EDEXCEL GCSE MATHS · FOUNDATION & HIGHER
Solving simple simultaneous equations
Equations and inequalities · Lesson 5 of 7
Teacher copy - includes the notes for whoever is teaching from it.
Warm-up
Answer each one, then check.
1. Solve 2x + 3 = 11.
x = 4
2. What is 7 − (−3)?
10
3. Substitute x = 2 into y = 3x − 1.
y = 5
4. What is the y-intercept of y = 2x + 5?
5
5. Solve x − 4 = −1.
x = 3
Learning Objectives
1. Explain what simultaneous equations are.
2. Solve by elimination when a pair of coefficients match.
3. Solve by substitution.
4. Set up simultaneous equations from a problem and read solutions from graphs.
Simultaneous Equations
Simultaneous equations have two unknowns, and the solution is the pair of values that makes both equations true at the same time.
On a graph, the solution is where the two lines cross.
Where Two Lines Meet
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The crossing point is the only pair of values on both lines. |
The lines x plus y equals 7 and x minus y equals 1 crossing at the point 4, 3.
Solving by Elimination
Make one unknown disappear.
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1 Label the equations Call them (1) and (2) |
2 Match the coefficients If they are already equal or opposite, go on; if not, multiply |
3 Add or subtract Add if the signs are opposite; subtract if they are the same |
4 Solve for one unknown You now have a one-variable equation |
5 Substitute back Find the other unknown, then check in the other equation |
Elimination by Adding
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Solve x + y = 7 and x − y = 1. |
1. The y terms have opposite signs, so add
(x + y) + (x − y) = 7 + 1
2. Simplify
2x = 8, so x = 4
3. Substitute into x + y = 7
4 + y = 7, so y = 3
4. Check in x − y = 1
4 − 3 = 1
Answer: x = 4 and y = 3
Elimination by Subtracting
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Solve 2x + y = 11 and x + y = 7. |
1. The y terms have the same sign, so subtract
(2x + y) − (x + y) = 11 − 7
2. Simplify
x = 4
3. Substitute into x + y = 7
y = 3
4. Check
2 × 4 + 3 = 11
Answer: x = 4 and y = 3
Solving by Substitution
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Solve y = 2x − 1 and x + y = 8. |
1. Substitute y = 2x − 1 into the second equation
x + (2x − 1) = 8
2. Simplify
3x − 1 = 8, so 3x = 9 and x = 3
3. Find y
y = 2 × 3 − 1 = 5
Answer: x = 3 and y = 5
A Word Problem
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2 adult tickets and 3 child tickets cost £27. 1 adult ticket and 2 child tickets cost £15. Find the cost of each ticket. |
1. Let a be an adult ticket and c a child ticket
2a + 3c = 27 and a + 2c = 15
2. Rearrange the second
a = 15 − 2c
3. Substitute
2(15 − 2c) + 3c = 27, so 30 − 4c + 3c = 27
4. Solve
c = 3, so a = 15 − 6 = 9
Answer: An adult ticket costs £9 and a child ticket costs £3.
Elimination or Substitution?
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ELIMINATION |
SUBSTITUTION |
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▸ Best when the equations are both in the form ax + by = c. ▸ Add or subtract to remove one unknown. ▸ Needs matching coefficients. |
▸ Best when one equation already gives y = … or x = …. ▸ Replace the unknown in the other equation. ▸ Watch brackets and signs. |
Key Terms
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Simultaneous equations Two or more equations that are true at the same time. |
Unknown A letter standing for a number to be found. |
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Elimination Removing one unknown by adding or subtracting equations. |
Substitution Replacing an unknown with an expression. |
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Solution The pair of values that satisfies both equations. |
Coefficient The number multiplying an unknown. |
Your Task: Solve and Check
12 minutes
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Solve each pair of simultaneous equations, then check your answer in both equations. (a) x + y = 10, x − y = 4 (b) 3x + y = 14, x + y = 8 (c) y = x + 2, 2x + y = 11. 1. Choose elimination or substitution. 2. Solve. 3. Check in both equations. |
A good answer shows: (a) x = 7, y = 3. (b) 2x = 6, so x = 3, y = 5. (c) 2x + x + 2 = 11, so x = 3, y = 5.
Note: Encourage students to write each step under the equation numbers.
Can I...?
☐ Explain what simultaneous equations are.
☐ Eliminate by adding.
☐ Eliminate by subtracting.
☐ Substitute one equation into the other.
☐ Find both unknowns.
☐ Check my solution.
☐ Set up equations from a problem.
☐ Read the solution from a graph.
Summary
✓ Same signs: subtract. Opposite signs: add.
✓ Substitute back to find the second unknown.
✓ Always check in both equations.
✓ On a graph, the solution is the crossing point.
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EXAM FOCUS 2 adult tickets and 3 child tickets cost £27. 1 adult ticket and 2 child tickets cost £15. Work out the cost of an adult ticket and the cost of a child ticket. (4 marks) Define letters, write two equations, then solve. Check that both original equations work with your answer. |
Exam Practice: Solving Simple Simultaneous Equations
Answer all questions. Show your working. · 30 minutes
▸ Question 1 · 3 marks · Non-calculator. Solve the simultaneous equations x + y = 7 and x − y = 1.
▸ Question 2 · 3 marks · Non-calculator. Solve the simultaneous equations 2x + y = 11 and x + y = 7.
▸ Question 3 · 3 marks · Non-calculator. Solve the simultaneous equations y = 2x − 1 and x + y = 8.
▸ Question 4 · 4 marks · Non-calculator. 2 adult tickets and 3 child tickets cost £27 in total. 1 adult ticket and 2 child tickets cost £15 in total. Work out the cost of an adult…
▸ Question 5 · 2 marks · Non-calculator. The graph shows the lines y = 2x − 3 and x + y = 6. Use the graph to solve the simultaneous equations y = 2x − 3 and x + y = 6.
▸ Question 6 · 3 marks · Non-calculator. Solve the simultaneous equations 3x + 2y = 16 and x − y = 2.
Question 1 · 3 marks · Non-calculator
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“Solve the simultaneous equations x + y = 7 and x − y = 1.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 3 marks, so plan before writing.
Question 1 · mark scheme
3 marks available. Award a mark for each point made.
▸ Adding or a correct elimination. M1
▸ x = 4. A1
▸ y = 3. A1
▸ Model answer. Adding gives 2x = 8, so x = 4. Then y = 3.
Question 2 · 3 marks · Non-calculator
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“Solve the simultaneous equations 2x + y = 11 and x + y = 7.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 3 marks, so plan before writing.
Question 2 · mark scheme
3 marks available. Award a mark for each point made.
▸ Subtracting the equations. M1
▸ x = 4. A1
▸ y = 3. A1
▸ Model answer. Subtracting gives x = 4. Then y = 7 − 4 = 3.
Question 3 · 3 marks · Non-calculator
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“Solve the simultaneous equations y = 2x − 1 and x + y = 8.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 3 marks, so plan before writing.
Question 3 · mark scheme
3 marks available. Award a mark for each point made.
▸ Substituting. M1
▸ x = 3. A1
▸ y = 5. A1
▸ Model answer. x + 2x − 1 = 8, so 3x = 9, x = 3 and y = 5.
Question 4 · 4 marks · Non-calculator
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“2 adult tickets and 3 child tickets cost £27 in total. 1 adult ticket and 2 child tickets cost £15 in total. Work out the cost of an adult ticket and the cost of a child ticket.” |
HOW TO ANSWER IT Command word: Non-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
▸ A correct elimination or substitution. M1
▸ One value correct. A1
▸ Both values with units. A1
▸ Model answer. 2a + 3c = 27 and a + 2c = 15. a = 15 − 2c, so 30 − 4c + 3c = 27, c = 3, a = 9. Adult £9, child £3.
Question 5 · 2 marks · Non-calculator
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The graph shows the lines y = 2x − 3 and x + y = 6. Use the graph to solve the simultaneous equations y = 2x − 3 and x + y = 6. (2 marks) |
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Question 5 · mark scheme
2 marks available. Award a mark for each point made.
▸ Reading the crossing point. M1
▸ x = 3 and y = 3. A1
▸ Model answer. The lines cross at (3, 3), so x = 3 and y = 3.
Question 6 · 3 marks · Non-calculator
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“Solve the simultaneous equations 3x + 2y = 16 and x − y = 2.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 3 marks, so plan before writing.
Question 6 · mark scheme
3 marks available. Award a mark for each point made.
▸ Substituting or multiplying to match. M1
▸ y = 2. A1
▸ x = 4. A1
▸ Model answer. x = y + 2, so 3(y + 2) + 2y = 16, 5y = 10, y = 2 and x = 4.