GCSE Rearranging formulae Practice Questions

Free GCSE Rearranging formulae practice questions with full step-by-step worked solutions. Covers changing the subject, inverse operations, negative terms, science formula. Practise exam-style problems and check your method.

changing the subjectinverse operationsnegative termsscience formulaspeed formulaarea formula
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
Make xx the subject of y=x+8y = x + 8.
Show worked solution

Worked solution

  1. Write down the formula

    y=x+8y = x + 8

    Make xx the subject, starting from the given formula.

  2. Subtract 8 from both sides

    y−8=xy - 8 = x

    The inverse of +8+8 is −8-8; do it to both sides to keep the balance.

  3. State the answer

    x=y−8x = y - 8

    Now xx is the subject.

Answer
x=y−8x = y - 8
Question 2
1 markeasy
Make pp the subject of q=p+20q = p + 20.
Show worked solution

Worked solution

  1. Write down the formula

    q=p+20q = p + 20

    Make pp the subject.

  2. Subtract 20 from both sides

    q−20=pq - 20 = p

    The inverse of +20+20 is −20-20.

  3. State the answer

    p=q−20p = q - 20

    Now pp is the subject.

Answer
p=q−20p = q - 20
Question 3
2 marksintermediate
Make xx the subject of y=2x+15y = \frac{2x + 1}{5}.
Show worked solution

Worked solution

  1. Write down the formula

    y=2x+15y = \frac{2x + 1}{5}

    Make xx the subject.

  2. Plan the approach

    ×5,  −1,  ÷2\times 5, \; - 1, \; \div 2

    Clear the denominator, then undo the numerator.

  3. Multiply both sides by 5

    5y=2x+15y = 2x + 1

    This clears the denominator.

  4. Subtract 1, then divide by 2

    x=5y−12x = \frac{5y - 1}{2}

    Undo the +1+1 and the ×2\times 2.

  5. State the rearranged formula

    x=5y−12x = \frac{5y - 1}{2}

    Now xx is the subject.

  6. Check with numbers

    x=2⇒y=1;  5(1)−12=2x=2 \Rightarrow y=1; \; \frac{5(1) - 1}{2}=2

    The rearrangement reverses correctly.

Answer
x=5y−12x = \frac{5y - 1}{2}
Question 4
4 markshard
Make xx the subject of ax=b+cxax = b + cx.
Show worked solution

Worked solution

  1. Write down the formula

    ax=b+cxax = b + cx

    Make xx the subject.

  2. Notice x appears twice

    ax and cxax \text{ and } cx

    Collect the xx terms on one side.

  3. Plan the approach

    collect x, factorise, divide\text{collect } x, \text{ factorise, divide}

    Bring the xx terms together first.

  4. Subtract cx from both sides

    ax−cx=bax - cx = b

    All xx terms are now on the left.

  5. Factorise out x

    x(a−c)=bx(a - c) = b

    Take out the common factor xx.

  6. Divide both sides by (a - c)

    x=ba−cx = \frac{b}{a - c}

    This isolates xx.

  7. State the rearranged formula

    x=ba−cx = \frac{b}{a - c}

    Now xx is the subject.

  8. Substitute test values

    a=5,c=2,b=6⇒x=63=2a=5, c=2, b=6 \Rightarrow x = \frac{6}{3} = 2

    Choose easy values to check.

  9. Confirm both sides match

    5(2)=10 and 6+2(2)=105(2) = 10 \text{ and } 6 + 2(2) = 10

    Both sides equal 10.

  10. Avoid the common error

    factorise before dividing\text{factorise before dividing}

    Do not divide by aa while cxcx is still on the right.

Answer
x=ba−cx = \frac{b}{a - c}
Question 5
6 markschallenging
Make xx the subject of y=2x−3x+1y = \frac{2x - 3}{x + 1}.
Show worked solution

Worked solution

  1. Write down the formula

    y=2x−3x+1y = \frac{2x - 3}{x + 1}

    Make xx the subject.

  2. Notice x appears twice

    top and bottom\text{top and bottom}

    xx is in both numerator and denominator.

  3. Plan the approach

    clear the fraction, collect x, factorise\text{clear the fraction, collect } x, \text{ factorise}

    A clear route through the problem.

  4. Multiply both sides by (x + 1)

    y(x+1)=2x−3y(x + 1) = 2x - 3

    This clears the denominator.

  5. Expand the left side

    yx+y=2x−3yx + y = 2x - 3

    Multiply out the bracket.

  6. Subtract 2x from both sides

    yx−2x+y=−3yx - 2x + y = -3

    Bring the xx terms together.

  7. Subtract y from both sides

    yx−2x=−3−yyx - 2x = -3 - y

    Move the constant across.

  8. Look at the left side

    yx−2xyx - 2x

    Both terms contain xx.

  9. Factorise out x

    x(y−2)=−(y+3)x(y - 2) = -(y + 3)

    Take out the common factor xx.

  10. Divide both sides by (y - 2)

    x=−(y+3)y−2x = \frac{-(y + 3)}{y - 2}

    This isolates xx.

  11. Tidy the signs

    x=y+32−yx = \frac{y + 3}{2 - y}

    Multiply top and bottom by −1-1.

  12. State the rearranged formula

    x=y+32−yx = \frac{y + 3}{2 - y}

    Now xx is the subject.

  13. Substitute a test value

    x=1⇒y=−12x=1 \Rightarrow y = \frac{-1}{2}

    Since 2−31+1=−12\frac{2 - 3}{1 + 1} = -\tfrac{1}{2}.

  14. Check it reverses

    −12+32+12=2.52.5=1\frac{-\frac{1}{2} + 3}{2 + \frac{1}{2}} = \frac{2.5}{2.5} = 1

    The formula recovers x=1x=1.

  15. State the final answer

    x=y+32−yx = \frac{y + 3}{2 - y}

    The completed rearrangement.

Answer
x=y+32−yx = \frac{y + 3}{2 - y}

Unlock 65 more Rearranging formulae questions

Create a free account to work through every GCSE Rearranging formulae question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Rearranging formulae practice

Hard QuestionsWorked Solutions

Related Algebra topics

Algebraic fractionsEquation of a circleCoordinatesCubic and reciprocal graphsExponential and trig graphsExpanding binomials