Lesson 8, Topic 2
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# Change of Subject of Formulae (Transposition of Formulae)

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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

100I  = $$\frac{PRT}{100} \scriptsize \; \times \; 100$$

100I  = PRT

Divide both sides by RT

⇒ $$\frac{100I}{RT} = \frac{PRT}{RT}$$

P = $$\frac{100I}{RT}$$

Example 1

Make x the subject of formulae of questions a – d

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

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

⇒ $$\scriptsize \sqrt {x^2} = x$$

∴ 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

⇒ $$\frac{nRT}{nR}= \frac{PV}{nR}$$

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

Take the square of both sides

T2 = $$\scriptsize2^2 \pi^2 \normalsize \frac{L}{g}$$

T2 = $$\scriptsize 4\pi^2 \normalsize \frac{L}{g}$$

Multiply both sides by g

gT2 = $$\scriptsize 4 \pi^2 \normalsize \frac{L}{g}\scriptsize \; \times \; g$$

gT2 = $$\scriptsize 4 \pi^2 L$$

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

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

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

#### Responses

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