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Chapter 3 Class 10 Pair of Linear Equations in Two Variables
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Last updated at April 16, 2024 by Teachoo
Transcript
Question 8Solve the following pair of equations by reducing them to a pair of linear equations : 5/(π₯ β1) + 1/(π¦ β2) = 2 6/(π₯ β1) β 3/(π¦ β2) = 1 5/(π₯ β 1) + 1/(π¦ β 2) = 2 6/(π₯ β 1) β 3/(π¦ β 2) = 1So, our equations become5u + v = 26u β 3v = 1Thus, our equations are 5u + v = 2β¦(3)6u β 3v = 1β¦(4)From (3)5u + v = 2v = 2 β 5uPutting value of v in (4)6u β 3v = 16u β 3(2 β 5u) = 16u β 6 + 15u = 16u + 15u = 1 + 621u = 7u = 7/21u = 1/3Putting u = 1/3 in equation (3)5u + v = 25(1/3) + v = 25/3 + v = 2v = 2 β 5/3v = (2(3) β 5)/3v = (6 β 5)/3v = π/πHence, u = 1/3 & v = 1/3But we need to find x & y u = π/(π β π) 1/3 = 1/(π₯ β 1) x β 1 = 3 x = 3 + 1 x = 4v = π/(π β π) 1/3 = 1/(π¦ β2) y β 2 = 3 y = 3 + 2 y = 5So, x = 4, y = 5 is the solution of our equations
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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.
