Last updated at Aug. 2, 2021 by Teachoo

Transcript

Ex 3.4 , 1 (Elimination) Solve the following pair of linear equations by the elimination method and the substitution method : (ii) 3x + 4y = 10 and 2x – 2y = 2 3x + 4y = 10 2x – 2y = 2 We multiply equation (2) by 2 2(2x – 2y) = 2 × 2 4x – 4y = 4 Using elimination with equations (3) & (1) 7x = 14 x = 14/7 x = 2 Putting x = 2 in (2) 2x – 2y = 2 2(2) – 2y = 2 4 – 2y = 2 –2y = 2 – 4 –2y = –2 y = (−2)/(−2) y = 1 Thus, x = 2, y = 1 is the solution of the given equations Ex 3.4, 1 (Substitution) Solve the following pair of linear equations by the elimination method and the substitution method : (ii) 3x + 4y = 10 and 2x – 2y = 2 3x + 4y = 10 2x – 2y = 2 From (1) 3x + 4y = 10 3x = 10 – 4y x = ((𝟏𝟎 − 𝟒𝒚)/𝟑) Putting value of x in (2) 2x – 2y = 2 2((10 − 4𝑦)/3)−2𝑦=2 (2(10 − 4𝑦))/3−2𝑦=2 (2(10 − 4𝑦) − 2𝑦 × 3)/3 =2 2(10 – 4y) – 6y = 2 × 3 20 – 8y – 6y = 6 –8y – 6y = 6 – 20 –14y = –14 y = (−14)/(−14) y = 1 Putting y = 1 in (2) 2x – 2y = 2 2x – 2(1) = 2 2x – 2 = 2 2x = 2 + 2 2x = 4 x = 4/2 x = 2 Therefore, x = 2, y = 1 are the solution of the given equations.

About the Author

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.