Check sibling questions

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

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

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

Ask a doubt
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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.