Ex 3.4, 2 (i) - If we add 1 to the numerator and subtract 1 from the

Ex 3.4, 2 (i) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Ex 3.4, 2 (i) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3
Ex 3.4, 2 (i) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4

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

Transcript

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’š Given that, If 1 is added to numerator and 1 is subtracted from the denominator, fraction becomes 1. (๐‘๐‘ข๐‘›๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘œ๐‘Ÿ + 1)/(๐ท๐‘’๐‘›๐‘œ๐‘š๐‘–๐‘›๐‘Ž๐‘ก๐‘œ๐‘Ÿ โˆ’1)= 1 (๐‘ฅ + 1)/(๐‘ฆ โˆ’ 1)= 1 (x + 1) = (y โ€“ 1) x โ€“ y = โ€“1 โ€“ 1 x โ€“ y = โ€“2 Also, If we add 1 to the denominator, fraction becomes 1/2. ๐‘๐‘ข๐‘š๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘œ๐‘Ÿ/(๐ท๐‘’๐‘›๐‘œ๐‘š๐‘–๐‘›๐‘Ž๐‘ก๐‘œ๐‘Ÿ + 1)=1/2 ๐‘ฅ/(๐‘ฆ +1)=1/2 2x = y + 1 2x โ€“ y = 1 Hence, our equations are x โ€“ y = โ€“2 โ€ฆ(1) 2x โ€“ y = 1 โ€ฆ(2) We use elimination method with equation (1) & (2) โ€“x = โ€“3 x = 3 Putting x = 3 in equation (1) x โ€“ y = โ€“2 3 โ€“ y = โ€“2 โ€“y = โ€“2 โ€“ 3 โ€“y = โ€“5 y = 5 So, x = 3, y = 5 is the solution of our equation โˆด Numerator = x = 3 & Denominator = y = 5 Hence, Original Fraction = ๐‘๐‘ข๐‘š๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘œ๐‘Ÿ/๐ท๐‘’๐‘›๐‘œ๐‘š๐‘–๐‘›๐‘Ž๐‘ก๐‘œ๐‘Ÿ "= " ๐‘ฅ/๐‘ฆ=๐Ÿ‘/๐Ÿ“

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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.