Slide20.JPG

Slide21.JPG


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

Go Ad-free
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.