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Exam questions · Maths · Algebra

Solving Linear Equations and Inequalities

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

    Solve \(3x - 8 = 19\).

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    Model answer

    Add 8 to both sides to get \(3x = 27\), then divide by 3 to get \(x = 9\).

    Mark scheme

    • \(3x = 27\) or \(\dfrac{19 + 8}{3}\) — M1
    • \(x = 9\) — A1
  2. 2 Solve [3 marks]

    Solve \(6x + 1 = 2x + 25\).

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    Model answer

    Subtract \(2x\) to get \(4x + 1 = 25\). Subtract 1 to get \(4x = 24\). So \(x = 6\).

    Mark scheme

    • \(4x + 1 = 25\) or \(6x - 2x\) used correctly — M1
    • \(4x = 24\) — M1
    • \(x = 6\) — A1
  3. 3 Solve [3 marks]

    Solve \(4(2x - 1) = 3(x + 7)\).

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    Model answer

    Expand both brackets: \(8x - 4 = 3x + 21\). Subtract \(3x\): \(5x - 4 = 21\). Add 4: \(5x = 25\). So \(x = 5\).

    Mark scheme

    • Expands both brackets correctly: \(8x - 4 = 3x + 21\) — M1
    • \(5x = 25\) — M1
    • \(x = 5\) — A1
  4. 4 Solve [3 marks]

    Solve \(\dfrac{2x}{3} - 1 = 7\).

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    Model answer

    Add 1 to both sides to get \(\dfrac{2x}{3} = 8\). Multiply by 3 to get \(2x = 24\). Divide by 2 to get \(x = 12\).

    Mark scheme

    • \(\dfrac{2x}{3} = 8\) — M1
    • \(2x = 24\) — M1
    • \(x = 12\) — A1
  5. 5 Work out [4 marks]

    A rectangle has length \((3x + 2)\) cm and width \((x + 4)\) cm. The perimeter of the rectangle is 44 cm. Work out the area of the rectangle.

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    Model answer

    Perimeter \(= 2(3x + 2 + x + 4) = 2(4x + 6) = 44\), so \(4x + 6 = 22\) and \(4x = 16\), giving \(x = 4\). The length is \(3 \times 4 + 2 = 14\) cm and the width is \(4 + 4 = 8\) cm. The area is \(14 \times 8 = 112\) cm\(^2\).

    Mark scheme

    • \(2(3x + 2 + x + 4) = 44\) or equivalent — M1
    • \(x = 4\) — A1
    • Length 14 and width 8 — M1
    • 112 (cm\(^2\)) — A1
  6. 6 Solve [3 marks]

    Solve \(7x + 4 < 3x - 12\).

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    Model answer

    Subtract \(3x\) from both sides to get \(4x + 4 < -12\). Subtract 4 to get \(4x < -16\). Divide by 4 to get \(x < -4\).

    Mark scheme

    • \(4x + 4 < -12\) or \(4x < -16\) — M1
    • \(4x < -16\) or \(x = -4\) found — M1
    • \(x < -4\) — A1
  7. 7 Write down [3 marks]

    \(n\) is an integer. Write down all the values of \(n\) such that \(-5 \leq 2n < 7\).

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    Model answer

    Divide every part by 2: \(-2.5 \leq n < 3.5\). The integers in this range are \(-2, -1, 0, 1, 2, 3\).

    Mark scheme

    • \(-2.5 \leq n < 3.5\) or equivalent — M1
    • At least 4 correct integers, or a list that is only wrong at one end — M1
    • \(-2, -1, 0, 1, 2, 3\) — 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.