Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

 

 

 

  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
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Transcript

Ex 3.5,4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction Let numerator be x and denominator be y So, fraction is ๐‘ฅ/๐‘ฆ Given that, if 1 is subtracted from numerator fraction becomes 1/3. (๐‘๐‘ข๐‘›๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘œ๐‘Ÿ โˆ’1)/(๐ท๐‘’๐‘›๐‘œ๐‘š๐‘–๐‘›๐‘Ž๐‘ก๐‘œ๐‘Ÿ )=1/3 (๐‘ฅ โˆ’ 1)/(๐‘ฆ )=1/3 3(x โ€“ 1) = 1(y) 3x โ€“ 3 = y 3x โ€“ y = 3 Also, if 8 is added to the denominator, fraction becomes 1/4. (๐‘๐‘ข๐‘š๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘œ๐‘Ÿ )/(๐ท๐‘’๐‘›๐‘œ๐‘š๐‘–๐‘›๐‘Ž๐‘ก๐‘œ๐‘Ÿ + 8)=1/4 (๐‘ฅ )/(๐‘ฆ + 8)=1/4 4(x) = 1(y + 8) 4x = y + 8 4x โ€“ y = 8 From (1) 3x โ€“ y = 3 3x = y + 3 x = ((๐‘ฆ + 3)/3) Putting value of x in (2) 4x โ€“ y โ€“ 8 = 0 4 ((๐‘ฆ + 3)/3 ) โ€“ y โ€“ 8 = 0 Multiplying the equation by 3 3 ร—"4 (" (๐‘ฆ + 3)/3 ")"โˆ’3ร—๐‘ฆโˆ’3ร—8=0ร—8 4(y + 3 ) โ€“ 3y โ€“ 24 = 0 4y + 12 โ€“ 3y โ€“ 24 = 0 4y โ€“ 3y โ€“ 24 + 12 = 0 y โ€“ 12 = 0 y = 12 Putting y = 12 in equation (1) 3x โ€“ y = 3 3x โ€“ 12 = 3 3x = 12 + 3 3x = 15 x = 15/3 x = 5 Therefore x = 5, y = 12 is the solution So, Numerator = x = 5 Denominator = y = 12 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=5/12

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.