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Ex 2.5, 10 - Simplify and solve 0.25(4f - 3) = 0.05(10f - 9) - Teachoo

Ex 2.5, 10 - Chapter 2 Class 8 Linear Equations in One Variable - Part 2


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Ex 2.5, 10 Simplify and solve the following linear equations. 0.25 (4f – 3) = 0.05 (10f – 9)0.25 (4f – 3) = 0.05 (10f – 9) 25/100 (4𝑓 −3)=5/100(10𝑓 − 9) 1/4 (4𝑓 −3)=1/20 (10𝑓 −9) (4𝑓 − 3)/4=(10𝑓 − 9)/20 20 (4f − 3) = 4(10f − 9) 80f − 60 = 4(10f − 9) 80f − 60 = 40f − 36 80f − 40f − 60 = −36 40f − 60 = − 36 40f = −36 + 60 40f = 24 f = 24/40 f = 𝟔/𝟏𝟎 = 0.6 Check:- L.H.S 0.25 (4f − 3) = 0.25 (4 × 0.6 − 3) = 0.25 (4 × 6/10 − 3) = 0.25 (24/10 – 3) = 0.25 ((24 − 30)/10) = 0.25 ((−6)/10) = 25/100 × ((−6)/10) = (−150)/1000 = (−15)/100 = − 0.15 R.H.S 0.05 (10f − 9) = 0.05 (10 × 0.6 − 9) = 0.05 (10 × 6/10 − 9) = 0.05 (6 − 9) = 0.05 × (−3) = 5/100 × (−3) = (−15)/100 = − 0.15 ∴ 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 12 years. He provides courses for Maths and Science at Teachoo.