- Books Name
- CBSE Class 7 Mathematics Book
- Publication
- Param Publication
- Course
- CBSE Class 7
- Subject
- Mathmatics
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.