Ex 2.2

Chapter 2 Class 8 Linear Equations in One Variable
Serial order wise

### Transcript

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

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.