Name of Lecturer: Mr. J.Agius
Course: HVAC1
Lesson 35
Chapter 6: Algebra

Transposition of Formulae
Formulae play a very important part in mathematics, science and engineering, as by them it
is possible to give a concise, accurate and generalised statement of laws.
From the previous section it can be seen that some formulae contain several symbols.
Usually, one of the symbols is isolated on one side of the equation and is called the subject
of the formula. Sometimes a symbol other than the subject is required to be calculated. In
such circumstances it is usually easiest to rearrange the formula to make a new subject
before numbers are substituted for symbols. This rearranging process is called transposing
the formula or simply transposition. Transposition should be treated carefully because an
error can result in a statement that is not true, with possibly serious effects in certain
branches of industry.
Basically the rules used for transposition of formulae are the same as the rules used for the
solution of simple equations.
Worked problems on transposition of formulae
In the following graded worked problems four types of transposition have been categorised
(Case A to Case D); however, it will become evident that many practical examples will
require the application of several of these categories in order to transpose a formula.
The objective, in transposition, is to obtain the required new subject on its own on the lefthand side (L.H.S.) of the equation. Therefore, as a first step in any transposition, the
equation, if necessary, should be changed around so that the side containing the required
new subject is on the left.
Case A. Symbols connected by a plus or minus sign.
Example 1
Transpose a = b + c + d to make c the subject.
Answer
To leave the letter c alone, we have to put the letters, b and d on the other side of the
equation. In order to do this we have to change their sign since no multiplication is involved
here.
So
Implies
6 Algebra
a
a–b–d
=b+c+d
=c
Page 1
Name of Lecturer: Mr. J.Agius
Course: HVAC1
Example 2
If x + y = a – b + c express b as the subject.
Answer
Same reason applies here (note there is no multiplication or division involved here)
So
x+y
=a–b+c
Implies
x+y–a–c
=–b
Putting b on this side to change sign gives
b
=a+c–x–y
Case B. Formulae involving products
Example 3
Transpose V = IR to make I the subject.
Answer
Here we only have multiplication signs. So since R is multiplied with I, in order to remove it
we have to put it in a division position.
i.e.
Implies
V
=IR
VR =I
Or simply
I
V
R
Example 4
If a body falls freely through a height h the velocity is given by the formula V2 = 2gh.
Express this formula with h as the subject.
Answer
Same reason as previous question
So
Implies
V2 = 2  g  h
V2  2  g = h
Or simply
V2
h
2g
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Page 2
Name of Lecturer: Mr. J.Agius
Course: HVAC1
Case C. Formulae containing fractions.
Example 5
M
, rearrange the equation to make M the subject.
V
If K 
Answer
In this case we have to throw V on the other side of the equation. Note that V is on the
division side, so, on the other side of the equation it becomes multiplication.
K
So
M
V
Implies
Or simply
K=MV
KV=M
or
M = KV
Example 6
If a 
F
rearrange the equation to make m the subject.
m
Answer
We apply the same method as in the previous question.
F
m
Rearranging , we obtain
a
So
or simply
=Fm
a
am =F
And putting a on the other side we obtain
m
Or simply
m
=Fa
F
a
Example 7
Transpose R 
l
a
to make (i) “a” and (ii) “” the subject.
Answer
i)
Applying same method as previous, we send “a” from the division side to the
multiplication side.
Implies
aR=l
Then leaving “a” as subject of the formula, we obtain a 
ii)
6 Algebra
Proceeding as usual we obtain

l
R
aR
l
Page 3
Name of Lecturer: Mr. J.Agius
Course: HVAC1
Example 8
If v = u +
ft
transpose the equation to make t the subject.
m
Answer
v=u+
ft
m
Look carefully at the equation. The first thing to arrange is to send the letter “u” on the other
side of the equation since it is alone and with a +ve sign in front of it.
So
v–u=
ft
m
Hence by proceeding as usual we obtain; (note. If you put v – u in a bracket it would be
simpler to solve.)
ft
(v – u) =
m
t
mv  u 
f
Case D. Formulae containing the required new subject in a bracket
Example 9
Transpose the equation s = ½ (u + v)t to make v the subject.
Answer
s = ½ (u + v)t
In this case we have to put ½ and “t” first on the other side of the equation since all two are
in a multiplication state with the bracket.
So
2s
 u  v 
t
Continue as usual
v
2s
u
t
or
v
2s  ut
t
Both of the above answers for v are correct, the latter being the neater version.
6 Algebra
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Name of Lecturer: Mr. J.Agius
Course: HVAC1
Transposition of Formulae
Q1
Make the symbol indicated in round brackets, the subject of each of the formulae shown in
problems 1-54 and express each in its simplest form.
1.
k+l=m–n+p
(m)
2.
a + 2b = c
(b)
3.
2abc = d
(c)
4.
c = 2r
(r)
5.
5 – 2bc = p + q
(c)
6.
y = mx + c
(m)
7.
pV = c
(p)
8.
A = rl
(l)
9.
I = PRT
(R)
10.
I
V
R
(R)
11.
P
RT
V
(V)
12.
S
a
1 r
(r)
(h)
14.
wL2
a
4d
(d)

13.
V
15.
9
F  C  32
5
(C)
16.
T
17.
Y
Fl
Ax
(l)
18.
a  b 3p

b
q
(q)
19.
A = 2r(r + h)
(h)
20.
A = B – 5.6C + 2
(C)
21.
W = aq(x – t)
(x)
22.
P
N a
4b
(N)
23.
8D  d 
B
L
(d)
24.
I
bd 3
12
(b)
25.
y
Ml 2
8EI
(I)
26.
R2 = R1(1 + t)
(t)
27.
1
1
1


R R1 R2
(R1)
28.
I
E e
Rr
(r)
6 Algebra
12
2
d h
 x  C 
C
(x)
Page 5
Name of Lecturer: Mr. J.Agius
Course: HVAC1
Q2
If P is the safe load which may be carried by a steel plate weekend by rivet holes then P =
f(b – nd)t. Make f, the safe working stress in the steel and then n, the number of rivet holes,
the subject of the formula.
Q3
The modulus of elasticity of a structural material (E) is given by the formula E 
Wl
. Make
Ax
x the subject of the formula.
Q4
The viscosity coefficient of a liquid () is given by the equation  
 Pr 4 t
. Make v the
8l
subject of the formula.
6 Algebra
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