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Last updated at June 18, 2018 by Teachoo

Transcript

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 โ 5๐ฆ))/8 Putting value of x in (2) 3x + 2y = 4 3 (((9 โ 5๐ฆ))/8) + 2y = 4 (3(9 โ 5๐ฆ))/8 + 2y = 4 (3(9 โ 5๐ฆ) +8(2๐ฆ) )/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 = โ6 x = (โ6)/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 ๐ฅ/(5 ร(โ4) โ 2 ร(โ9) ) ๐ฅ/((โ20) + 18 ) ๐ฅ/(โ2 ) = ๐ฆ/((โ9) ร 3 โ (โ4) ร 8 ) = ๐ฆ/((โ27) + 32 ) = ๐ฆ/(5 ) = 1/(8 ร 2 โ 3 ร 5 ) = 1/(16 โ 15) = 1/1 Now, ๐ฅ/(โ2 ) = 1/(1 ) x = โ2 ร 1 x = โ2 ๐ฆ/(5 ) = 1/(1 ) y = 5 ร 1 x = 5 Hence, x = โ 2 , y = 5 is the solution

About the Author

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.