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  1. Chapter 2 Class 8 Linear Equations in One Variable
  2. Serial order wise
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Ex 2.5, 10 Simplify and solve the following linear equations. 0.25 (4f โ€“ 3) = 0.05 (10f โ€“ 9)0.25 (4f โ€“ 3) = 0.05 (10f โ€“ 9) 25/100 (4๐‘“ โˆ’3)=5/100(10๐‘“ โˆ’ 9) 1/4 (4๐‘“ โˆ’3)=1/20 (10๐‘“ โˆ’9) (4๐‘“ โˆ’ 3)/4=(10๐‘“ โˆ’ 9)/20 20 (4f โˆ’ 3) = 4(10f โˆ’ 9) 80f โˆ’ 60 = 4(10f โˆ’ 9) 80f โˆ’ 60 = 40f โˆ’ 36 80f โˆ’ 40f โˆ’ 60 = โˆ’36 40f โˆ’ 60 = โˆ’ 36 40f = โˆ’36 + 60 40f = 24 f = 24/40 f = ๐Ÿ”/๐Ÿ๐ŸŽ = 0.6 Check:- L.H.S 0.25 (4f โˆ’ 3) = 0.25 (4 ร— 0.6 โˆ’ 3) = 0.25 (4 ร— 6/10 โˆ’ 3) = 0.25 (24/10 โ€“ 3) = 0.25 ((24 โˆ’ 30)/10) = 0.25 ((โˆ’6)/10) = 25/100 ร— ((โˆ’6)/10) = (โˆ’150)/1000 = (โˆ’15)/100 = โˆ’ 0.15 R.H.S 0.05 (10f โˆ’ 9) = 0.05 (10 ร— 0.6 โˆ’ 9) = 0.05 (10 ร— 6/10 โˆ’ 9) = 0.05 (6 โˆ’ 9) = 0.05 ร— (โˆ’3) = 5/100 ร— (โˆ’3) = (โˆ’15)/100 = โˆ’ 0.15 โˆด LHS = RHS , Hence Verified

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.