40% off

Early-bird ends 15 Aug, 9am

Lock in £90

Rearranging Formulae to Change the Subject

Updated August 2025

Rearranging formulae is a core algebraic skill for the TMUA that involves isolating a specific variable as the subject. By applying inverse operations like addition, division, or squaring, students can transform complex equations. Mastering this technique allows for the direct calculation of unknown quantities in scientific and mathematical contexts.

Core concept

Changing the subject of a formula means using inverse operations to isolate a specific variable so that it stands alone on one side of the equals sign, represented as x=f(y,z,)x = f(y, z, \dots).

Changing the Subject of a Formula

Changing the subject of a formula is the process of expressing one variable in terms of all the other variables in the equation. To do this, you must systematically rearrange the formula using the standard rules of arithmetic and algebra to isolate the new subject. teaching this effectively requires following strict algebraic balance: any operation performed on one side of the equals sign must be performed on the other.

The Fundamental Rules for Rearranging

According to the official guide, there are five key operations you can use to transform a formula:

  1. Add or subtract the same quantity or term from each side.
  2. Multiply or divide both sides by the same non-zero quantity.
  3. Invert both sides of a formula if both sides are single fractions with non-zero denominators.
  4. Square both sides of a formula or raise both sides to the same non-zero power.
  5. Take the square root of both sides of a formula.

Changing the Subject Using Subtraction

In many cases, the variable you wish to isolate appears in multiple terms. The goal is to gather all terms containing that variable on one side of the equation while moving all other terms to the opposite side.

Worked Example: Making pp the Subject

Consider the formula: q+2p=p+4q6rq + 2p = p + 4q - 6r

  1. Subtract pp from both sides to move all terms involving pp to the left: q+2pp=4q6rq + 2p - p = 4q - 6r.
  2. Collect like terms: q+p=4q6rq + p = 4q - 6r.
  3. Subtract qq from both sides to isolate the pp term: p=4q6rqp = 4q - 6r - q.
  4. Simplify by collecting the remaining like terms: p=3q6rp = 3q - 6r.
  5. Factorise by taking out the common factor of 3 to provide a clean final form: p=3(q2r)p = 3(q - 2r).

Rearranging with Multiple Functions

When a formula includes brackets and fractions, it is usually most efficient to expand the brackets and clear the denominators early in the process.

Worked Example: Making xx the Subject

Consider the formula: y=3(x2)+x2y = 3(x - 2) + \frac{x}{2}

  1. Multiply out the bracket first: y=3x6+x2y = 3x - 6 + \frac{x}{2}.
  2. Multiply both sides of the entire formula by 2 to eliminate the fraction: 2y=6x12+x2y = 6x - 12 + x.
  3. Collect the xx terms on the right-hand side: 2y=7x122y = 7x - 12.
  4. Add 12 to both sides to further isolate the xx term: 2y+12=7x2y + 12 = 7x.
  5. Finally, divide by 7 to make xx the subject: x=2y+127x = \frac{2y + 12}{7}.

Variables in the Denominator

If the required subject is part of a denominator, you must use multiplication or inversion to bring it to the numerator. This often requires combining terms into a single fraction first.

Worked Example: Making uu the Subject

Consider the formula: 1v1u=1p\frac{1}{v} - \frac{1}{u} = \frac{1}{p}

  1. Rearrange the equation to isolate the term containing uu: 1v1p=1u\frac{1}{v} - \frac{1}{p} = \frac{1}{u}.
  2. To prepare for inversion, combine the left-hand side into a single fraction using a common denominator of vpvp: pvvp=1u\frac{p - v}{vp} = \frac{1}{u}.
  3. Invert both sides of the equation (turn the fractions upside down): vppv=u1=u\frac{vp}{p - v} = \frac{u}{1} = u.

Formulae Involving Square Roots

When the subject is trapped inside a square root, you must isolate the root and then square the entire side of the equation.

Worked Example: Making xx the Subject

Consider the formula: y=3x+54y = \sqrt{\frac{3x + 5}{4}}

  1. Square both sides of the formula to remove the radical: y2=3x+54y^2 = \frac{3x + 5}{4}.
  2. Multiply both sides by 4: 4y2=3x+54y^2 = 3x + 5.
  3. Subtract 5 from both sides: 4y25=3x4y^2 - 5 = 3x.
  4. Divide the whole expression by 3 to isolate xx: x=4y253x = \frac{4y^2 - 5}{3}.

Key takeaways

  • Always apply the same operation to both sides of the formula to keep the equation balanced.
  • If the subject appears in more than one term, group those terms on one side and factorise the variable out.
  • To remove a square root, square the entire side: to remove a squared term, take the square root of the entire side.
  • When the subject is in a denominator, rearrange to create single fractions on each side, then invert both sides.
Tips

For the TMUA, always check your rearranged formula by substituting simple numbers into both the original and the new versions to ensure they give the same result.

Cautions

A common mistake is forgetting to multiply or divide every single term in the equation when clearing fractions or isolating variables. Ensure that operations like squaring or dividing apply to the whole side, not just one term.

Insight

Rearranging formulae is effectively the process of finding the inverse of a composite function. This explains why we perform operations in the reverse order of the original expression.

Worked Examples

Example 1
The variables xx and yy are related by the equation:
x=52y3+112y3x = 5 - \frac{2y^3 + 1}{1-2y^3}

Which of the following is a rearrangement to make
yy the subject?
A:
y=x48x483y = \sqrt[3]{\frac{x-4}{8x-48}}
B:
y=x68x323y = \sqrt[3]{\frac{x-6}{8x-32}}
C:
y=x2x63y = \sqrt[3]{\frac{x-2}{x-6}}
D:
y=x3x43y = \sqrt[3]{\frac{x-3}{x-4}}
E:
y=x42x123y = \sqrt[3]{\frac{x-4}{2x-12}}
F:
y=x62x83y = \sqrt[3]{\frac{x-6}{2x-8}}

Practice Questions

Practice Question 1
A solid cylinder has radius rr cm and height hh cm.

A cube has side length
3r3r cm.

The total surface area of the cylinder is equal to four times the total surface area of the cube.

Which of the following is an expression for
hh in terms of rr?
A:(18π2)r(\frac{18}{\pi} - 2)r
B:(18π1)r(\frac{18}{\pi} - 1)r
C:
27rπ\frac{27r}{\pi}
D:(27π1)r(\frac{27}{\pi} - 1)r
E:(274π1)r(\frac{27}{4\pi} - 1)r
F:
108rπ\frac{108r}{\pi}
G:(108π1)r(\frac{108}{\pi} - 1)r
H:(108π12)r(\frac{108}{\pi} - \frac{1}{2})r

Frequently asked questions

Can I invert the terms individually if there is more than one fraction on a side?

No. You can only invert both sides when each side is a single fraction. For example, in 1v1p=1u\frac{1}{v} - \frac{1}{p} = \frac{1}{u}, you must combine the left-hand side into pvvp\frac{p - v}{vp} before inverting.

What happens if I take the square root of a side with multiple terms?

You must take the square root of the entire side. For example, if x2=a+bx^2 = a + b, then x=a+bx = \sqrt{a + b}, not a+b\sqrt{a} + \sqrt{b}.

How do I decide which operation to do first?

Generally, you should work in the reverse order of operations (BIDMAS) to 'undo' the functions surrounding the variable you wish to isolate.

Ready to test your knowledge?

You've reached the end of this section. Start a practice session to solidify your understanding and master this topic.