Page 81 - Mathematics_F1

Therefore, the value of 3mn − 6mp + 2 n p is – 3916.
2
3
(b) m n − p n + 7nmp
2
3
2
2

2
2
3
2
= (2) × (4) − (− 5) × (4) + 7 × 4 × 2 × − 5
= 64 − 1600 − 280
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= − 1816.
Therefore, the value of m n – p n + 7nmp is − 1816.
2 3
2 2
Example 5.4
Tanzania Institute of Education
Simplify 8 m n − 6 a c + 14 − 4 m n + 3 a c − 8 and state:
2
2
2
2
(a) The coefficient of m n
2
(b) The number of terms
(c) The constant term
Solution
8 m n − 6 a c + 14 − 4 m n + 3 a c − 8
2
2
2
2

= 8 m n − 4 m n + 3 a c − 6 a c + 14 − 8
2
2
2
2

= 4 m n − 3 a c + 6.
2
2

Therefore, 8 m n − 6 a c + 14 − 4 m n + 3 a c − 8 = 4 m n − 3 a c + 6.
2
2
2
2
2
2
(a) The coefficient of m n is 4.
2
(b) There are 3 terms.
(c) The constant term is 6.
Exercise 5.1
1. Simplify each of the following algebraic expressions:
(a) n + n + n + n + n + k + k + k + x + x
(b) 3x + 4y – 7z + 3x – 7y + 2z
Mathematics Form One (c) 3 x + 7x – x
1 _
1 _
2
4

Then state:
(i) The number of terms
(i) The coefficient of each of the terms

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Mathematics form 1.indd 74 21/10/2025 10:50:33
Mathematics form 1.indd 74