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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 2.2, 1 Solve the following linear equations. ๐‘ฅ/2โˆ’1/5=๐‘ฅ/3+1/4 ๐‘ฅ/2โˆ’1/5=๐‘ฅ/3+1/4 (๐Ÿ“๐’™ โˆ’ ๐Ÿ)/๐Ÿ๐ŸŽ=(๐Ÿ’๐’™ + ๐Ÿ‘)/๐Ÿ๐Ÿ 12 (5x โˆ’ 2) = 10 (4x + 3) 60x โˆ’ 24 = 10 (4x + 3) 60x โˆ’ 24 = 40x + 30 60x โˆ’ 40x โˆ’ 24 = 30 20x โˆ’ 24 = 30 20x = 30 + 24 20x = 54 x = 54/20 x = ๐Ÿ๐Ÿ•/๐Ÿ๐ŸŽ Check:- โˆด LHS = RHS Hence Verified. L.H.S ๐‘ฅ/2 โˆ’1/5 =27/10 ร—1/2โˆ’1/5=(27 โˆ’ 4)/20=23/20 R.H.S ๐‘ฅ/3+ 1/4 =27/10ร—1/3+1/4=(54 + 15)/60=69/60=23/20

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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.