Page 140 - ICSE Math 7

method. Alternatively, write the polynomials to be added one below the other such that the like terms
(with signs) are in the same vertical column and then add them column wise. This method is known
as column method. Alternatively,
Example 3: Add 2x + 4y – 3z, 5y + 2z and –6x – 9y + 3z. By column method:
Solution: (2x + 4y – 3z) + (5y + 2z) + (–6x – 9y + 3z) 2x + 4y – 3z

= (2 – 6)x + (4 + 5 – 9)y + (–3 + 2 + 3)z = –4x + 2z 5y + 2z

Subtraction – 6x – 9y + 3z
– 4x + 0y + 2z
Subtraction of like terms
The difference of two like terms is a like term whose numerical coefficient is the difference of the
numerical coefficients of the given terms. Example:
2
2
2
(a) 3x – 5x = (3 – 5)x = –2x (b) 18ab – 4ab = (18 – 4)ab = 14ab 2
Subtraction of unlike terms
Unlike terms cannot be subtracted to get a single term. So, we just write the unlike terms as the terms
2
of the algebraic expression with a subtraction sign. Example: if we subtract 5a from 9xy, it gives
2
9xy – 5a .
Subtraction of polynomials
Write the polynomial to be subtracted in a bracket with a minus sign preceding the bracket. Now,
remove the bracket by changing the sign of each term in the bracket. Then combine the like terms
and retain the unlike terms. Alternatively, polynomials can be subtracted using column method. Write
the polynomial to be subtracted below the other polynomial such that the like terms are in the same
vertical column. Change the sign of each term of the lower line and then combine them column wise.
2
2
2
2
Example 4: Subtract 9pq + 7p – 8q from 12p – 15q + 8p.
2
2
2
2
2
2
2
2
Solution: (12p – 15q + 8p) – (9pq + 7p – 8q ) = 12p – 15q + 8p – 9pq – 7p + 8q
(Changing the sign of each term in the bracket)
2
2
= (12 – 7)p + (–15 + 8)q + 8p – 9pq (Combining the like terms)
2
2
= 5p – 7q + 8p – 9pq
Alternatively,
By column method:
2
2
12p – 15q + 8p
2
2
7p – 8q + 9pq

– + – (Changing the sign)
2
2
5p – 7q + 8p – 9pq
3
2
3
2
Example 5: What should be added to 3a – 10ab + 11ab to get –2a + 5a b + 9ab – 5?
3
2
Solution: To get the required algebraic expression, subtract 3a – 10ab + 11ab from
2
3
–2a + 5a b + 9ab – 5.
3
2
– 2a + 5a b + 9ab – 5
3
3a + 11ab – 10ab 2
– – + (Changing the sign)
3

2
– 5a + 5a b – 2ab – 5 + 10ab 2
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