# Ex 3.5, 3 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

Last updated at June 18, 2018 by Teachoo

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.