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Exam questions · Maths · Further Algebra

Simultaneous Equations

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

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

    Two straight lines, A and B, that cross at the point (2, 4).
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    Model answer

    The lines cross at \((2, 4)\), so \(x = 2\) and \(y = 4\).

    Mark scheme

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

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

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    Model answer

    Adding gives \(3x = 15\), so \(x = 5\). Then \(y = 13 - 10 = 3\). Check: \(5 - 3 = 2\).

    Mark scheme

    • \(3x = 15\) or another correct elimination — M1
    • \(x = 5\) — A1
    • \(y = 3\) — A1
  3. 3 Solve [3 marks]

    Solve the simultaneous equations \(5x + 2y = 16\) and \(3x + 2y = 8\).

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    Model answer

    Subtracting gives \(2x = 8\), so \(x = 4\). Then \(20 + 2y = 16\), so \(y = -2\). Check: \(12 - 4 = 8\).

    Mark scheme

    • \(2x = 8\) — M1
    • \(x = 4\) — A1
    • \(y = -2\) — A1
  4. 4 Solve [4 marks]

    Solve the simultaneous equations \(3x + 2y = 18\) and \(2x - y = 5\).

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    Model answer

    From the second equation \(y = 2x - 5\). Substituting gives \(3x + 2(2x - 5) = 18\), so \(7x - 10 = 18\) and \(x = 4\). Then \(y = 3\).

    Mark scheme

    • \(y = 2x - 5\), or the equations made to match — M1
    • \(3x + 2(2x - 5) = 18\) — M1
    • \(x = 4\) — A1
    • \(y = 3\) — A1
  5. 5 Work out [4 marks]

    3 pens and 2 pencils cost 190p. 2 pens and 1 pencil cost 110p. Work out the cost of one pen and the cost of one pencil.

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    Model answer

    \(3p + 2q = 190\) and \(2p + q = 110\). Then \(q = 110 - 2p\), so \(3p + 220 - 4p = 190\) and \(p = 30\), \(q = 50\). A pen costs 30p and a pencil costs 50p.

    Mark scheme

    • Two correct equations — M1
    • A correct method to eliminate a letter — M1
    • Pen 30p — A1
    • Pencil 50p — A1
  6. 6 Solve [4 marks]

    Solve the simultaneous equations \(x^2 + y^2 = 20\) and \(y = 2x\).

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    Model answer

    Substituting gives \(x^2 + 4x^2 = 20\), so \(5x^2 = 20\), \(x^2 = 4\) and \(x = 2\) or \(x = -2\). Then \(y = 4\) or \(y = -4\). The solutions are \((2, 4)\) and \((-2, -4)\).

    Mark scheme

    • \(x^2 + (2x)^2 = 20\) — M1
    • \(x = 2\) and \(x = -2\) — A1
    • One correct \(y\)-value — M1
    • \((2, 4)\) and \((-2, -4)\) — 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\(x = 10\), \(y = -1\)
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    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
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    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\(x = 4\), \(y = 4\)
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    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\(x = 2\), \(y = 3\)
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    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\(x = 5\), \(y = 3\)
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    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
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    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\(\pounds 6\)
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    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 \((-2, 4)\)
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    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 \((-1, 0)\)
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    C: \((3, 4)\) and \((-4, -3)\)

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