Check sibling questions


Transcript

Ex 8.4, 3 Simplify. (iv) (๐‘Ž+๐‘) (๐‘โˆ’๐‘‘)+(๐‘Žโˆ’๐‘) (๐‘+๐‘‘)+2 (๐‘Ž๐‘+๐‘๐‘‘)Here, there are 3 expressions 1st expression = (๐‘Ž+๐‘) (๐‘โˆ’๐‘‘) 2nd expression = (๐‘Žโˆ’๐‘) (๐‘+๐‘‘) 3rd expression = 2 (๐‘Ž๐‘+๐‘๐‘‘) Solving 1st expression (๐’‚+๐’ƒ) (๐’„โˆ’๐’…) = ๐‘Ž(๐‘โˆ’๐‘‘)+๐‘ (๐‘โˆ’๐‘‘) = (๐‘Ž ร— ๐‘)โˆ’(๐‘Ž ร— ๐‘‘)+(๐‘ ร— ๐‘)โˆ’(๐‘ ร— ๐‘‘) = ๐’‚๐’„โˆ’๐’‚๐’…+๐’ƒ๐’„โˆ’๐’ƒ๐’… Solving 2nd expression (๐’‚โˆ’๐’ƒ) (๐’„+๐’…) = ๐‘Ž(๐‘+๐‘‘)โˆ’๐‘ (๐‘+๐‘‘) = (๐‘Žร—๐‘)+(๐‘Žร—๐‘‘)โˆ’(๐‘ร—๐‘)โˆ’(๐‘โˆ’๐‘‘) = ๐’‚๐’„+๐’‚๐’…โˆ’๐’ƒ๐’„โˆ’๐’ƒ๐’… Solving 3rd expression 2 (๐‘Ž๐‘+๐‘๐‘‘) = (2ร—๐‘Ž๐‘)+(2ร—๐‘๐‘‘) = ๐Ÿ๐’‚๐’„+๐Ÿ๐’ƒ๐’… Now, our expression becomes (๐’‚+๐’ƒ) (๐’„โˆ’๐’…)+(๐’‚โˆ’๐’ƒ) (๐’„+๐’…)+๐Ÿ (๐’‚๐’„+๐’ƒ๐’…) = ๐‘Ž๐‘โˆ’๐‘Ž๐‘‘+๐‘๐‘โˆ’๐‘๐‘‘+๐‘Ž๐‘+๐‘Ž๐‘‘โˆ’๐‘๐‘โˆ’๐‘๐‘‘+2๐‘Ž๐‘+2๐‘๐‘‘ = (๐‘Ž๐‘+๐‘Ž๐‘+2๐‘Ž๐‘)+(โˆ’๐‘Ž๐‘‘+๐‘Ž๐‘‘)+(๐‘๐‘โˆ’๐‘๐‘)+(โˆ’๐‘๐‘‘โˆ’๐‘๐‘‘+2๐‘๐‘‘) = (1+1+2)๐‘Ž๐‘+(โˆ’1+1)๐‘Ž๐‘‘+(1โˆ’1)๐‘๐‘+(โˆ’1โˆ’1+2)๐‘๐‘‘ = 4๐‘Ž๐‘+(0)๐‘Ž๐‘‘+(0)๐‘๐‘+(0)๐‘๐‘‘ = ๐Ÿ’๐’‚๐’„

  1. Chapter 8 Class 8 Algebraic Expressions and Identities
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo