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Exam questions · Maths · Algebra

Solving Linear Equations and Inequalities

  • 7 exam questions
  • 22 marks
  • 10 quick checks
  1. 1 Solve [2 marks]

    Solve \(4x - 9 = 15\).

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    Model answer

    Add 9 to both sides to get \(4x = 24\), then divide by 4 to get \(x = 6\).

    Mark scheme

    • \(4x = 24\) or \(\dfrac{15 + 9}{4}\) — M1
    • \(x = 6\) — A1
  2. 2 Solve [3 marks]

    Solve \(7x - 4 = 3x + 20\).

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    Model answer

    Subtract \(3x\) from both sides to get \(4x - 4 = 20\). Add 4 to get \(4x = 24\). Divide by 4 to get \(x = 6\).

    Mark scheme

    • Correct first step, such as \(4x - 4 = 20\) or \(7x - 3x\) used correctly — M1
    • \(4x = 24\) — M1
    • \(x = 6\) — A1
  3. 3 Solve [3 marks]

    Solve \(5(x - 2) = 3x + 8\).

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    Model answer

    Expand: \(5x - 10 = 3x + 8\). Subtract \(3x\): \(2x - 10 = 8\). Add 10: \(2x = 18\). So \(x = 9\).

    Mark scheme

    • Expands the bracket correctly: \(5x - 10\) — M1
    • \(2x = 18\) — M1
    • \(x = 9\) — A1
  4. 4 Solve [3 marks]

    Solve \(\dfrac{2x + 1}{3} = \dfrac{x + 4}{2}\).

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    Model answer

    Multiply both sides by 6: \(2(2x + 1) = 3(x + 4)\). Expand: \(4x + 2 = 3x + 12\). Subtract \(3x\) and 2: \(x = 10\). Check: \(\dfrac{21}{3} = 7\) and \(\dfrac{14}{2} = 7\).

    Mark scheme

    • \(2(2x + 1) = 3(x + 4)\) or an equivalent equation without fractions — M1
    • \(4x + 2 = 3x + 12\) — M1
    • \(x = 10\) — A1
  5. 5 Work out [4 marks]

    The three angles of a triangle are \(x^\circ\), \((2x + 10)^\circ\) and \((3x - 10)^\circ\). Work out the size of the largest angle.

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    Model answer

    The angles in a triangle add up to \(180^\circ\), so \(x + 2x + 10 + 3x - 10 = 180\). That gives \(6x = 180\), so \(x = 30\). The angles are \(30^\circ\), \(70^\circ\) and \(80^\circ\), so the largest is \(80^\circ\).

    Mark scheme

    • Sets the sum of the angles equal to 180 — M1
    • \(6x = 180\) — M1
    • \(x = 30\) — A1
    • \(80^\circ\) — A1
  6. 6 Solve [4 marks]

    (a) Solve \(3x + 2 < 14\). (2 marks) (b) Write down all the integer values of \(x\) that satisfy \(-2 \leq x < 4\). (2 marks)

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    Model answer

    (a) Subtract 2 to get \(3x < 12\), then divide by 3 to get \(x < 4\). (b) \(-2\) is included and 4 is not, so the integers are \(-2, -1, 0, 1, 2, 3\).

    Mark scheme

    • (a) \(3x < 12\) — M1
    • (a) \(x < 4\) — A1
    • (b) At least 4 correct integers and no more than one incorrect — M1
    • (b) \(-2, -1, 0, 1, 2, 3\) — A1
  7. 7 Solve [3 marks]

    Solve \(3 - 2x \geq 11\).

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    Model answer

    Subtract 3 from both sides to get \(-2x \geq 8\). Divide both sides by \(-2\) and reverse the inequality sign, so \(x \leq -4\).

    Mark scheme

    • \(-2x \geq 8\) — M1
    • Divides by \(-2\) and reverses the sign, or shows \(x = -4\) as the boundary — M1
    • \(x \leq -4\) — A1

Quick check

  1. 1

    Three consecutive integers add up to 72. What is the smallest of them?

    1. A24
    2. B23
    3. C25
    4. D21
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    B: 23

    Let the numbers be \(n\), \(n + 1\) and \(n + 2\). Then \(3n + 3 = 72\), so \(n = 23\).

  2. 2

    Which inequality means "at most 20"?

    1. A\(x > 20\)
    2. B\(x \geq 20\)
    3. C\(x < 20\)
    4. D\(x \leq 20\)
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    D: \(x \leq 20\)

    "At most 20" means 20 or less, which is \(x \leq 20\).

  3. 3

    Solve \(2x + 7 = 19\).

    1. A\(x = 13\)
    2. B\(x = 12\)
    3. C\(x = 6\)
    4. D\(x = 5\)
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    C: \(x = 6\)

    Subtract 7 to get \(2x = 12\), then divide by 2.

  4. 4

    Solve \(5x - 3 = 2x + 9\).

    1. A\(x = 6\)
    2. B\(x = 12\)
    3. C\(x = 2\)
    4. D\(x = 4\)
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    D: \(x = 4\)

    Subtract \(2x\) to get \(3x - 3 = 9\), add 3 to get \(3x = 12\), and divide by 3.

  5. 5

    Solve \(3(x + 2) = 18\).

    1. A\(x = 16\)
    2. B\(x = 4\)
    3. C\(x = 5\)
    4. D\(x = 8\)
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    B: \(x = 4\)

    Expand to get \(3x + 6 = 18\), so \(3x = 12\) and \(x = 4\).

  6. 6

    Solve \(\dfrac{x}{5} - 2 = 3\).

    1. A\(x = 5\)
    2. B\(x = 15\)
    3. C\(x = 1\)
    4. D\(x = 25\)
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    D: \(x = 25\)

    Add 2 to get \(\dfrac{x}{5} = 5\), then multiply both sides by 5.

  7. 7

    Solve \(-2x > 6\).

    1. A\(x < 3\)
    2. B\(x > 3\)
    3. C\(x > -3\)
    4. D\(x < -3\)
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    D: \(x < -3\)

    Dividing by \(-2\) reverses the inequality sign, so \(x < -3\).

  8. 8

    Which list shows all the integers that satisfy \(-2 \leq x < 2\)?

    1. A\(-2, -1, 0, 1\)
    2. B\(-2, -1, 0, 1, 2\)
    3. C\(-1, 0, 1\)
    4. D\(-1, 0, 1, 2\)
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    A: \(-2, -1, 0, 1\)

    \(-2\) is included because of \(\leq\), but 2 is not included because of \(<\).

  9. 9

    Which inequality is shown by an open circle at 5 with an arrow pointing to the right?

    1. A\(x > 5\)
    2. B\(x < 5\)
    3. C\(x \geq 5\)
    4. D\(x \leq 5\)
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    A: \(x > 5\)

    An open circle means 5 is not included, and the arrow to the right means larger numbers, so \(x > 5\).

  10. 10

    I think of a number, double it and subtract 3. The answer is 11. Which equation represents this?

    1. A\(x - 6 = 11\)
    2. B\(2x - 3 = 11\)
    3. C\(2(x - 3) = 11\)
    4. D\(2x + 3 = 11\)
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    B: \(2x - 3 = 11\)

    Doubling gives \(2x\) and subtracting 3 gives \(2x - 3\), which equals 11.