Page 84 - Mathematics_F1

Example 5.7

Re-write –5a(m + 2n – 2) without brackets.
Solution
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–5a(m + 2n – 2) = (–5a)m + (–5a × 2n) + (–5a × –2) Tanzania Institute of Education
= –5am – 10an + 10a.
Therefore, –5a(m + 2n – 2) = –5am – 10an + 10a.

Example 5.8

Divide 4ax + 6ay – 10az by 2a.
Solution
4ax +  6ay − 10az 
az ÷
(4ax+ 6ay− 10 ) 2a =  
 2a 
 4ax 6ay 10a 
=  + − 
 2a 2a 2a 
= 2x + 3y – 5z.
Therefore, 4ax + 6ay – 10az divided by 2a is 2x + 3y – 5z.

Exercise 5.2

Multiply each of the following algebraic expressions:
1. 5x + 6m + 4y by 5 6. 3x – 6y – 2z by a
2. 3m + 7n – r by 2 7. 5x + 3y – 3z by 3a
3. 2p – 5q + 3 by 2 8. 7d + 10e – 5f by – 3a
4. 4m – 8n – 5p by – 4 9. 6x – 2y – 7z by 2ab

5. a – 7b + 9c by r 10. –13x + y – 6z by – 2m

Simplify each of the following algebraic expressions where possible:
11. 5x (2a + b + 3c) 16. –5hk (2r – 3d – 1)
Mathematics Form One
12. 3m (4x + 3y – 1) 17. 4uv (3a – 7)
13. –2n (5p – 8q – 3) 18. a(6m – n)
14. 9x (–2a – 3b – c) 19. (3x – b) (–2a)
15. 8mn (a – 2b + 5c) 20. 0.5m (3a – 2b)

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25/09/2025 15:01:02
Mathematics form 1.indd 77
Mathematics form 1.indd 77 25/09/2025 15:01:02