Slide5.JPG

Slide6.JPG

  1. Chapter 2 Class 8 Linear Equations in One Variable
  2. Serial order wise
Ask Download

Transcript

Ex 2.5, 3 Solve the following linear equations. x + 7 โˆ’ 8๐‘ฅ/3=17/6โˆ’5๐‘ฅ/2x + 7 โˆ’ 8๐‘ฅ/3=17/6โˆ’5๐‘ฅ/2 (3(๐‘ฅ + 7) โˆ’ 8๐‘ฅ)/3=(17 โˆ’ 5๐‘ฅ ร— 3)/6 (3๐‘ฅ + 21 โˆ’ 8๐‘ฅ)/3=(17 โˆ’ 15๐‘ฅ)/6 (3๐‘ฅ โˆ’ 8๐‘ฅ + 21)/3=(17 โˆ’ 15๐‘ฅ)/6 (โˆ’5๐‘ฅ + 21)/3=(17 โˆ’ 15๐‘ฅ)/6 6 (โˆ’5x + 21) = 3 (17 โˆ’ 15x) โˆ’30x + 126 = 51 โˆ’ 45x โˆ’30x = โˆ’45x + 51 โˆ’ 126 โˆ’30x = โˆ’45x โˆ’ 75 โˆ’30x + 45x = โˆ’75 15x = โˆ’ 75 x = (โˆ’75)/15 x = โˆ’5 Check:- โˆด LHS = RHS Hence Verified. L.H.S ๐‘ฅ+7โˆ’8๐‘ฅ/3 =โˆ’5+7โˆ’(8 (โˆ’5))/3 =2+40/3 =(6 + 40)/3=46/3 R.H.S 17/6โˆ’5๐‘ฅ/2 =17/6โˆ’5(โˆ’5)/2=17/6+25/2 =(17 +3(25))/6 =(17 + 75)/6=92/6=46/3

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.