Question
Subtract: [2 MARKS]
(i) 3a−4b+5c from 4a−b+6c
(ii) 5a−3b+2c from a−4b−2c
Answer: Option A
:
Solution: 1 Mark each
(i) (4a−b+6c)−(3a−4b+5c)
= 4a−b+6c−3a+4b−5c
= 4a−3a−b+4b+6c−5c
= a+3b+c
(ii) (a−4b−2c)−(5a−3b+2c)
= a−4b−2c−5a+3b−2c
= a−5a−4b+3b−2c−2c
= −4a−b−4c
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:
Solution: 1 Mark each
(i) (4a−b+6c)−(3a−4b+5c)
= 4a−b+6c−3a+4b−5c
= 4a−3a−b+4b+6c−5c
= a+3b+c
(ii) (a−4b−2c)−(5a−3b+2c)
= a−4b−2c−5a+3b−2c
= a−5a−4b+3b−2c−2c
= −4a−b−4c
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