# Example 5 - Chapter 9 Class 8 Algebraic Expressions and Identities

Last updated at Dec. 24, 2018 by Teachoo

Last updated at Dec. 24, 2018 by Teachoo

Transcript

Example 5 Simplify the expressions and evaluate them as directed: (i) π₯ (π₯β3)+2 for π₯=1 π₯ (π₯β3)+2=π₯Γπ₯βπ₯Γ3+2 =π₯^2β3π₯+2 For x = 1 Putting x = 1 in expression π₯^2β3π₯+2 = 1^2β3Γ1+2 = 1β3+2 = (1+2)β3 = 3β3 = π Example 5 Simplify the expressions and evaluate them as directed: (ii) 3π¦(2π¦β7)β3(π¦β4)β63 for π¦=β2 3π¦(2π¦β7)β3(π¦β4)β63 =3π¦Γ2π¦β3π¦Γ7β3Γπ¦+3Γ4β63 =6π¦^2β21π¦β3π¦+12β63 =6π¦^2β24π¦β51 For y = βπ Putting π¦=β2 in expression 6π¦^2β24π¦β51 = 6(β2)^2β24(β2)β51 =6Γ4+24Γ2β51 = 24+48β51 = 72β51 = ππ

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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.