Change of Subject of a Formula: Steps and Worked Examples

By Engr. Henry Akpan

Rearranging Formulas

The formula v = u + at gives the final velocity of an accelerating car. But what if you know the velocities and the time, and you want the acceleration? You could substitute the numbers and solve every time, or you could rearrange the formula once, so that a stands alone on one side. This skill is called change of subject, and it is used constantly in physics, chemistry, engineering and finance. This tutorial covers change of subject of a formula as tested in WAEC, NECO and JAMB mathematics: the balancing principle, inverse operations, formulas with fractions, squares and roots, cases where the new subject appears more than once, and applications in science, with worked examples.

The Basic Idea

The subject of a formula is the variable that stands alone on one side, usually the left. In A = πr², the subject is A.

To change the subject, treat the formula like an equation and use inverse operations, doing the same thing to both sides, until the new subject is alone:

  • Addition is undone by subtraction, and subtraction by addition.
  • Multiplication is undone by division, and division by multiplication.
  • Squaring is undone by taking the square root, and a square root by squaring.

Undo the operations in the reverse order to the order in which they were applied to the new subject. It is like taking off socks and shoes: the last thing put on is the first thing taken off.

Simple Rearrangements

Worked example 1

Make a the subject of v = u + at.

  1. Subtract u from both sides: v − u = at.
  2. Divide both sides by t: a = (v − u) ÷ t.

A car accelerating from 10 m/s to 25 m/s in 5 s has a = (25 − 10) ÷ 5 = 3 m/s².

Worked example 2

Make C the subject of F = 9C/5 + 32, the formula for converting Celsius to Fahrenheit.

  1. Subtract 32: F − 32 = 9C/5.
  2. Multiply by 5: 5(F − 32) = 9C.
  3. Divide by 9: C = 5(F − 32) ÷ 9.

Check: F = 212 gives C = 5 × 180 ÷ 9 = 100, the boiling point of water.

Formulas with Fractions

Clear fractions early by multiplying both sides by the denominator.

Worked example 3

Make h the subject of V = ⅓πr²h, the volume of a cone.

Multiply by 3: 3V = πr²h. Divide by πr²: h = 3V ÷ (πr²).

Worked example 4: the lens formula

Make v the subject of 1/f = 1/u + 1/v.

  1. Subtract 1/u: 1/v = 1/f − 1/u.
  2. Combine the fractions: 1/v = (u − f) ÷ (uf).
  3. Take the reciprocal of both sides: v = uf ÷ (u − f).

For an object 30 cm from a lens of focal length 10 cm: v = 30 × 10 ÷ 20 = 15 cm.

Note that you cannot simply "flip" each term: 1/f = 1/u + 1/v does not mean f = u + v. Combine the fractions first.

Formulas with Squares and Roots

Worked example 5

Make r the subject of A = πr².

Divide by π: r² = A ÷ π. Take the square root: r = √(A ÷ π). Since r is a length, only the positive root is needed.

Worked example 6: the simple pendulum

Make l the subject of T = 2π√(l/g).

  1. Divide by 2π: T ÷ (2π) = √(l/g).
  2. Square both sides: T² ÷ (4π²) = l/g.
  3. Multiply by g: l = gT² ÷ (4π²).

A pendulum with a period of 2.0 s, where g = 9.8 m/s², has a length of 9.8 × 4 ÷ (4π²) ≈ 0.993 m, close to 1 m.

Worked example 7

Make r the subject of V = (4/3)πr³, the volume of a sphere.

Multiply by 3: 3V = 4πr³. Divide by 4π: r³ = 3V ÷ (4π). Take the cube root: r = ∛(3V ÷ (4π)).

Worked example 8

Make x the subject of s = √((x − a) ÷ b).

Square both sides: s² = (x − a) ÷ b. Multiply by b: bs² = x − a. Add a: x = a + bs².

When the New Subject Appears More Than Once

If the new subject appears in more than one term, the strategy is:

  1. Clear any fractions.
  2. Expand any brackets.
  3. Collect all the terms containing the new subject on one side, and everything else on the other.
  4. Factorise to take the new subject out as a common factor.
  5. Divide by the bracket.

Worked example 9

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

  1. Multiply by (x − 3): y(x − 3) = x + 2.
  2. Expand: xy − 3y = x + 2.
  3. Collect the x terms: xy − x = 2 + 3y.
  4. Factorise: x(y − 1) = 3y + 2.
  5. Divide: x = (3y + 2) ÷ (y − 1).

This is exactly how the inverse of a function is found.

Worked example 10

Make b the subject of a = (b + c) ÷ (b − c).

a(b − c) = b + c, so ab − ac = b + c. Collecting the b terms gives ab − b = ac + c, so b(a − 1) = c(a + 1), and b = c(a + 1) ÷ (a − 1).

Applications in Science

Worked example 11: energy

The total energy of a moving object at a height is E = ½mv² + mgh. Make v the subject, then find v when E = 100 J, m = 2 kg, g = 10 m/s² and h = 3 m.

  1. Subtract mgh: ½mv² = E − mgh.
  2. Multiply by 2 and divide by m: v² = 2(E − mgh) ÷ m.
  3. Square root: v = √(2(E − mgh) ÷ m).

v = √(2(100 − 60) ÷ 2) = √40 ≈ 6.32 m/s.

Worked example 12: simple interest

Make R the subject of I = PRT ÷ 100, and find the rate if ₦50,000 earns ₦12,000 interest in 3 years.

R = 100I ÷ (PT) = 100 × 12,000 ÷ (50,000 × 3) = 8% per annum.

Common Mistakes with Change of Subject

  • Doing an operation to only one term, such as dividing only part of one side. Every operation applies to the whole of each side.
  • Taking the square root term by term: √(a² + b²) is not a + b.
  • Flipping each term in a sum of fractions. Combine the fractions first, then take the reciprocal.
  • Leaving the new subject on both sides. If it still appears on the right, collect terms and factorise.
  • Undoing operations in the wrong order. Reverse the order in which they were applied.

Change of Subject: Practice Questions

  1. Make w the subject of P = 2(l + w).
  2. Make x the subject of y = mx + c.
  3. Make a the subject of s = ut + ½at².
  4. Make k the subject of T = 2π√(m/k).
  5. Make u the subject of v² = u² + 2as.
  6. Make t the subject of p = (3t − 1) ÷ (t + 2).
  7. Make R the subject of 1/R = 1/R₁ + 1/R₂, then find R when R₁ = 6 Ω and R₂ = 3 Ω.

Answers: 1. w = P/2 − l. 2. x = (y − c) ÷ m. 3. a = 2(s − ut) ÷ t². 4. k = 4π²m ÷ T². 5. u = √(v² − 2as). 6. p(t + 2) = 3t − 1, so pt − 3t = −1 − 2p, and t = (1 + 2p) ÷ (3 − p). 7. R = R₁R₂ ÷ (R₁ + R₂) = 18 ÷ 9 = 2 Ω.

Conclusion

Changing the subject of a formula means isolating a chosen variable by applying inverse operations to both sides, undoing the operations in reverse order. Clear fractions by multiplying, undo squares with square roots (and roots by squaring), and combine fractions before taking reciprocals. When the new subject appears more than once, collect those terms on one side and factorise. Rearranging formulas once saves work in physics, finance and every science, and it is the same technique used to find inverse functions.

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