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  1. Chapter 2 Class 8 Linear Equations in One Variable
  2. Serial order wise
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Ex 2.5, 1 Solve the following linear equations. π‘₯/2βˆ’1/5=π‘₯/3+1/4 π‘₯/2βˆ’1/5=π‘₯/3+1/4 (5π‘₯ βˆ’ 2)/10=(4π‘₯ + 3)/12 12 (5x βˆ’ 2) = 10 (4x + 3) 60x βˆ’ 24 = 10 (4x + 3) 60x βˆ’ 24 = 40x + 30 60x βˆ’ 40x βˆ’ 24 = 30 20x βˆ’ 24 = 30 20x = 30 + 24 20x = 54 x = 54/20 x = πŸπŸ•/𝟏𝟎 Check:- ∴ LHS = RHS Hence Verified. L.H.S π‘₯/2 βˆ’1/5 =27/10 Γ—1/2βˆ’1/5=(27 βˆ’ 4)/20=23/20 R.H.S π‘₯/3+ 1/4 =27/10Γ—1/3+1/4=(54 + 15)/60=69/60=23/20

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.