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Maths · Equations and inequalities

More simultaneous equations

Solve simultaneous equations when neither pair of coefficients matches, by multiplying one or both equations first, and solve worded problems.

  • 6 key terms
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Teacher resources

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Student handouts

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Warm-up

Answer each one, then check.

  1. 1

    Solve \(x + y = 5\) and \(x - y = 1\).

    Show answerHide answer

    \(x = 3\), \(y = 2\)

  2. 2

    Multiply \(2x + 3y = 5\) by 3.

    Show answerHide answer

    \(6x + 9y = 15\)

  3. 3

    What is the lowest common multiple of 2 and 3?

    Show answerHide answer

    6

  4. 4

    Solve \(4y = 12\).

    Show answerHide answer

    \(y = 3\)

  5. 5

    What is \(-6 + 6\)?

    Show answerHide answer

    0

Learning Objectives

  1. 1Multiply one equation to match coefficients.
  2. 2Multiply both equations to match coefficients.
  3. 3Solve worded problems using two equations.
  4. 4Check that answers satisfy both equations.

Multiplying Before Eliminating

When no pair of coefficients matches.

  1. 1 Choose the unknown to eliminate

    Look for the easiest pair of coefficients
  2. 2 Find the lowest common multiple

    For 3 and 2 it is 6
  3. 3 Multiply each equation

    So the coefficients of that unknown match
  4. 4 Add or subtract

    Same signs subtract, opposite signs add
  5. 5 Solve, substitute, check

    Two values, checked in both equations

Multiplying Both Equations

Solve \(2x + 3y = 13\) and \(5x - 2y = 4\).

Show the solutionHide the solution
  1. 1 Multiply (1) by 2 and (2) by 3 to match the y terms \(4x + 6y = 26\) and \(15x - 6y = 12\)
  2. 2 The y terms have opposite signs, so add \(19x = 38\)
  3. 3 Solve \(x = 2\)
  4. 4 Substitute into (1) \(4 + 3y = 13\), so \(y = 3\)
  5. 5 Check in (2) \(5 \times 2 - 2 \times 3 = 4\)

Answer\(x = 2\) and \(y = 3\)

Another Pair

Solve \(3x + 2y = 12\) and \(5x - 3y = 1\).

Show the solutionHide the solution
  1. 1 Multiply (1) by 3 and (2) by 2 \(9x + 6y = 36\) and \(10x - 6y = 2\)
  2. 2 Add \(19x = 38\), so \(x = 2\)
  3. 3 Substitute into (1) \(6 + 2y = 12\), so \(y = 3\)
  4. 4 Check in (2) \(10 - 9 = 1\)

Answer\(x = 2\) and \(y = 3\)

A Number Problem

The sum of two numbers is 25. Twice the first number plus three times the second number is 62. Find the numbers.

Show the solutionHide the solution
  1. 1 Let the numbers be \(x\) and \(y\) \(x + y = 25\) and \(2x + 3y = 62\)
  2. 2 Multiply the first by 2 \(2x + 2y = 50\)
  3. 3 Subtract from the second \(y = 12\)
  4. 4 Find x \(x = 25 - 12 = 13\)

AnswerThe numbers are 13 and 12.

A Cost Problem

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.

Show the solutionHide the solution
  1. 1 Write the equations in pence \(3a + 2b = 180\) and \(2a + 5b = 230\)
  2. 2 Multiply (1) by 2 and (2) by 3 \(6a + 4b = 360\) and \(6a + 15b = 690\)
  3. 3 Subtract \(11b = 330\), so \(b = 30\)
  4. 4 Find a \(3a + 60 = 180\), so \(a = 40\)

AnswerAn 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.

Spot the Method

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 \(x = 3\), \(y = 2\).

Can I...?

  1. 1Multiply one equation to match coefficients.
  2. 2Multiply both equations.
  3. 3Choose add or subtract correctly.
  4. 4Find both unknowns.
  5. 5Check in both equations.
  6. 6Form equations from a worded problem.
  7. 7Work in consistent units.
  8. 8Explain each step.

Summary & Exam Focus

  • 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.

Exam focus

Solve the simultaneous equations \(2x + 3y = 13\) and \(5x - 2y = 4\). (4 marks) (4 marks)

Multiply both equations so that the y-coefficients match (6 and 6), then add because the signs are opposite.

Key terms

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

Simultaneous equations
Equations that must both be true for the same unknown values.
Elimination
Removing an unknown by adding or subtracting.
Lowest common multiple
The smallest number that is a multiple of two numbers.
Coefficient
The number multiplying an unknown.
Substitute
Replace an unknown by its value.
Check
Put the answers back into both equations.

Questions and answers

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

1. Exam question Non-calculator 4 marks Easier

Solve the simultaneous equations \(2x + 3y = 13\) and \(5x - 2y = 4\).

Mark scheme — 4 marks available

  • 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 \(y = 3\).

2. Exam question Non-calculator 4 marks Easier

Solve the simultaneous equations \(3x + 2y = 12\) and \(5x - 3y = 1\).

Mark scheme — 4 marks available

  • 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 \(y = 3\).

3. Exam question Non-calculator 4 marks Easier

The sum of two numbers is 25. Twice the first number plus three times the second number is 62. Work out the two numbers.

Mark scheme — 4 marks available

  • 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 \(x = 13\).

4. Exam question Calculator 4 marks Easier

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.

Mark scheme — 4 marks available

  • 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.

5. Exam question Non-calculator 3 marks Easier

Find the coordinates of the point where the lines \(y = 3x - 2\) and \(y = x + 4\) cross.

Mark scheme — 3 marks available

  • 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 \((3, 7)\).

6. Exam question Non-calculator 4 marks Easier

Solve the simultaneous equations \(4x + y = 14\) and \(2x - 3y = 0\).

Mark scheme — 4 marks available

  • 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.)

7. Multiple choice 1 mark Core

To match the y-coefficients in \(2x + 3y = 13\) and \(5x - 2y = 4\), which multipliers do you use?

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

Why: Multiply the first by 2 and the second by 3 to get \(6y\) and \(-6y\).

8. Multiple choice 1 mark Core

After matching, you have \(6y\) and \(-6y\). What do you do?

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

Why: The signs are opposite, so add the equations.

9. Multiple choice 1 mark Core

Solve \(x + y = 5\) and \(2x + y = 8\).

  1. A \(x = 2, y = 3\)
  2. B \(x = 5, y = 0\)
  3. C \(x = 3, y = 2\) Correct
  4. D \(x = 4, y = 1\)

Why: Subtracting gives \(x = 3\), then \(y = 2\).

10. Multiple choice 1 mark Core

Two numbers add to 10 and differ by 4. What are they?

  1. A 6 and 4
  2. B 5 and 5
  3. C 8 and 2
  4. D 7 and 3 Correct

Why: \(x + y = 10\) and \(x - y = 4\) give \(x = 7\) and \(y = 3\).

11. Multiple choice 1 mark Core

Why should you check your solution in both original equations?

  1. A The question requires it
  2. B To catch mistakes Correct
  3. C It makes the answer larger
  4. D To use up time

Why: A slip in the working could still satisfy one equation, so checking both catches mistakes.

12. Multiple choice 1 mark Stretch

Which pair of equations has no solution?

  1. A \(y = x\) and \(y = 2x\)
  2. B \(y = x + 1\) and \(y = -x + 1\)
  3. C \(y = 2x + 1\) and \(y = 2x + 5\) Correct
  4. D \(y = 3x\) and \(y = x + 2\)

Why: \(y = 2x + 1\) and \(y = 2x + 5\) are parallel, so they never meet.