mathlinE mathematics lesson
Transposition & Reasoning (suitable for years 8 to 10)
 Copyright 2003 mathlinE
Free download from www.thereisnosquare.com/mathline/
Transposition of formulas
A mathematical formula is a relationship among
quantities. Usually it is written in the format that
one quantity is expressed in terms of the other(s),
i.e. the quantity is the subject of the formula.
The followings are some of the formulas that you
have learnt.
4
(a) C  2r (b) A  r 2 (c) V  r 3
3
1
1
(d) A  4r 2 (e) A  a  b h (f) V  r 2 h
2
3
(g) A  2r r  s  (h) A  ss  a s  bs  c 
The process of changing the subject of a formula is
called transposition. It involves the transfer of
terms or factors from one side of the formula to the
other. A term remains a term and a factor remains
a factor after each transfer.
Example 1 Transpose formulas (a), (b), (c), (d)
and (f) to make r the subject in each case.
Solutions: C  2r , r 
A  r 2 , r 2 
A

, r
C
.
2
A
.

3V
4
3V
V  r 3 , r 3 
, r 3
.
3
4
4
A
A
1 A
A  4r 2 , r 2 
, r
, r
.
4
4
2 
Not to be copied by any means
Example 3 For the formula A  2r r  s  , express
s in terms of the variables A and r.
A
A
r.
Solution: A  2r r  s  , r  s 
, s
2r
2r
Example 4 Transpose formula (h) to make s  b
the subject.
Solution: A  ss  a s  bs  c  ,
ss  a s  bs  c   A 2 , s  b 
A2
.
ss  a s  c 
Mathematical reasoning involving inequalities
and transposition
Example 1
Show that if a  b , then a  b  0 .
Solution: a  b , transpose b to the left side to give
a b  0.
Example 2
Show that if
y
x
 1 , then
 1.
x
y
y
 1 , take the reciprocal of both sides,
x
this requires the reversal of the inequality sign,
x
 1.
y
Solution:
Example 3
Show that if x 
1
1
and y  , then
2
2
1
 1.
x y
3V
1
3V
V  r 2 h , r 2 
, r
.
3
h
h
Solution: Since x 
Example 2
subject.
1 1
 ,  x  y  1 . Take the reciprocal of
2 2
1
both sides to yield
 1 . (Note: the inequality
x y
sign is reversed).
x y 
Transpose formula (e) to make b the
Solution: A 
1
a  b h , a  b  2 A , b  2 A  a .
2
h
h
1
1
and y  , then
2
2
mathlinE mathematics lesson
Transposition & Reasoning (suitable for years 8 to 10)
 Copyright 2003 mathlinE
Example 4
Show that if
then x  y  1 .
Free download from www.thereisnosquare.com/mathline/
y
 1 and x  1 ,
x 1
y
 1 , since x  1 , x 1 is positive.
x 1
Transpose x 1 to the right side to give y  x  1
which is equivalent to x 1  y . Transpose 1 and y
to obtain x  y  1 .
Exercise C
1) Transpose each of the followings to make the
pro-numeral in [ ] the subject of the formula.
Solution:
[R]
(g)
I  PRT
1
A  bh
2
A  2r r  s 
1
V  r 2 h
3
V
R
I
1
s  u  v t
2
v  u  at
(h)
y  1 x
[x]
(a)
(b)
(c)
(d)
Example 5
Show that if
x  y 1.
y
 1 and x  1, then
x 1
(e)
(f)
Solution: Since x  1, x 1 is negative.
Transpose x 1 to the right and reverse the
inequality sign because x 1 is negative.
Now the inequality becomes y  x  1 or x 1  y .
Transpose 1 and y to obtain x  y  1 .
Example 6
Show that a 2  b 2  2ab always.
Solution: Since a perfect square is always positive,
2
 a  b   0 .
Expand the left side, a 2  2ab  b 2  0 . Transpose
2ab to obtain a 2  b 2  2ab .
Solution: Start from
left side,
 a
2

2 a
  0 . Expand the
b   b   0 , simplify to
a b
2
2
a  2 ab  b  0 . Transpose 2 ab to the right
side, a  b  2 ab and then 2 to the left to obtain
ab
 ab .
2
2
[h]
[s]
[h]
[I]
[t]
[a]
2)
Show that if a  b , the a  b  0 .
3)
Show that if x  1 and y  1 , then
4)
Show that if
y  x 1 .
5)
Example 7 Show that if a  0 and b  0 , then
ab
 ab .
2
Not to be copied by any means
Show that if
y  x  1.
1
1
 .
x y 2
y 1
 1 and x  0 , then
x
y 1
 1 and x  0 , then
x
6)
Show that if a and b are both positive or both
a b
 2 .
negative values, then
b a
7)
Prove that if a and b are integers of opposite
a b
  2 .
signs, then
b a
8)
Prove that x  4  4 x for x  0 .