Last updated at May 29, 2018 by Teachoo

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Ex 3.6 , 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (i) 1/2๐ฅ + 1/3๐ฆ = 2 1/3๐ฅ + 1/2๐ฆ = 13/6 1/2๐ฅ + 1/3๐ฆ = 2 1/3๐ฅ + 1/2๐ฆ = 13/6 So, our equations become Our equations are 3u + 2v = 12 โฆ(3) 2u + 3v = 13 โฆ(4) From (3) 3u + 2v = 12 3u = 12 โ 2v u = (12 โ2๐ฃ)/3 Putting value of u in (4) 2u + 3v = 13 2((12 โ2๐ฃ)/3) + 3v = 13 Multiplying both sides by 3 3 ร 2((12 โ2๐ฃ)/3) + 3 ร 3v = 3 ร 13 2(12 โ 2v) + 9v = 39 24 โ 4v + 9v = 39 โ 4v + 9v = 39 โ 24 5v = 15 v = 15/5 v = 3 Putting v = 3 in (3) 3u + 2v = 12 3u + 2(3) = 12 3u + 6 = 12 3u = 12 โ 6 3u = 6 u = 6/3 u = 2 Hence v = 3, u = 2 But we have to find x & y We know that So, x = 1/2 , y = 1/3 is the solution of the given equation Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (ii) 2/โ๐ฅ + 3/โ๐ฆ = 2 4/โ๐ฅ โ 9/โ๐ฆ = โ1 2/โ๐ฅ + 3/โ๐ฆ = 2 4/โ๐ฅ โ 9/โ๐ฆ = -1 Our equations 2u + 3v = 2 โฆ(3) 4u โ 9v = โ1 โฆ(4) From (3) 2u + 3v = 2 2u = 2 โ 3v u = (2 โ3๐ฃ)/2 Putting value of u in (4) 4u โ 9v = โ 1 4((2 โ3๐ฃ)/2) โ 9v = โ 1 2(2 โ 3v) โ 9v = โ 1 4 โ 6v โ 9v = โ 1 โ 6v โ 9v = โ 1 โ 4 โ 15v = โ 5 v = (โ5)/(โ15) v = 1/3 Putting v = 1/3 in (3) 2u + 3v = 2 2u + 3(1/3) = 2 2u + 1 = 2 2u = 2 โ 1 2u = 1 u = 1/2 Hence u = 1/2 & v = 1/3 But, we need to find x & y We need to find x & y Therefore, x = 4, y = 9 is the solution of the given equation

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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.