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Maths · Equations and inequalities

Solving simple simultaneous equations

Solve pairs of linear equations with two unknowns by elimination and substitution, and read the solution from a graph.

  • 6 key terms
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Student handouts

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Warm-up

Answer each one, then check.

  1. 1

    Solve \(2x + 3 = 11\).

    Show answerHide answer

    \(x = 4\)

  2. 2

    What is \(7 - (-3)\)?

    Show answerHide answer

    10

  3. 3

    Substitute \(x = 2\) into \(y = 3x - 1\).

    Show answerHide answer

    \(y = 5\)

  4. 4

    What is the y-intercept of \(y = 2x + 5\)?

    Show answerHide answer

    5

  5. 5

    Solve \(x - 4 = -1\).

    Show answerHide answer

    \(x = 3\)

Learning Objectives

  1. 1Explain what simultaneous equations are.
  2. 2Solve by elimination when a pair of coefficients match.
  3. 3Solve by substitution.
  4. 4Set 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.

Solving by Elimination

Make one unknown disappear.

  1. 1 Label the equations

    Call them (1) and (2)
  2. 2 Match the coefficients

    If they are already equal or opposite, go on; if not, multiply
  3. 3 Add or subtract

    Add if the signs are opposite; subtract if they are the same
  4. 4 Solve for one unknown

    You now have a one-variable equation
  5. 5 Substitute back

    Find the other unknown, then check in the other equation

Elimination by Adding

Solve \(x + y = 7\) and \(x - y = 1\).

Show the solutionHide the solution
  1. 1 The y terms have opposite signs, so add \((x + y) + (x - y) = 7 + 1\)
  2. 2 Simplify \(2x = 8\), so \(x = 4\)
  3. 3 Substitute into \(x + y = 7\) \(4 + y = 7\), so \(y = 3\)
  4. 4 Check in \(x - y = 1\) \(4 - 3 = 1\)

Answer\(x = 4\) and \(y = 3\)

Elimination by Subtracting

Solve \(2x + y = 11\) and \(x + y = 7\).

Show the solutionHide the solution
  1. 1 The y terms have the same sign, so subtract \((2x + y) - (x + y) = 11 - 7\)
  2. 2 Simplify \(x = 4\)
  3. 3 Substitute into \(x + y = 7\) \(y = 3\)
  4. 4 Check \(2 \times 4 + 3 = 11\)

Answer\(x = 4\) and \(y = 3\)

Solving by Substitution

Solve \(y = 2x - 1\) and \(x + y = 8\).

Show the solutionHide the solution
  1. 1 Substitute \(y = 2x - 1\) into the second equation \(x + (2x - 1) = 8\)
  2. 2 Simplify \(3x - 1 = 8\), so \(3x = 9\) and \(x = 3\)
  3. 3 Find y \(y = 2 \times 3 - 1 = 5\)

Answer\(x = 3\) and \(y = 5\)

A Word Problem

2 adult tickets and 3 child tickets cost £27. 1 adult ticket and 2 child tickets cost £15. Find the cost of each ticket.

Show the solutionHide the solution
  1. 1 Let \(a\) be an adult ticket and \(c\) a child ticket \(2a + 3c = 27\) and \(a + 2c = 15\)
  2. 2 Rearrange the second \(a = 15 - 2c\)
  3. 3 Substitute \(2(15 - 2c) + 3c = 27\), so \(30 - 4c + 3c = 27\)
  4. 4 Solve \(c = 3\), so \(a = 15 - 6 = 9\)

AnswerAn adult ticket costs £9 and a child ticket costs £3.

Elimination or Substitution?

Elimination

  • Best when the equations are both in the form \(ax + by = c\).
  • Add or subtract to remove one unknown.
  • Needs matching coefficients.

Substitution

  • Best when one equation already gives \(y = \dots\) or \(x = \dots\).
  • Replace the unknown in the other equation.
  • Watch brackets and signs.

Solve and Check

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 \(x = 3\), \(y = 5\).

Can I...?

  1. 1Explain what simultaneous equations are.
  2. 2Eliminate by adding.
  3. 3Eliminate by subtracting.
  4. 4Substitute one equation into the other.
  5. 5Find both unknowns.
  6. 6Check my solution.
  7. 7Set up equations from a problem.
  8. 8Read the solution from a graph.

Summary & Exam Focus

  • 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.

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) (4 marks)

Define letters, write two equations, then solve. Check that both original equations work with your answer.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Simultaneous equations
Two or more equations that are true at the same time.
Unknown
A letter standing for a number to be found.
Elimination
Removing one unknown by adding or subtracting equations.
Substitution
Replacing an unknown with an expression.
Solution
The pair of values that satisfies both equations.
Coefficient
The number multiplying an unknown.

Questions and answers

12 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Non-calculator 3 marks Easier

Solve the simultaneous equations \(x + y = 7\) and \(x - y = 1\).

Mark scheme — 3 marks available

  • Adding or a correct elimination — M1
  • \(x = 4\) — A1
  • \(y = 3\) — A1

Model answer

Adding gives \(2x = 8\), so \(x = 4\). Then \(y = 3\).

2. Exam question Non-calculator 3 marks Easier

Solve the simultaneous equations \(2x + y = 11\) and \(x + y = 7\).

Mark scheme — 3 marks available

  • Subtracting the equations — M1
  • \(x = 4\) — A1
  • \(y = 3\) — A1

Model answer

Subtracting gives \(x = 4\). Then \(y = 7 - 4 = 3\).

3. Exam question Non-calculator 3 marks Easier

Solve the simultaneous equations \(y = 2x - 1\) and \(x + y = 8\).

Mark scheme — 3 marks available

  • Substituting — M1
  • \(x = 3\) — A1
  • \(y = 5\) — A1

Model answer

\(x + 2x - 1 = 8\), so \(3x = 9\), \(x = 3\) and \(y = 5\).

4. Exam question Non-calculator 4 marks Easier

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.

Mark scheme — 4 marks available

  • 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.

5. Exam question Non-calculator 2 marks Easier

The graph shows the lines \(y = 2x - 3\) and \(x + y = 6\). Use the graph to solve the simultaneous equations \(y = 2x - 3\) and \(x + y = 6\).

The lines y equals 2x minus 3 and x plus y equals 6 drawn on a grid, crossing at the point 3, 3.

Mark scheme — 2 marks available

  • Reading the crossing point — M1
  • \(x = 3\) and \(y = 3\) — A1

Model answer

The lines cross at \((3, 3)\), so \(x = 3\) and \(y = 3\).

6. Exam question Non-calculator 3 marks Easier

Solve the simultaneous equations \(3x + 2y = 16\) and \(x - y = 2\).

Mark scheme — 3 marks available

  • 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 \(x = 4\).

7. Multiple choice 1 mark Easier

Solve \(x + y = 10\) and \(x - y = 2\).

  1. A \(x = 4, y = 6\)
  2. B \(x = 5, y = 5\)
  3. C \(x = 6, y = 4\) Correct
  4. D \(x = 8, y = 2\)

Why: Adding gives \(2x = 12\), so \(x = 6\) and \(y = 4\).

8. Multiple choice 1 mark Core

To eliminate y from \(3x + y = 10\) and \(x + y = 6\), you...

  1. A Add the equations
  2. B Subtract the equations Correct
  3. C Multiply both by 3
  4. D Divide the equations

Why: The y terms have the same sign, so subtract.

9. Multiple choice 1 mark Core

On a graph, the solution of two simultaneous equations is...

  1. A The y-intercept of one line
  2. B The x-intercept of one line
  3. C The gradient of the lines
  4. D The crossing point of the lines Correct

Why: The point where the two lines cross.

10. Multiple choice 1 mark Core

If \(y = x + 2\) and \(x + y = 8\), what is \(x\)?

  1. A 3 Correct
  2. B 5
  3. C 6
  4. D 2

Why: \(x + x + 2 = 8\), so \(2x = 6\) and \(x = 3\).

11. Multiple choice 1 mark Core

You solve simultaneous equations and find \(x = 2\). What should you do next?

  1. A Stop; you are done
  2. B Substitute back to find y Correct
  3. C Divide by 2
  4. D Multiply by 2

Why: Substitute back to find y, then check in both equations.

12. Multiple choice 1 mark Stretch

How many solutions do two parallel lines have?

  1. A One
  2. B Two
  3. C None Correct
  4. D Infinitely many

Why: Parallel lines never meet, so there is no solution.