Brackets
Quantities are enclosed within brackets to indicate that they
are to be treated as a single entity. If we wish to subtract,
say, 3a  2b from 4a  5b then we do this as follows.
Example 1
(a)
(4a  5b ) 
(3a  2b ) 
= 4a  5b  3a  ( 2b ) =
4a  5b  3a + 2b
= 4a  3a  5b + 2b
= a  3b

and similarly
(b)
(7 x + 5 y ) 
(2 x  3 y ) 
= 7 x + 5 y  2 x  (  3 y ) = 7 x +
5 y  2 x + 3 y
= 7 x  2 x + 5 y + 3 y
= 5 x + 8 y .

When there is more than one bracket it is usually necessary to
begin with the inside bracket and work outwards.
Example 2
Simplify the following expressions by removing the brackets.
(a) 3 a  c + (5 a  2 b  [3 a  c + 2 b]),
(b) {3 y  (2 x  3 y) + (3 x  2 y)} + 2 x.
Solution
(a) We have
3 a  c + (5 a
 2 b  [3 a  c + 2 b]) 
= 3 a  c + (5 a  2 b  3 a + c  2 b) =
3 a  c + (2 a  4 b + c)
= 3 a  c + 2 a  4 b + c
= 3 a + 2 a  4 b  c + c
= 5 a  4 b .

(b) Similarly we have
3 y  (2 x 
3 y) + (3 x  2 y) + 2 x 
= 3 y  2 x + 3 y + 3 x  2 y + 2 x =
3 y + 3 y  2 y + 3 x  2 x + 2 x
= 4 y + x + 2 x
= 4 y  x + 2 x
= x  4 y .

Exercise
Remove the brackets from each of the
following expressions and simplify as far as
possible.
(a) x  ( y  z ) + x + ( y  z )  ( z + x )
,
(b) 2 x  (5 y + [3 z  x] )  (5 x  [ y + z
] ),
(c) (3 /a) + b + (7 /a)  2 b,
(d) a  ( b + [ c  {a  b}] ) .
Solution
(a)
x  ( y  z ) + x + ( y  z )  ( z + x )
= x  y + z + x + y  z  z  x
= x + x  x  y + y + z  z  z
= x  z .
(b)
2 x  (5 y + [3 z  x ])  (5 x  [ y + z ])
= 2 x  (5 y + 3 z  x )  (5 x  y  z )
= 2 x  5 y  3 z + x  5 x + y + z
= 2 x + x  5 x  5 y + y  3 z + z
=  2 x  4 y  2 z .
(c)
(d)
a  ( b + [ c  { a  b } ])
= a  ( b + [ c  a + b ])
= a  ( b + c  a + b )
= a  (2 b + c  a )
= a  2 b  c + a
= 2 a  2 b  c .
