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Exam questions · Maths · Algebra

Substitution and Rearranging Formulae

  • 7 exam questions
  • 20 marks
  • 10 quick checks
  1. 1 Work out [2 marks]

    Work out the value of \(4x - 3y\) when \(x = -2\) and \(y = 5\).

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    Model answer

    \(4 \times (-2) - 3 \times 5 = -8 - 15 = -23\).

    Mark scheme

    • \(-8\) or \(-15\) seen — M1
    • \(-23\) — A1
  2. 2 Work out [2 marks]

    Work out the value of \((a - 3)^2\) when \(a = -1\).

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    Model answer

    \((-1 - 3)^2 = (-4)^2 = 16\).

    Mark scheme

    • \(-1 - 3 = -4\) — M1
    • 16 — A1
  3. 3 Work out [4 marks]

    The formula \(F = \dfrac{9C}{5} + 32\) converts a temperature in degrees Celsius, \(C\), to degrees Fahrenheit, \(F\). (a) Work out \(F\) when \(C = -10\). [2 marks] (b) Make \(C\) the subject of the formula. [2 marks]

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    Model answer

    (a) \(F = \dfrac{9 \times (-10)}{5} + 32 = -18 + 32 = 14\). (b) Subtract 32 to get \(F - 32 = \dfrac{9C}{5}\). Multiply by 5 to get \(5(F - 32) = 9C\). Divide by 9 to get \(C = \dfrac{5(F - 32)}{9}\).

    Mark scheme

    • (a) \(\dfrac{9 \times (-10)}{5}\) or \(-18\) — M1
    • (a) 14 — A1
    • (b) \(F - 32 = \dfrac{9C}{5}\) or \(5(F - 32) = 9C\) — M1
    • (b) \(C = \dfrac{5(F - 32)}{9}\) — A1
  4. 4 Make [2 marks]

    Make \(x\) the subject of \(y = 2(x + 3)\).

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    Model answer

    Divide both sides by 2 to get \(\dfrac{y}{2} = x + 3\), then subtract 3 to get \(x = \dfrac{y}{2} - 3\). Alternatively, \(x = \dfrac{y - 6}{2}\).

    Mark scheme

    • \(\dfrac{y}{2} = x + 3\) or \(y = 2x + 6\) — M1
    • \(x = \dfrac{y}{2} - 3\) or \(x = \dfrac{y - 6}{2}\) — A1
  5. 5 Make [3 marks]

    Make \(b\) the subject of \(a = \dfrac{3b + 1}{c}\).

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    Model answer

    Multiply both sides by \(c\) to get \(ac = 3b + 1\). Subtract 1 to get \(ac - 1 = 3b\). Divide by 3 to get \(b = \dfrac{ac - 1}{3}\).

    Mark scheme

    • \(ac = 3b + 1\) — M1
    • \(ac - 1 = 3b\) — M1
    • \(b = \dfrac{ac - 1}{3}\) — A1
  6. 6 Make [3 marks]

    Make \(y\) the subject of \(x^2 + y^2 = r^2\).

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    Model answer

    Subtract \(x^2\) from both sides to get \(y^2 = r^2 - x^2\). Take the square root of both sides to get \(y = \sqrt{r^2 - x^2}\).

    Mark scheme

    • \(y^2 = r^2 - x^2\) — M1
    • Takes the square root of the whole right-hand side — M1
    • \(y = \sqrt{r^2 - x^2}\) — A1
  7. 7 Make [4 marks]

    Make \(x\) the subject of \(y = \dfrac{3x + 1}{x - 2}\).

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    Model answer

    Multiply by \((x - 2)\): \(y(x - 2) = 3x + 1\). Expand: \(xy - 2y = 3x + 1\). Collect the \(x\) terms: \(xy - 3x = 1 + 2y\). Factorise: \(x(y - 3) = 1 + 2y\). So \(x = \dfrac{1 + 2y}{y - 3}\).

    Mark scheme

    • \(y(x - 2) = 3x + 1\) — M1
    • \(xy - 2y = 3x + 1\) — M1
    • \(xy - 3x = 1 + 2y\) and \(x(y - 3)\) — M1
    • \(x = \dfrac{1 + 2y}{y - 3}\) — A1

Quick check

  1. 1

    Make \(a\) the subject of \(P = 2(a + b)\).

    1. A\(a = P - 2b\)
    2. B\(a = 2P - b\)
    3. C\(a = \dfrac{P}{2} - b\)
    4. D\(a = \dfrac{P}{2} + b\)
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    C: \(a = \dfrac{P}{2} - b\)

    Divide both sides by 2 to get \(\dfrac{P}{2} = a + b\), then subtract \(b\).

  2. 2

    Use \(A = \tfrac{1}{2}(a + b)h\) to work out \(A\) when \(a = 7\), \(b = 11\) and \(h = 5\).

    1. A45
    2. B18
    3. C35
    4. D90
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    A: 45

    \(\tfrac{1}{2} \times 18 \times 5 = 45\).

  3. 3

    Work out the value of \(2a + b\) when \(a = 4\) and \(b = -3\).

    1. A\(-5\)
    2. B5
    3. C11
    4. D1
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    B: 5

    \(2 \times 4 + (-3) = 8 - 3 = 5\).

  4. 4

    Work out the value of \(x^2\) when \(x = -5\).

    1. A25
    2. B\(-10\)
    3. C\(-25\)
    4. D10
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    A: 25

    \((-5) \times (-5) = 25\), because a negative multiplied by a negative is positive.

  5. 5

    Work out the value of \(3x^2\) when \(x = -2\).

    1. A\(-36\)
    2. B36
    3. C12
    4. D\(-12\)
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    C: 12

    The power is done first: \((-2)^2 = 4\), then \(3 \times 4 = 12\).

  6. 6

    Make \(x\) the subject of \(y = x + 7\).

    1. A\(x = 7 - y\)
    2. B\(x = y + 7\)
    3. C\(x = 7y\)
    4. D\(x = y - 7\)
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    D: \(x = y - 7\)

    Subtract 7 from both sides to get \(x = y - 7\).

  7. 7

    Make \(x\) the subject of \(y = 4x - 1\).

    1. A\(x = \dfrac{y + 1}{4}\)
    2. B\(x = 4(y + 1)\)
    3. C\(x = \dfrac{y - 1}{4}\)
    4. D\(x = \dfrac{y}{4} + 4\)
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    A: \(x = \dfrac{y + 1}{4}\)

    Add 1 to both sides to get \(y + 1 = 4x\), then divide by 4.

  8. 8

    Make \(t\) the subject of \(v = u + at\).

    1. A\(t = v - u - a\)
    2. B\(t = \dfrac{v - u}{a}\)
    3. C\(t = \dfrac{v + u}{a}\)
    4. D\(t = a(v - u)\)
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    B: \(t = \dfrac{v - u}{a}\)

    Subtract \(u\) from both sides to get \(v - u = at\), then divide by \(a\).

  9. 9

    What is the value of \((2x)^2\) when \(x = 3\)?

    1. A12
    2. B36
    3. C6
    4. D18
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    B: 36

    \(2x = 6\), and \(6^2 = 36\). Note that \(2x^2\) would be \(2 \times 9 = 18\).

  10. 10

    Make \(r\) the subject of \(A = \pi r^2\).

    1. A\(r = \dfrac{A^2}{\pi}\)
    2. B\(r = \dfrac{\sqrt{A}}{\pi}\)
    3. C\(r = \dfrac{A}{\pi}\)
    4. D\(r = \sqrt{\dfrac{A}{\pi}}\)
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    D: \(r = \sqrt{\dfrac{A}{\pi}}\)

    Divide by \(\pi\) to get \(r^2 = \dfrac{A}{\pi}\), then take the square root of both sides.