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

Transcript

Example 10 Simplify (a + b) (2a – 3b + c) – (2a – 3b) c. Here there are two expressions 1st expression = (π‘Ž+𝑏) (2π‘Žβˆ’3𝑏+𝑐) 2nd expression = (2π‘Žβˆ’3𝑏) 𝑐 Solving 1st expression (π‘Ž+𝑏) (2π‘Žβˆ’3𝑏+𝑐) = π‘Ž(2π‘Žβˆ’3𝑏+𝑐)+𝑏(2π‘Žβˆ’3𝑏+𝑐) = (π‘ŽΓ—2π‘Ž)βˆ’(π‘ŽΓ—3𝑏)+(π‘ŽΓ—π‘)+(𝑏×2π‘Ž)βˆ’(𝑏×3𝑏)+(𝑏×𝑐) = 2π‘Ž^2βˆ’3π‘Žπ‘+π‘Žπ‘+2π‘Žπ‘βˆ’3𝑏^2+𝑏𝑐 = 2π‘Ž^2βˆ’3𝑏^2+π‘Žπ‘+π‘π‘βˆ’3π‘Žπ‘ +2π‘Žπ‘ = 2π‘Ž^2βˆ’3𝑏^2+π‘Žπ‘+π‘π‘βˆ’π‘Žπ‘ Solving 2nd expression (2π‘Žβˆ’3𝑏)𝑐 = 2π‘Žπ‘βˆ’3𝑏𝑐 Now, our expression is (π‘Ž+𝑏)(2π‘Žβˆ’3𝑏+𝑐)βˆ’(2π‘Žβˆ’3𝑏)𝑐 = 2π‘Ž^2βˆ’3𝑏^2+π‘Žπ‘+π‘π‘βˆ’π‘Žπ‘βˆ’(2π‘Žπ‘βˆ’3𝑏𝑐) = 2π‘Ž^2βˆ’3𝑏^2+π‘Žπ‘+π‘π‘βˆ’π‘Žπ‘βˆ’2π‘Žπ‘+3𝑏𝑐 = 2π‘Ž^2βˆ’3𝑏^2βˆ’π‘Žπ‘+(π‘Žπ‘βˆ’2π‘Žπ‘)+(𝑏𝑐+3𝑏𝑐) = πŸπ’‚^πŸβˆ’πŸ‘π’ƒ^πŸβˆ’π’‚π’ƒβˆ’π’‚π’„+πŸ’π’ƒπ’„

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.