[discia] GCSE Maths › Algebra › Solving One Step Equations › Practise
  1. Either paper 1 mark

    \(9x = 54\)

    Work out the value of \(x\).

    Show answer
    \(6\)

    Both sides divide by 9:

    \(x = \frac{ 54 }{ 9 } = 6\)

  2. Either paper 1 mark

    \(x - 6 = -14\)

    Work out the value of \(x\).

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    \(-8\)

    Add 6 to both sides:

    \(x = -14 - (-6) = -8\)

  3. Either paper 1 mark

    Solve the equation \(\frac{x}{ 10 } = 5\)

    Show answer
    \(50\)

    \(x\) has been divided by 10, so multiply both sides by 10:

    \(x = 5 \times 10 = 50\)

  4. Either paper 2 marks

    Solve the equation \(-8x = 56\)

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    \(-7\)

    Divide both sides by \(-8\):

    \(x = \frac{ 56 }{ -8 } = -7\)

    Check: \(-8 \times -7 = 56\)

    • M1divides both sides by \(-8\)
    • A1\(x = -7\)
  5. Either paper 2 marks

    Solve the equation \(28 - x = 14\)

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    \(14\)

    Add \(x\) to both sides so that the \(x\) term is positive:

    \(28 = 14 + x\)

    Subtract 14 from both sides:

    \(x = 28 - (14) = 14\)

    • M1adds \(x\) to both sides: \(28 = 14 + x\)
    • A1\(x = 14\)
  6. Either paper 2 marks

    Solve \(x + 0.6 = 15.4\)

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    \(14.8\)

    Subtract 0.6 from both sides:

    \(x = 15.4 - 0.6 = 14.8\)

    • M1subtracts 0.6 from both sides: \(x = 15.4 - 0.6\)
    • A1\(x = 14.8\)
  7. Either paper 3 marks

    A shop sells paint brushes. Each paint brush costs \(c\) pounds.

    (a)

    Naomi buys 3 paint brushes. She pays £39 altogether.

    Write down an equation in terms of \(c\).

    (b)

    Solve your equation to find the cost of one paint brush.

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    (a) \(3c = 39\)
    (b) \(\text{£}13\)

    3 paint brushes at \(c\) pounds each cost \(3c\) pounds, and that came to £39:

    \(3c = 39\)

    Divide both sides by 3:

    \(c = \frac{ 39 }{ 3 } = 13\)

    One paint brush costs £13.

    • (b) M1divides both sides by 3: \(c = \frac{ 39 }{ 3 }\)
    • (b) A1cost of one paint brush £13
  8. Either paper 2 marks

    The solution of the equation

    \(\frac{x}{ a } = 3\)

    is \(x = 27\).

    Calculate the value of \(a\).

    Show answer
    \(9\)

    Substitute \(x = 27\).

    \(\frac{ 27 }{ a } = 3\)

    Dividing 27 by \(a\) gives 3, so \(a\) is 27 divided by 3.

    \(a = 27 \div 3 = 9\)

    • M1substitutes \(x = 27\) to get \(\frac{ 27 }{ a } = 3\), or 27 \(\div\) 3
    • A1\(a = 9\)
  9. Either paper 1 mark

    \(6x = 66\)

    Work out the value of \(x\).

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    \(11\)

    Both sides divide by 6:

    \(x = \frac{ 66 }{ 6 } = 11\)

  10. Either paper 1 mark

    Solve \(x - 6 = 2\)

    Show answer
    \(8\)

    Add 6 to both sides:

    \(x = 2 - (-6) = 8\)