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Edexcel

Maths · Further Algebra

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Simultaneous Equations

Solving pairs of equations by elimination, substitution and graphs, including a linear and a quadratic equation.

  • 9 key terms
  • All boards

Learning Objectives

  1. 1Solve a pair of linear simultaneous equations by elimination and by substitution.
  2. 2Solve simultaneous equations graphically, by reading the point where two lines cross.
  3. 3Form simultaneous equations from a problem and solve them.
  4. 4Solve a pair where one equation is quadratic or a circle (Higher tier).

Two equations, one answer

Simultaneous equations are two equations that must both be true at the same time. With two unknowns, such as \(x\) and \(y\), one equation is not enough, but two give a single pair of values. They appear on almost every Paper 1, and a typical question gives you the numbers so that nothing is awkward to work out. The marks are for a clear method, and the final check, putting your answers in the other equation, protects you from silly errors.

Elimination

Add or subtract the equations to get rid of one letter. It is the quickest method when the equations are in the same form.

  • Make a pair of terms match

    If the \(y\)-terms are equal, subtract the equations. If they are opposites, add them.

  • Multiply if you need to

    Multiply one or both equations so the numbers in front of one letter match.

  • Solve for one letter, then the other

    Substitute your first answer back into one of the original equations.

  • Always check

    Put both answers in the equation you did not use.

Elimination by subtracting

Solve \(3x + 2y = 16\) and \(x + 2y = 8\).

Show the solutionHide the solution
  1. 1 Match the y terms Both equations have \(2y\), so subtract the second from the first.
  2. 2 Subtract \((3x - x) + (2y - 2y) = 16 - 8\), so \(2x = 8\) and \(x = 4\).
  3. 3 Find y Put \(x = 4\) into \(x + 2y = 8\): \(4 + 2y = 8\), so \(y = 2\).
  4. 4 Check \(3 \times 4 + 2 \times 2 = 16\). Correct.

Answer\(x = 4\), \(y = 2\)

Elimination after multiplying

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

Show the solutionHide the solution
  1. 1 Match the y terms Multiply the second equation by 3: \(9x - 3y = 9\).
  2. 2 Add \(2x + 3y + 9x - 3y = 13 + 9\), so \(11x = 22\) and \(x = 2\).
  3. 3 Find y \(3 \times 2 - y = 3\), so \(y = 3\).
  4. 4 Check \(2 \times 2 + 3 \times 3 = 13\). Correct.

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

Substitution

When one equation already gives a letter on its own, replace it in the other equation.

  • Spot the easy equation

    \(y = 2x + 1\) has \(y\) on its own.

  • Replace

    In \(3x + y = 16\), write \(3x + (2x + 1) = 16\).

  • Solve

    \(5x + 1 = 16\), so \(x = 3\). Then \(y = 2 \times 3 + 1 = 7\).

  • Use brackets

    When you substitute an expression, put it in brackets to avoid sign mistakes.

Problems with simultaneous equations

In a word problem, define the letters, write one equation for each statement, and solve.

  • Define the letters

    "Let \(a\) be the cost of an adult ticket and \(c\) the cost of a child ticket."

  • Write two equations

    "2 adult and 3 child tickets cost £19" gives \(2a + 3c = 19\), and "3 adult and 1 child ticket cost £18" gives \(3a + c = 18\).

  • Solve them

    From the second, \(c = 18 - 3a\). Then \(2a + 3(18 - 3a) = 19\), so \(2a + 54 - 9a = 19\), giving \(-7a = -35\) and \(a = 5\), \(c = 3\).

  • Answer in words

    An adult ticket costs £5 and a child ticket £3.

A linear and a quadratic equation (Higher tier)

Substitute the linear equation into the quadratic one. You will get a quadratic, and so two pairs of solutions.

  • Example

    Solve \(y = x^2\) and \(y = x + 6\). Set them equal: \(x^2 = x + 6\), so \(x^2 - x - 6 = 0\) and \((x - 3)(x + 2) = 0\).

  • Find both y-values

    \(x = 3\) gives \(y = 9\), and \(x = -2\) gives \(y = 4\). The solutions are \((3, 9)\) and \((-2, 4)\).

  • A circle

    For \(x^2 + y^2 = 25\) and \(y = x + 1\): \(x^2 + (x + 1)^2 = 25\) gives \(2x^2 + 2x - 24 = 0\), so \(x^2 + x - 12 = 0\) and \(x = 3\) or \(x = -4\), giving \((3, 4)\) and \((-4, -3)\).

  • Pair the answers

    Each \(x\) goes with its own \(y\), so use the linear equation to find them.

Test yourself

  1. 1

    When do you add the equations in elimination?

    Show answerHide answer

    When the terms in one letter are opposites, such as \(+3y\) and \(-3y\).

  2. 2

    What does the crossing point of two lines mean?

    Show answerHide answer

    The solution of the pair of equations.

  3. 3

    What is the first step in solving \(y = 2x + 1\) and \(3x + y = 16\)?

    Show answerHide answer

    Substitute \(2x + 1\) for \(y\) in the second equation.

  4. 4

    How do you check the solution?

    Show answerHide answer

    Put both values into the equation you did not use.

  5. 5

    How many solution pairs can a line and a parabola have (Higher tier)?

    Show answerHide answer

    Up to two.

Exam technique: simultaneous equations

The method is worth most of the marks, so show it clearly.

  • Label the equations

    Call them (1) and (2) so you can say what you are adding or subtracting.

  • Subtract with care

    Watch the signs, especially when subtracting a negative term.

  • Find both letters

    Stopping after finding \(x\) loses the marks for \(y\).

  • Check your answer

    Use the equation you did not substitute into.

Summary and exam focus

  • Simultaneous equations are solved by elimination, substitution or by drawing both lines.
  • Make the number in front of one letter match, then add or subtract to eliminate it.
  • Find both \(x\) and \(y\), and check in the other equation.
  • With a quadratic or a circle (Higher tier), substitute and solve a quadratic, giving two solution pairs.

Exam focus

Solve the simultaneous equations \(x + y = 9\) and \(x - y = 1\). (3 marks) (3 marks)

Adding the two equations eliminates \(y\), giving \(2x = 10\) and \(x = 5\). Then \(y = 4\). Check with \(5 - 4 = 1\). Writing the check in your answer shows you have verified it.

Key terms

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

Simultaneous equations
Equations that are true for the same values of the unknowns.
Elimination
Adding or subtracting equations to remove one unknown.
Substitution
Replacing a letter in one equation with its value or expression from another.
Unknown
A letter standing for a number that has to be found.
Solution pair
The values of \(x\) and \(y\) that make both equations true.
Intersection
The point where two lines or curves cross.
Linear equation
An equation whose graph is a straight line.
Coefficient
The number in front of a letter.
Equation (1) and (2)
Labels used to refer to each equation when you add or subtract them.

Questions and answers

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

1. Exam question Use 2 marks Easier

The graph shows the straight lines \(A\) and \(B\). Use the graph to solve the simultaneous equations \(y = x - 1\) and \(2x + y = 11\).

Two straight lines, A and B, that cross at the point (4, 3).

Mark scheme — 2 marks available

  • \(x = 4\) — B1
  • \(y = 3\) — B1

Model answer

The lines cross at \((4, 3)\), so \(x = 4\) and \(y = 3\).

2. Exam question Solve 3 marks Core

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

Mark scheme — 3 marks available

  • \(3y = 9\) or another correct elimination — M1
  • \(y = 3\) — A1
  • \(x = 5\) — A1

Model answer

Subtracting gives \(3y = 9\), so \(y = 3\). Then \(x = 5\). Check: \(5 - 3 = 2\).

3. Exam question Solve 3 marks Core

Solve the simultaneous equations \(4x + y = 19\) and \(3x - y = 9\).

Mark scheme — 3 marks available

  • \(7x = 28\) — M1
  • \(x = 4\) — A1
  • \(y = 3\) — A1

Model answer

Adding gives \(7x = 28\), so \(x = 4\). Then \(16 + y = 19\), so \(y = 3\). Check: \(12 - 3 = 9\).

4. Exam question Solve 4 marks Stretch

Solve the simultaneous equations \(4x + 3y = 17\) and \(2x - y = 1\).

Mark scheme — 4 marks available

  • \(y = 2x - 1\), or the equations made to match — M1
  • \(4x + 3(2x - 1) = 17\) — M1
  • \(x = 2\) — A1
  • \(y = 3\) — A1

Model answer

From the second equation \(y = 2x - 1\). Substituting gives \(4x + 3(2x - 1) = 17\), so \(10x - 3 = 17\) and \(x = 2\). Then \(y = 3\).

5. Exam question Calculate 4 marks Stretch

2 large boxes and 3 small boxes have a total mass of 29 kg. 1 large box and 2 small boxes have a total mass of 17 kg. Calculate the mass of a large box and the mass of a small box.

Mark scheme — 4 marks available

  • Two correct equations — M1
  • A correct elimination step — M1
  • Small box 5 kg — A1
  • Large box 7 kg — A1

Model answer

\(2l + 3s = 29\) and \(l + 2s = 17\). Doubling the second gives \(2l + 4s = 34\). Subtracting the first gives \(s = 5\). Then \(l = 17 - 10 = 7\). A large box is 7 kg and a small box is 5 kg.

6. Exam question Solve 4 marks Stretch

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

Mark scheme — 4 marks available

  • \(x^2 = 3x + 4\) — M1
  • \((x - 4)(x + 1)\) — M1
  • \(x = 4\) and \(x = -1\) — A1
  • \((4, 16)\) and \((-1, 1)\) — A1

Model answer

Setting them equal gives \(x^2 = 3x + 4\), so \(x^2 - 3x - 4 = 0\) and \((x - 4)(x + 1) = 0\). So \(x = 4\) or \(x = -1\), and \(y = 16\) or \(y = 1\). The solutions are \((4, 16)\) and \((-1, 1)\).

7. Multiple choice 1 mark Easier

Solve \(x + y = 9\) and \(x - y = 1\).

  1. A \(x = 4\), \(y = 5\)
  2. B \(x = 8\), \(y = 1\)
  3. C \(x = 5\), \(y = 4\) Correct
  4. D \(x = 10\), \(y = -1\)

Why: Adding gives \(2x = 10\), so \(x = 5\) and \(y = 4\).

8. Multiple choice 1 mark Easier

When do you add the two equations in elimination?

  1. A When the terms in one letter are the same
  2. B When the terms in one letter are opposites Correct
  3. C When there are no brackets
  4. D When one equation has a fraction

Why: Opposites such as \(+3y\) and \(-3y\) cancel when added.

9. Multiple choice 1 mark Core

Solve \(3x + 2y = 16\) and \(x + 2y = 8\).

  1. A \(x = 4\), \(y = 2\) Correct
  2. B \(x = 2\), \(y = 4\)
  3. C \(x = 8\), \(y = 0\)
  4. D \(x = 4\), \(y = 4\)

Why: Subtract: \(2x = 8\), so \(x = 4\). Then \(4 + 2y = 8\) gives \(y = 2\).

10. Multiple choice 1 mark Core

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

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

Why: Multiply the second equation by 3 and add: \(11x = 22\), so \(x = 2\), \(y = 3\).

11. Multiple choice 1 mark Core

Solve \(y = 2x + 1\) and \(3x + y = 16\).

  1. A \(x = 3\), \(y = 5\)
  2. B \(x = 7\), \(y = 3\)
  3. C \(x = 3\), \(y = 7\) Correct
  4. D \(x = 5\), \(y = 3\)

Why: \(3x + 2x + 1 = 16\), so \(x = 3\) and \(y = 7\).

12. Multiple choice 1 mark Core

Two straight lines are parallel. How many solutions do their simultaneous equations have?

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

Why: Parallel lines never meet, so there is no point on both.

13. Multiple choice 1 mark Stretch

2 adult and 3 child tickets cost \(\pounds 19\). 3 adult and 1 child ticket cost \(\pounds 18\). What does an adult ticket cost?

  1. A \(\pounds 5\) Correct
  2. B \(\pounds 3\)
  3. C \(\pounds 4\)
  4. D \(\pounds 6\)

Why: \(2a + 3c = 19\) and \(3a + c = 18\) give \(a = 5\), \(c = 3\).

14. Multiple choice 1 mark Stretch

Solve \(y = x^2\) and \(y = x + 6\).

  1. A \((3, 9)\) only
  2. B \((2, 4)\) and \((-3, 9)\)
  3. C \((3, 9)\) and \((-3, 9)\)
  4. D \((3, 9)\) and \((-2, 4)\) Correct

Why: \(x^2 = x + 6\) gives \((x - 3)(x + 2) = 0\).

15. Multiple choice 1 mark Stretch

Solve \(x^2 + y^2 = 25\) and \(y = x + 1\).

  1. A \((4, 3)\) and \((-3, -4)\)
  2. B \((3, 4)\) only
  3. C \((3, 4)\) and \((-4, -3)\) Correct
  4. D \((0, 1)\) and \((-1, 0)\)

Why: \(x^2 + (x + 1)^2 = 25\) gives \(x^2 + x - 12 = 0\), so \(x = 3\) or \(x = -4\).