Ex 12.2, 3 - Chapter 12 Class 7 Algebraic Expressions - Part 7

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Ex 12.2, 3 - Chapter 12 Class 7 Algebraic Expressions - Part 8

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  1. Chapter 12 Class 7 Algebraic Expressions
  2. Serial order wise

Transcript

Ex 12.2, 3 (Method 1) Subtract: (vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2 (3ab – 2a2 – 2b2) − (5a2 – 7ab + 5b2 ) = 3ab – 2a2 – 2b2 − 5a2 + 7ab − 5b2 = −2a2 − 5a2 − 2b2 − 5b2 + 3ab + 7ab = a2 (−2 − 5) + b2 (−2 − 5) + ab (3 + 7) = a2 (−7) + b2 (−7) + ab (10) = −7a2 − 7b2 + 10ab Ex 12.2, 3 (Method 2) Subtract: (vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2 We need to subtract ((5a2 – 7ab + 5b2 ) from (3ab – 2a2 – 2b2) 5a2 − 7ab + 5b2 from 3ab − 2a2 − 2b2 Answer is 10ab – 7a2 – 7b2

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.