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

Transcript

Ex 9.4, 3 Simplify. (vii) (1.5π‘₯βˆ’4𝑦)(1.5π‘₯+4𝑦+3)βˆ’4.5π‘₯+12𝑦 Here, there are 2 expressions: First expression = (1.5π‘₯βˆ’4𝑦)(1.5π‘₯+4𝑦+3) Second expression = 4.5π‘₯+12𝑦 Solving First expression (1.5π‘₯βˆ’4𝑦)(1.5π‘₯+4𝑦+3) = 1.5π‘₯(1.5π‘₯+4𝑦+3)βˆ’4𝑦(1.5π‘₯+4𝑦+3) = (1.5π‘₯Γ—1.5π‘₯)+(1.5π‘₯Γ—4𝑦)+(1.5π‘₯Γ—3)βˆ’(4𝑦×1.5π‘₯) βˆ’(4𝑦×4𝑦)+(4𝑦×3) = (15/10Γ—15/10) π‘₯^2+(15/10Γ—4)π‘₯𝑦+(15/10Γ—3)π‘₯βˆ’(4Γ—15/10)𝑦π‘₯ βˆ’16𝑦^2+12𝑦 = (225/100) π‘₯^2+(60/10)π‘₯𝑦+(45/10)π‘₯βˆ’(60/10)𝑦π‘₯βˆ’16𝑦^2+12𝑦 = 2.25π‘₯^2+6π‘₯𝑦+4.5π‘₯βˆ’6𝑦π‘₯βˆ’16𝑦^2+12𝑦 = 2.25π‘₯^2βˆ’16𝑦^2+4.5π‘₯βˆ’12𝑦+6π‘₯π‘¦βˆ’6π‘₯𝑦 = 2.25π‘₯^2βˆ’16𝑦^2+4.5π‘₯βˆ’12𝑦 Now, our equation becomes (1.5π‘₯βˆ’4𝑦)(1.5π‘₯+4𝑦+3)βˆ’4.5π‘₯+12𝑦 = 2.25π‘₯^2βˆ’16𝑦^2+4.5π‘₯βˆ’12π‘¦βˆ’4.5π‘₯+12𝑦 = 2.25π‘₯^2βˆ’16𝑦^2+4.5π‘₯βˆ’4.5π‘₯βˆ’12𝑦+12𝑦 = 𝟐.πŸπŸ“π’™^πŸβˆ’πŸπŸ”π’š^𝟐

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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.