. Then 6y + y = 14, so y = 2 and x = 3. (Or multiply the first by 3 and add.)