Addition or subtraction of like terms
The sum or difference of several like terms is another like term whose coefficient is the sum or difference of those like terms.
Illustration 1
Add the following : 3xy, 10xy and 5xy.
Solution
The sum of the numerical coefficients of the given like terms is 3 + 10 + 5 = 18.
Hence, 3xy + 10xy + 5xy = 18xy.

Illustration  2
Add the following : – 2p2q,  – 9p2q,  – 14p2q and  – 5p2q.
Solution
The sum of the numerical coefficients (without negative sign) is : 2 + 9 + 14 + 5 = 30
Hence, –2p2q – 9p2q – 14p2q – 5p2q = – 30p2q.

Illustration  3
Add the following : 3x + y + 4 and 4x + 3y + 7.
Solution
Horizontal Method
(3x + y + 4) + (4x + 3y + 7)
=     (3x + 4x) + (y + 3y) + (4 + 7)
=     (3 + 4)x + (1 + 3)y + (4 + 7)
=     7x + 4y + 11

Or Illustration 4
Add the following : 3x + 4y + 5z and 2x – 3y – 4z.
Solution
Horizontal Method
(3x + 4y + 5z) + (2x – 3y – 4z)
=     (3x + 2x) + (4y – 3y) + (5z – 4z)
=     (3 + 2)x + (4 – 3)y + (5 – 4)z
=     5x + y + z

Or NOTE : To subtract an expression from another, we change the sign (from' + ' to ' – ' and from' – ' to ' + ') of each term of the expression to be subtracted and then add the two expressions.

Illustration 5
Subtract :
(i) 3p from 7p        (ii) – 8x from 9x    (iii) – 3a from 7a    (iv) – 9b from – 2b
Solution
(i)     7p – 3p = (7 – 3)p = 4p
(ii)     9x – (–8x) = 9x + 8x =(9 + 8) x = 17x
(iii)     7a – (–3a) = 7a + 3a = (7 + 3)a = 10a
(iv)     – 2b – (– 9b) = – 2b + 9b = (– 2 + 9)b = 7b

Illustration 6
What should be subtracted from 2p3 – 4p2 + 5p – 6 to obtain p2 – 2p + 1 ?
Solution
Let X denote the required expression.
Then, (2p3 – 4p2 + 5p – 6) – X = p2 – 2p + 1
Hence, required expression
X = (2p3 – 4p2 + 5p – 6) – (p2 – 2p + 1)
X = 2p3 – 4p2 + 5p – 6 – p2 + 2p – 1
X = 2p3 – 4p2 – p2 + 5p + 2p – 7
X = 2p3 – 5p2 + 7p – 7

NOTE : When a grouping symbol preceded by, ‘–' sign is removed or inserted, then the sign of each term of the corresponding expression is changed (from '+' to '–' and from '–'  to  '+').

Illustration 7
Simplify : 2x – {4y – (3x – 5y)}.
Solution
We first remove the innermost grouping symbol ( ) and then braces { }.
Thus, we have
2x – {4y – (3x – 5y)}
= 2x – {4y – 3x + 5y}        [Removing ( )]
= 2x – {9y – 3x}
= 2x – 9y + 3x
= 2x + 3x – 9y
= 5x – 9y.

Illustration 8
Simplify and find the value of the following expression when a = 2 and b = 3 :
4(a2 + b2 + 2ab) – [4(a2 + b2 – 2ab) – {– b3 + 4(a – 3)}]
Solution
Proceeding outward from the innermost bracket,
4(a2 + b2 + 2ab) – [4(a2 + b2 – 2ab) – {– b3 + 4(a – 3)}]
= 4(a2 + b2 + 2ab) – [4(a2 + b2 – 2ab) – {– b3 + 4a – 12}]
= 4a2 + 4b2 + 8ab – [4a2 + 4b2 – 8ab + b3 – 4a + 12]
= 4a2 + 4b2 + 8ab – 4a2 – 4b2 + 8ab – b3 + 4a – 12
= 4a2 – 4a2 + 4b2 – 4b2 + 8ab + 8ab – b3 + 4a – 12
=  (4 – 4)a2 + (4 – 4)b2 + (8 + 8)ab – b3 + 4a – 12
= 16ab – b3 +  4a – 12
Thus va    lue of this expression for a = 2 and b = 3 is :
16 × 2 × 3 – (3)3 + 4 × 2 – 12  = 96 – 27 + 8 – 12 = 65.