Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 17 Solve the pair of equations: 2/๐ฅ+ 3/๐ฆ=13 5/๐ฅโ4/๐ฆ=โ2 2/๐ฅ+ 3/๐ฆ=13 5/๐ฅโ4/๐ฆ=โ2 Hence, our equations are 2u + 3v = 13 โฆ(3) 5u โ 4v = โ 2 โฆ(4) From (3) 2u + 3v = 13 2u = 13 โ 3V u = (13 โ 3๐ฃ)/2 Putting value of u (4) 5u โ 4v = - 2 5((13โ3๐ฃ)/2)โ4๐ฃ=โ2 Multiplying 2 both sides 2 ร 5((13โ3๐ฃ)/2)โ"2 ร" 4๐ฃ="2 ร"โ2 5(13 โ 3v) โ 8v = โ4 65 โ 15v โ 8v = โ4 โ 15v โ 8v = โ 4 โ 65 โ 23v = โ 69 v = (โ69)/(โ23) v = 3 Putting v = 3 in (3) 2u + 3v = 13 2u + 3(3) = 13 2u + 9 = 13 2u = 13 โ 9 2u = 4 u = 4/2 u = 2 Hence, u = 2, v = 3 is the solution But we have to find x & y we know that u = 1/๐ฅ 2 = 1/๐ฅ x = 1/2 Hence, x = 1/2 , y = 1/3 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.