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Example 10 - Chapter 9 Class 8 Algebraic Expressions and Identities
Last updated at March 31, 2023 by Teachoo
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Transcript
Example 10 Simplify (a + b) (2a β 3b + c) β (2a β 3b) c. Here there are two expressions 1st expression = (π+π) (2πβ3π+π) 2nd expression = (2πβ3π) π Solving 1st expression (π+π) (2πβ3π+π) = π(2πβ3π+π)+π(2πβ3π+π) = (πΓ2π)β(πΓ3π)+(πΓπ)+(πΓ2π)β(πΓ3π)+(πΓπ) = 2π^2β3ππ+ππ+2ππβ3π^2+ππ = 2π^2β3π^2+ππ+ππβ3ππ +2ππ = 2π^2β3π^2+ππ+ππβππ Solving 2nd expression (2πβ3π)π = 2ππβ3ππ Now, our expression is (π+π)(2πβ3π+π)β(2πβ3π)π = 2π^2β3π^2+ππ+ππβππβ(2ππβ3ππ) = 2π^2β3π^2+ππ+ππβππβ2ππ+3ππ = 2π^2β3π^2βππ+(ππβ2ππ)+(ππ+3ππ) = ππ^πβππ^πβππβππ+πππ
