Last updated at July 24, 2021 by Teachoo

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Ex 3.4, 2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction? Let Numerator be x and Denominator be y So, Fraction is ๐/๐ Given that, If 1 is added to numerator and 1 is subtracted from the denominator, fraction becomes 1. (๐๐ข๐๐๐๐๐ก๐๐ + 1)/(๐ท๐๐๐๐๐๐๐๐ก๐๐ โ1)= 1 (๐ฅ + 1)/(๐ฆ โ 1)= 1 (x + 1) = (y โ 1) x โ y = โ1 โ 1 x โ y = โ2 Also, If we add 1 to the denominator, fraction becomes 1/2. ๐๐ข๐๐๐๐๐ก๐๐/(๐ท๐๐๐๐๐๐๐๐ก๐๐ + 1)=1/2 ๐ฅ/(๐ฆ +1)=1/2 2x = y + 1 2x โ y = 1 Hence, our equations are x โ y = โ2 โฆ(1) 2x โ y = 1 โฆ(2) We use elimination method with equation (1) & (2) โx = โ3 x = 3 Putting x = 3 in equation (1) x โ y = โ2 3 โ y = โ2 โy = โ2 โ 3 โy = โ5 y = 5 So, x = 3, y = 5 is the solution of our equation โด Numerator = x = 3 & Denominator = y = 5 Hence, Original Fraction = ๐๐ข๐๐๐๐๐ก๐๐/๐ท๐๐๐๐๐๐๐๐ก๐๐ "= " ๐ฅ/๐ฆ=๐/๐

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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.