Question 3 - Solving difficult equations - Word Problems - Chapter 2 Class 8 Linear Equations in One Variable
Last updated at April 16, 2024 by Teachoo
Solving difficult equations - Word Problems
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Solving difficult equations - Word Problems
Last updated at April 16, 2024 by Teachoo
Question 3 Solve the following equations. 𝑧/(𝑧 + 15)=4/99z = 4 (z + 15) 9z = 4z + 60 9z − 4z = 60 5z = 60 z = 60/5 z = 12 Check:- L.H.S = 𝑧/(𝑧 + 15) =12/(12 + 15)=12/27 =4/9 = R.H.S ∴ LHS = RHS Hence Verified.