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  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise
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Ex 3.5 ,3 (Substitution method) Solve the following pair of linear equations by the substitution and cross-multiplication methods : 8x + 5y = 9 3x + 2y = 4 8x + 5y = 9 3x + 2y = 4 From (1) 8x + 5y = 9 8x = 9 โ€“ 5y x = ((9 โˆ’ 5๐‘ฆ))/8 Putting value of x in (2) 3x + 2y = 4 3 (((9 โˆ’ 5๐‘ฆ))/8) + 2y = 4 (3(9 โˆ’ 5๐‘ฆ))/8 + 2y = 4 (3(9 โˆ’ 5๐‘ฆ) +8(2๐‘ฆ) )/8 = 4 3(9 โ€“ 5y) + 8(2y) = 4 ร— 8 27 โ€“ 15y + 16y = 32 27 + y = 32 y = 27 โ€“ 32 y = 5 Putting y = 5 in (2) 3x + 2y = 4 3x + 2(5) = 4 3x + 10 = 4 3x = 4 โ€“ 10 3x = โ€“6 x = (โˆ’6)/3 x = โ€“ 2 Therefore, x= โ€“ 2 & y = 5 is the solution of the given pair of linear equations Ex 3.5 ,3 (Cross โ€“ multiplication method) Solve the following pair of linear equations by the substitution and cross-multiplication methods : 8x + 5y = 9 3x + 2y = 4 For cross-multiplication 8x + 5y โ€“ 9 = 0 3x + 2y โ€“ 4 = 0 ๐‘ฅ/(5 ร—(โˆ’4) โˆ’ 2 ร—(โˆ’9) ) ๐‘ฅ/((โˆ’20) + 18 ) ๐‘ฅ/(โˆ’2 ) = ๐‘ฆ/((โˆ’9) ร— 3 โˆ’ (โˆ’4) ร— 8 ) = ๐‘ฆ/((โˆ’27) + 32 ) = ๐‘ฆ/(5 ) = 1/(8 ร— 2 โˆ’ 3 ร— 5 ) = 1/(16 โˆ’ 15) = 1/1 Now, ๐‘ฅ/(โˆ’2 ) = 1/(1 ) x = โˆ’2 ร— 1 x = โˆ’2 ๐‘ฆ/(5 ) = 1/(1 ) y = 5 ร— 1 x = 5 Hence, x = โ€“ 2 , y = 5 is the solution

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.