Last updated at Dec. 18, 2020 by Teachoo

Transcript

Example 18 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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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.