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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 2.2, 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.