Last updated at Dec. 13, 2024 by Teachoo
Question 8 Solve the following pair of equations by reducing them to a pair of linear equations : 5/(๐ฅ โ1) + 1/(๐ฆ โ2) = 2 6/(๐ฅ โ1) โ 3/(๐ฆ โ2) = 1 5/(๐ฅ โ 1) + 1/(๐ฆ โ 2) = 2 6/(๐ฅ โ 1) โ 3/(๐ฆ โ 2) = 1 So, our equations become 5u + v = 2 6u โ 3v = 1 Thus, our equations are 5u + v = 2 โฆ(3) 6u โ 3v = 1 โฆ(4) From (3) 5u + v = 2 v = 2 โ 5u Putting value of v in (4) 6u โ 3v = 1 6u โ 3(2 โ 5u) = 1 6u โ 6 + 15u = 1 6u + 15u = 1 + 6 21u = 7 u = 7/21 u = 1/3 Putting u = 1/3 in equation (3) 5u + v = 2 5(1/3) + v = 2 5/3 + v = 2 v = 2 โ 5/3 v = (2(3) โ 5)/3 v = (6 โ 5)/3 v = ๐/๐ Hence, u = 1/3 & v = 1/3 But we need to find x & y u = ๐/(๐ โ ๐) 1/3 = 1/(๐ฅ โ 1) x โ 1 = 3 x = 3 + 1 x = 4 v = ๐/(๐ โ ๐) 1/3 = 1/(๐ฆ โ2) y โ 2 = 3 y = 3 + 2 y = 5 So, x = 4, y = 5 is the solution of our equations
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About the Author
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