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Question 8 Solve the following pair of equations by reducing them to a pair of linear equations : 5/(๐‘ฅ โˆ’1) + 1/(๐‘ฆ โˆ’2) = 2 6/(๐‘ฅ โˆ’1) โ€“ 3/(๐‘ฆ โˆ’2) = 1 5/(๐‘ฅ โˆ’ 1) + 1/(๐‘ฆ โˆ’ 2) = 2 6/(๐‘ฅ โˆ’ 1) โ€“ 3/(๐‘ฆ โˆ’ 2) = 1 So, our equations become 5u + v = 2 6u โ€“ 3v = 1 Thus, our equations are 5u + v = 2 โ€ฆ(3) 6u โ€“ 3v = 1 โ€ฆ(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) โˆ’ 5)/3 v = (6 โˆ’ 5)/3 v = ๐Ÿ/๐Ÿ‘ Hence, u = 1/3 & v = 1/3 But we need to find x & y u = ๐Ÿ/(๐’™ โˆ’ ๐Ÿ) 1/3 = 1/(๐‘ฅ โˆ’ 1) x โ€“ 1 = 3 x = 3 + 1 x = 4 v = ๐Ÿ/(๐’š โˆ’ ๐Ÿ) 1/3 = 1/(๐‘ฆ โˆ’2) y โ€“ 2 = 3 y = 3 + 2 y = 5 So, x = 4, y = 5 is the solution of our equations

  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise

About the Author

Davneet Singh

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