What is the solution of the pair of linear equations 37x + 43y = 123, 43x + 37y = 117?

(a) x = 2, y = 1     (b) x = −1, y = 2  

(c) x = −2, y = 1   (d) x = 1, y = 2

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  1. Class 10
  2. Solutions of Sample Papers for Class 10 Boards

Transcript

Question 40 What is the solution of the pair of linear equations 37x + 43y = 123, 43x + 37y = 117? (a) x = 2, y = 1 (b) x = โˆ’1, y = 2 (c) x = โˆ’2, y = 1 (d) x = 1, y = 2 Given equations 37๐‘ฅ + 43๐‘ฆ = 123 โ€ฆ(1) 43๐‘ฅ + 37๐‘ฆ = 117 โ€ฆ(2) Adding both equations (1) and (2) (37๐’™ + 43๐’š) + (43๐’™ + 37๐’š) = 123 + 117 (37๐‘ฅ + 43๐‘ฅ) + (43๐‘ฆ + 37๐‘ฆ) = 240 80๐‘ฅ + 80๐‘ฆ = 240 80(๐’™ + ๐’š) = 240 ๐‘ฅ + ๐‘ฆ = 240/80 ๐‘ฅ + ๐‘ฆ = 3 Now, subtracting (1) and (2) (37๐’™ + 43๐’š) โˆ’ (43๐’™ + 37๐’š) = 123 โˆ’ 117 (37๐‘ฅ โˆ’ 43๐‘ฅ) + (43๐‘ฆ โˆ’ 37๐‘ฆ) = 8 โˆ’8๐‘ฅ + 8๐‘ฆ = 8 8(โˆ’๐’™ + ๐’š) = 8 โˆ’๐‘ฅ + ๐‘ฆ = 8/8 โˆ’๐‘ฅ + ๐‘ฆ = 1 Adding (3) and (4) (๐‘ฅ + ๐‘ฆ) + (โˆ’๐‘ฅ + ๐‘ฆ) = 3 + 1 2y = 4 y = 4/2 y = 2 Putting y = 2 in (3) ๐‘ฅ + ๐‘ฆ = 3 ๐‘ฅ + 2 = 3 ๐‘ฅ = 3 โˆ’ 2 ๐‘ฅ = 1 โˆด x = 1, y = 2 So, the correct answer is (d)

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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.