OpenRevise

Exam questions · Maths · Further Algebra

Simultaneous Equations

  • 6 exam questions
  • 21 marks
  • 9 quick checks
  1. 1 Use [2 marks]

    Here are the graphs of the straight lines \(A\) and \(B\). Use the graphs to solve the simultaneous equations \(y = x + 2\) and \(x + y = 10\).

    Two straight lines, A and B, that cross at the point (4, 6).
    Show answerHide answer

    Model answer

    The lines cross at \((4, 6)\), so \(x = 4\) and \(y = 6\).

    Mark scheme

    • \(x = 4\) — B1
    • \(y = 6\) — B1
  2. 2 Solve [3 marks]

    Solve the simultaneous equations \(x + y = 11\) and \(x - y = 3\).

    Show answerHide answer

    Model answer

    Adding gives \(2x = 14\), so \(x = 7\). Then \(y = 11 - 7 = 4\). Check: \(7 - 4 = 3\).

    Mark scheme

    • Adds or eliminates \(y\) correctly — M1
    • \(x = 7\) — A1
    • \(y = 4\) — A1
  3. 3 Solve [3 marks]

    Solve the simultaneous equations \(3x + y = 14\) and \(2x - y = 6\).

    Show answerHide answer

    Model answer

    Adding gives \(5x = 20\), so \(x = 4\). Then \(12 + y = 14\), so \(y = 2\). Check: \(8 - 2 = 6\).

    Mark scheme

    • \(5x = 20\) — M1
    • \(x = 4\) — A1
    • \(y = 2\) — A1
  4. 4 Solve [4 marks]

    Solve the simultaneous equations \(2x + 3y = 12\) and \(4x - y = 10\).

    Show answerHide answer

    Model answer

    Multiply the second equation by 3: \(12x - 3y = 30\). Adding to the first gives \(14x = 42\), so \(x = 3\). Then \(6 + 3y = 12\), so \(y = 2\). Check: \(12 - 2 = 10\).

    Mark scheme

    • Multiplies to match the coefficients, such as \(12x - 3y = 30\) — M1
    • Adds or subtracts correctly, \(14x = 42\) — M1
    • \(x = 3\) — A1
    • \(y = 2\) — A1
  5. 5 Work out [4 marks]

    3 adults and 2 children go to the cinema. The total cost is \(\pounds 29\). 2 adults and 4 children go to the cinema. The total cost is \(\pounds 30\). Work out the cost of an adult ticket and the cost of a child ticket.

    Show answerHide answer

    Model answer

    \(3a + 2c = 29\) and \(2a + 4c = 30\), so \(a + 2c = 15\). Subtracting this from the first equation gives \(2a = 14\), so \(a = 7\). Then \(2c = 8\) and \(c = 4\). An adult ticket costs \(\pounds 7\) and a child ticket \(\pounds 4\).

    Mark scheme

    • Two correct equations — M1
    • A correct elimination step — M1
    • Adult ticket \(\pounds 7\) — A1
    • Child ticket \(\pounds 4\) — A1
  6. 6 Solve [5 marks]

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

    Show answerHide answer

    Model answer

    Setting them equal gives \(x^2 - 4 = 2x - 1\), so \(x^2 - 2x - 3 = 0\) and \((x - 3)(x + 1) = 0\). So \(x = 3\) or \(x = -1\). Then \(y = 5\) when \(x = 3\), and \(y = -3\) when \(x = -1\).

    Mark scheme

    • \(x^2 - 4 = 2x - 1\) — M1
    • \(x^2 - 2x - 3 = 0\) — M1
    • \(x = 3\) and \(x = -1\) — A1
    • One correct \(y\)-value — M1
    • \((3, 5)\) and \((-1, -3)\) — A1

Quick check

  1. 1

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

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

    C: \(x = 5\), \(y = 4\)

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

  2. 2

    When do you add the two equations in elimination?

    1. AWhen the terms in one letter are the same
    2. BWhen the terms in one letter are opposites
    3. CWhen there are no brackets
    4. DWhen one equation has a fraction
    Show answerHide answer

    B: When the terms in one letter are opposites

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

  3. 3

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

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

    A: \(x = 4\), \(y = 2\)

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

  4. 4

    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\(x = 2\), \(y = 3\)
    Show answerHide answer

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

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

  5. 5

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

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

    C: \(x = 3\), \(y = 7\)

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

  6. 6

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

    1. AOne
    2. BNone
    3. CTwo
    4. DInfinitely many
    Show answerHide answer

    B: None

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

  7. 7

    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\(\pounds 5\)
    2. B\(\pounds 3\)
    3. C\(\pounds 4\)
    4. D\(\pounds 6\)
    Show answerHide answer

    A: \(\pounds 5\)

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

  8. 8

    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 \((-2, 4)\)
    Show answerHide answer

    D: \((3, 9)\) and \((-2, 4)\)

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

  9. 9

    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 \((-4, -3)\)
    4. D\((0, 1)\) and \((-1, 0)\)
    Show answerHide answer

    C: \((3, 4)\) and \((-4, -3)\)

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