Change of Subject of Formulae (Transposition of Formulae)

To change or transpose a formulae means to rearrange it so that a different letter or symbol becomes the subject. This process is exactly the same as solving equations.

Remember, what is done to the LHS is done to the RHS of the formulae at all times.

e.g.   I  = \( \frac{PRT}{100} \)

Make P the subject of the formula

Multiply both sides by 100

⇒ \( \scriptsize 100 \: \times \: I = \frac{PRT}{100} \scriptsize \; \times \; 100 \)

⇒ 100I  = PRT

Divide both sides by RT

⇒ \( \frac{100I}{RT} = \frac{PRT}{RT}\)

⇒ \( \frac{100I}{RT} = \frac{P\not{R}\not{T}}{\not{R}\not{T}}\)

P = \( \frac{100I}{RT} \)

Worked Example 8.2.1:

Make x the subject of formulae of Questions 8.2.1 a – d

a. y = x + 6
b.   x²  =  10 – y
c. \( \frac{x}{5} \scriptsize = a + 2x \)
d. y = \( \scriptsize 5 \left(\sqrt{x^2 \: + \: 2}\right) \)
e. Make T the subject of the formula: P = \( \frac{nRT}{V} \)
f. Make m the subject: w = \( \sqrt {\frac{K}{m}} \)
g. T = \(\scriptsize 2 \pi \sqrt {\normalsize \frac{L}{g}}\) make  g the subject

Solution

a. y = x + 6

Subtract 6 from both sides

y – 6 = x + 6 – 6

Move x (the subject) to the left-hand side

x + 6 – 6 = y – 6

x + 0 = y – 6

x = y – 6

b.   x²  =  10 – y

Take the square root of both sides

⇒ \( \scriptsize \sqrt {x^2} = \sqrt{10 \: – \: y} \)

\( \scriptsize \left(\sqrt {x^2} = x\right) \)

∴ x = \(\scriptsize \sqrt{10 \; – \; y} \)

c. \( \frac{x}{5} \scriptsize = a + 2x \)

Multiply both sides by 5

⇒ \(\scriptsize 5 \; \times \; \normalsize \frac{x}{5} \scriptsize = 5 \; \times \; a + 2x \)

⇒ \(\scriptsize x = 5(a + 2x) \)

⇒ \(\scriptsize x = 5a + 10x\)

Subtract 10x  from both sides

⇒ \(\scriptsize x \; – \; 10x = 5a + 10x \; – \; 10x\)

⇒ \(\scriptsize -9x = 5a \)

Divide both sides  by -9

⇒ \(\frac{-9x}{-9} = \frac{5a}{-9}\)

⇒ \(\frac{-\not{9}x}{-\not{9}} = \frac{5a}{-9}\)

\(\scriptsize x =\; – \; \normalsize \frac{5a}{9}\)

d. y = \( \scriptsize 5 \left(\sqrt{x^2 \: + \: 2}\right) \)

Find the square of both sides

y² = \( \scriptsize 5^2 \left(\sqrt{x^2 \: + \: 2}\right)^2 \)

y²  = 5² (x² + 2)

y²  =  5² x² + 5² × 2

y²  =  5² x² + 50

y²  =  25x² + 50

Subtract 50 from both sides

y² – 50  =  25x² + 50 – 50

y² – 50  =  25x² 

Divide both sides by 25

⇒ \(\frac{y^2 \; – \; 50}{25} = \frac{25x^2}{25}\)

Move x to the left-hand side

x² = \( \frac{y^2 \; – \; 50}{25}\)

⇒ \(\scriptsize x^2 = \normalsize \frac{y^2}{25} \; – \; \frac{50}{25}\)

i.e. \(\scriptsize x^2 = \normalsize \frac{y^2}{25} \; – \; \scriptsize 2\)

Take the square root of both sides

⇒ \(\scriptsize \sqrt{x^2} = \sqrt{\normalsize \frac{y^2}{25}} \; – \; \sqrt{\scriptsize 2}\)

\(\scriptsize x = \normalsize \frac{y}{5} \; – \; \sqrt{\scriptsize 2}\)

or

\(\scriptsize x = \normalsize \frac{y \; – \; 5\sqrt{2}}{5}\)

e. Make T the subject of the formula: P = \( \frac{nRT}{V} \)

Multiply both sides by V

⇒ \(\scriptsize P \: \times \: V = \normalsize \frac{nRT}{V} \scriptsize \; \times \; V \)

PV = nRT

Divide both sides by nR

⇒ \( \frac{PV}{nR} = \frac{nRT}{nR} \)

Move T to the left-hand side

⇒ \( \frac{nRT}{nR}= \frac{PV}{nR} \)

⇒ \( \frac{\not{n}\not{R}T}{\not{n}\not{R}}= \frac{PV}{nR} \)

T = \( \frac{PV}{nR} \)

f. Make m the subject: w = \( \sqrt {\frac{K}{m}} \)

Take the square of both sides

w² = \( \left( \sqrt{ \frac{K}{m} }\right)^2 \)

w² = \( \frac{K}{m}^{\frac{1}{2} \: \times \: 2} \)

w² = \(\frac{K}{m} \)

Cross multiply

w²m = K 

Divide both sides by w²

⇒ \( \frac{w^2m}{w^2} = \frac{K}{w^2} \)

⇒ \( \frac{\not{w^2}m}{\not{w^2}} = \frac{K}{w^2} \)

m = \( \frac{K}{w^2} \)

g. T = \(\scriptsize 2 \pi \sqrt {\normalsize \frac{L}{g}}:\) make  g the subject

This can also be written as: T² = \( \scriptsize2^2 \pi^2 \normalsize \frac{L}{g}^{\frac{1}{2}}\)

Take the square of both sides

⇒ T² = \( \scriptsize2^2 \pi^2 \normalsize \frac{L}{g}^{\left(\frac{1}{2}\scriptsize \: \times \: 2\right)}\)

⇒ T² = \( \scriptsize 4\pi^2 \normalsize \frac{L}{g}\)

Multiply both sides by g

⇒ \(\scriptsize g \: \times \: T^2 = 4 \pi^2 \normalsize \frac{L}{g}\scriptsize \; \times \; g\)

⇒ \(\scriptsize gT^2 = 4 \pi^2 L \)

Divide both sides by T²

⇒ \(\frac {g T^2}{T^2} = \frac{4 \pi^2 L}{T^2} \)

⇒ \(\frac {g\not{T^2}}{\not{T^2}} = \frac{4 \pi^2 L}{T^2} \)

g = \(\frac{4 \pi^2 L}{T^2} \)

h. f = \( \frac{1}{2 \pi LC}: \) make C the subject 

Multiply both sides by 2πLC

⇒ \(\scriptsize 2 \pi LC \: \times \: f = \normalsize \frac{1}{2 \pi LC} \scriptsize \; \times \; 2 \pi LC \)

⇒ \(\scriptsize 2 \pi LCf = 1 \)

Divide by 2πLf

⇒ \( \frac{2 \pi LCf}{2 \pi L f} = \frac{1}{2 \pi Lf} \)

⇒ \( \frac{\not{2} \pi \not{L}C\not{f}}{\not{2} \pi \not{L}\not{f}} = \frac{1}{2 \pi Lf} \)

∴ \( \scriptsize C = \normalsize \frac{1}{2 \pi Lf} \)