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Question
Add:
7a2bc, –3abc2, 3a2bc, 2abc2
Sum
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Solution
We have,
7a2bc + (–3abc2) + 3a2bc + 2abc2 = 7a2bc – 3abc2 + 3a2bc + 2abc2
= (7a2bc + 3a2bc) + (–3abc2 + 2abc2) ...[Grouping like terms]
= 10a2bc + (–abc2)
= 10a2bc – abc2
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Chapter 7: Algebraic Expression, Identities and Factorisation - Exercise [Page 230]
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