Last updated at Dec. 16, 2024 by Teachoo
Ex 2.2, 5 Solve the following linear equations. (3๐ก โ 2)/4โ(2๐ก + 3)/3=2/3โ๐ก (3๐ก โ 2)/4โ(2๐ก + 3)/3=2/3โ๐ก (๐(๐๐ โ ๐) โ ๐ (๐๐ + ๐))/๐๐=๐/๐โ๐ (9๐ก โ 6 โ 8๐ก โ 12)/12=2/3โ๐ก (9๐ก โ 8๐ก โ 6 โ 12)/12=2/3โ๐ก (๐ก โ 18)/12=2/3โ๐ก (๐ก โ 18)/12=(2 โ 3๐ก)/3 3 (t โ 18) = 12 (2 โ 3t) 3t โ 54 = 24 โ 36t 3t + 36t โ 54 = 24 39t โ 54 = 24 39t = 24 + 54 39t = 78 t = 78/39 t = 2 Check:- L.H.S (3๐ก โ 2)/4โ(2๐ก + 3)/3=(3(2) โ 2)/4 โ(2(2) + 3 )/3 = (6 โ 2)/4โ(4 + 3)/3=4/4โ 7/3=1โ7/3 = (3 โ 7)/3=(โ4)/3 R.H.S 2/3โ๐ก = 2/3โ2=(2 โ 6)/3=(โ4)/3
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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 and Computer Science at Teachoo