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Add: 7a2bc, –3abc2, 3a2bc, 2abc2 - Mathematics

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Question

Add:

7a2bc, –3abc2, 3a2bc, 2abc2 

Sum

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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NCERT Exemplar Mathematics [English] Class 8
Chapter 7 Algebraic Expression, Identities and Factorisation
Exercise | Q 81. (i) | Page 230
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