Ex 9.4, 3 (vii) - Simplify (1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y

Ex 9.4, 3 (vii) - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Ex 9.4, 3 (vii) - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3

  1. Chapter 9 Class 8 Algebraic Expressions and Identities
  2. Serial order wise

Transcript

Ex 9.4, 3 Simplify. (vii) (1.5๐‘ฅโˆ’4๐‘ฆ)(1.5๐‘ฅ+4๐‘ฆ+3)โˆ’4.5๐‘ฅ+12๐‘ฆ Here, there are 2 expressions: First expression = (1.5๐‘ฅโˆ’4๐‘ฆ)(1.5๐‘ฅ+4๐‘ฆ+3) Second expression = 4.5๐‘ฅ+12๐‘ฆ Solving First expression (1.5๐‘ฅโˆ’4๐‘ฆ)(1.5๐‘ฅ+4๐‘ฆ+3) = 1.5๐‘ฅ(1.5๐‘ฅ+4๐‘ฆ+3)โˆ’4๐‘ฆ(1.5๐‘ฅ+4๐‘ฆ+3) = (1.5๐‘ฅร—1.5๐‘ฅ)+(1.5๐‘ฅร—4๐‘ฆ)+(1.5๐‘ฅร—3)โˆ’(4๐‘ฆร—1.5๐‘ฅ) โˆ’(4๐‘ฆร—4๐‘ฆ)+(4๐‘ฆร—3) = (15/10ร—15/10) ๐‘ฅ^2+(15/10ร—4)๐‘ฅ๐‘ฆ+(15/10ร—3)๐‘ฅโˆ’(4ร—15/10)๐‘ฆ๐‘ฅ โˆ’16๐‘ฆ^2+12๐‘ฆ = (225/100) ๐‘ฅ^2+(60/10)๐‘ฅ๐‘ฆ+(45/10)๐‘ฅโˆ’(60/10)๐‘ฆ๐‘ฅโˆ’16๐‘ฆ^2+12๐‘ฆ = 2.25๐‘ฅ^2+6๐‘ฅ๐‘ฆ+4.5๐‘ฅโˆ’6๐‘ฆ๐‘ฅโˆ’16๐‘ฆ^2+12๐‘ฆ = 2.25๐‘ฅ^2โˆ’16๐‘ฆ^2+4.5๐‘ฅโˆ’12๐‘ฆ+6๐‘ฅ๐‘ฆโˆ’6๐‘ฅ๐‘ฆ = 2.25๐‘ฅ^2โˆ’16๐‘ฆ^2+4.5๐‘ฅโˆ’12๐‘ฆ Now, our equation becomes (1.5๐‘ฅโˆ’4๐‘ฆ)(1.5๐‘ฅ+4๐‘ฆ+3)โˆ’4.5๐‘ฅ+12๐‘ฆ = 2.25๐‘ฅ^2โˆ’16๐‘ฆ^2+4.5๐‘ฅโˆ’12๐‘ฆโˆ’4.5๐‘ฅ+12๐‘ฆ = 2.25๐‘ฅ^2โˆ’16๐‘ฆ^2+4.5๐‘ฅโˆ’4.5๐‘ฅโˆ’12๐‘ฆ+12๐‘ฆ = ๐Ÿ.๐Ÿ๐Ÿ“๐’™^๐Ÿโˆ’๐Ÿ๐Ÿ”๐’š^๐Ÿ

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Davneet Singh
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.