Hard GCSE Rearranging formulae Questions

Challenging, exam-style GCSE Rearranging formulae questions with worked solutions. Stretch yourself on the hardest changing the subject, square roots, circle formula, science formula problems.

changing the subjectsquare rootscircle formulascience formulasquaresPythagoras
GCSE Foundation34 questionsStep-by-step solutions
Question 1
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}
Question 2
6 markschallenging
The pendulum formula is T=2πlgT = 2\pi\sqrt{\frac{l}{g}}. Make gg the subject.
Show worked solution

Worked solution

  1. Write down the formula

    T=2πlgT = 2\pi\sqrt{\frac{l}{g}}

    Make gg the subject.

  2. Identify where g appears

    lg\sqrt{\frac{l}{g}}

    gg is inside the root, in the denominator.

  3. Plan the approach

    ÷2π, square, then free g\div 2\pi, \text{ square, then free } g

    A clear route through the problem.

  4. Divide both sides by 2π2\pi

    T2π=lg\frac{T}{2\pi} = \sqrt{\frac{l}{g}}

    This isolates the square root.

  5. Square both sides

    (T2π)2=lg\left(\frac{T}{2\pi}\right)^2 = \frac{l}{g}

    This removes the square root.

  6. Simplify the left side

    T24π2=lg\frac{T^2}{4\pi^2} = \frac{l}{g}

    Squaring the numerator and denominator.

  7. Multiply both sides by g

    gT24π2=l\frac{gT^2}{4\pi^2} = l

    This frees gg from the denominator.

  8. Multiply both sides by 4π24\pi^2

    gT2=4π2lgT^2 = 4\pi^2 l

    Clear the remaining denominator.

  9. Divide both sides by T2T^2

    g=4π2lT2g = \frac{4\pi^2 l}{T^2}

    This isolates gg.

  10. State the rearranged formula

    g=4π2lT2g = \frac{4\pi^2 l}{T^2}

    Now gg is the subject.

  11. Substitute test values

    l=10,g=10⇒T=2πl=10, g=10 \Rightarrow T = 2\pi

    Because 10/10=1\sqrt{10/10} = 1.

  12. Work out the check

    4π2(10)(2π)2=40π24π2=10\frac{4\pi^2 (10)}{(2\pi)^2} = \frac{40\pi^2}{4\pi^2} = 10

    The formula recovers g=10g=10.

  13. Avoid the common error

    g is in the denominator\text{g is in the denominator}

    After squaring, multiply up to get gg out of the bottom.

  14. Reflect on the method

    square before moving g\text{square before moving } g

    Remove the square root before touching gg.

  15. State the final answer

    g=4π2lT2g = \frac{4\pi^2 l}{T^2}

    The completed rearrangement.

Answer
g=4π2lT2g = \frac{4\pi^2 l}{T^2}
Question 3
6 markschallenging
Make xx the subject of a−xb=xc\frac{a - x}{b} = \frac{x}{c}.
Show worked solution

Worked solution

  1. Write down the formula

    a−xb=xc\frac{a - x}{b} = \frac{x}{c}

    Make xx the subject.

  2. Notice x appears twice

    left numerator and right\text{left numerator and right}

    xx is on both sides.

  3. Plan the approach

    cross-multiply, collect x, factorise\text{cross-multiply, collect } x, \text{ factorise}

    A clear route through the problem.

  4. Cross-multiply

    c(a−x)=bxc(a - x) = bx

    Multiply each numerator by the other denominator.

  5. Expand the left side

    ca−cx=bxca - cx = bx

    Multiply out the bracket.

  6. Add cx to both sides

    ca=bx+cxca = bx + cx

    Gather the xx terms on the right.

  7. Look at the right side

    bx+cxbx + cx

    Both terms contain xx.

  8. Factorise out x

    ca=x(b+c)ca = x(b + c)

    Take out the common factor xx.

  9. Divide both sides by (b + c)

    cab+c=x\frac{ca}{b + c} = x

    This isolates xx.

  10. State the rearranged formula

    x=cab+cx = \frac{ca}{b + c}

    Now xx is the subject.

  11. Substitute test values

    a=5,b=2,c=3a=5, b=2, c=3

    Choose easy values to check.

  12. Work out x

    x=3×52+3=155=3x = \frac{3 \times 5}{2 + 3} = \frac{15}{5} = 3

    The formula gives x=3x=3.

  13. Check the left side

    5−32=1\frac{5 - 3}{2} = 1

    Evaluate the original left side.

  14. Check the right side

    33=1\frac{3}{3} = 1

    Both sides equal 1.

  15. State the final answer

    x=cab+cx = \frac{ca}{b + c}

    The completed rearrangement.

Answer
x=cab+cx = \frac{ca}{b + c}
Question 4
5 markschallenging
Make xx the subject of x+a=b\sqrt{x + a} = b.
Show worked solution

Worked solution

  1. Write down the formula

    x+a=b\sqrt{x + a} = b

    Make xx the subject.

  2. Identify the operation on x

    x+a\sqrt{x + a}

    aa is added to xx, then square-rooted.

  3. Plan the approach

    square both sides, then −a\text{square both sides, then } - a

    Undo the root, then the addition.

  4. Square both sides

    x+a=b2x + a = b^2

    Squaring undoes the square root.

  5. Subtract a from both sides

    x=b2−ax = b^2 - a

    This isolates xx.

  6. State the rearranged formula

    x=b2−ax = b^2 - a

    Now xx is the subject.

  7. Note the condition

    b≥0b \ge 0

    A square root gives a non-negative result.

  8. Substitute test values

    a=1,x=3⇒4=2, so b=2a=1, x=3 \Rightarrow \sqrt{4} = 2, \text{ so } b = 2

    Choose easy values to check.

  9. Work out the check

    b2−a=4−1=3b^2 - a = 4 - 1 = 3

    The formula recovers x=3x=3.

  10. Try another value

    a=5,x=11⇒16=4a=5, x=11 \Rightarrow \sqrt{16} = 4

    A second check.

  11. Check it

    42−5=114^2 - 5 = 11

    The formula recovers x=11x=11.

  12. Avoid the common error

    x≠b−ax \ne b - a

    You must square the whole of bb, giving b2b^2.

  13. Reflect on the method

    square both sides fully\text{square both sides fully}

    Squaring is the inverse of the square root.

  14. Restate the working

    x+a=b2⇒x=b2−ax + a = b^2 \Rightarrow x = b^2 - a

    The key line of working.

  15. State the final answer

    x=b2−ax = b^2 - a

    The completed rearrangement.

Answer
x=b2−ax = b^2 - a
Question 5
6 markschallenging
The volume of a sphere is V=43πr3V = \frac{4}{3}\pi r^3. Make rr the subject.
Show worked solution

Worked solution

  1. Write down the formula

    V=43πr3V = \frac{4}{3}\pi r^3

    Make rr the subject.

  2. Identify where r appears

    r3r^3

    rr is cubed, multiplied by 43π\tfrac{4}{3}\pi.

  3. Plan the approach

    ×3,  ÷4π, cube root\times 3, \; \div 4\pi, \text{ cube root}

    Reverse each operation in turn.

  4. Multiply both sides by 3

    3V=4πr33V = 4\pi r^3

    This clears the 43\tfrac{4}{3} into 4π4\pi.

  5. Divide both sides by 4π4\pi

    3V4π=r3\frac{3V}{4\pi} = r^3

    This removes 4π4\pi.

  6. Write r3r^3 as the subject

    r3=3V4πr^3 = \frac{3V}{4\pi}

    Swap sides for clarity.

  7. Take the cube root

    r=3V4π3r = \sqrt[3]{\frac{3V}{4\pi}}

    Undo a cube with a cube root.

  8. State the rearranged formula

    r=3V4π3r = \sqrt[3]{\frac{3V}{4\pi}}

    Now rr is the subject.

  9. Note no plus-or-minus is needed

    cube roots keep the sign\text{cube roots keep the sign}

    Unlike square roots, a cube root has one real value.

  10. Substitute a test value

    r=3⇒V=43π(27)=36πr=3 \Rightarrow V = \frac{4}{3}\pi (27) = 36\pi

    Choose an easy radius to check.

  11. Work out the check

    3(36π)4π=108π4π=27\frac{3(36\pi)}{4\pi} = \frac{108\pi}{4\pi} = 27

    The inside of the root is 27.

  12. Cube root the check

    273=3\sqrt[3]{27} = 3

    The formula recovers r=3r=3.

  13. Avoid the common error

    r≠3V4πr \ne \sqrt{\frac{3V}{4\pi}}

    It is a cube, so use a cube root, not a square root.

  14. Reflect on the method

    undo powers last\text{undo powers last}

    Deal with ×\times and ÷\div before the power.

  15. State the final answer

    r=3V4π3r = \sqrt[3]{\frac{3V}{4\pi}}

    The completed rearrangement.

Answer
r=3V4π3r = \sqrt[3]{\frac{3V}{4\pi}}

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