Β

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Chapter 3 Class 10 Pair of Linear Equations in Two Variables

Serial order wise

Last updated at May 29, 2023 by Teachoo

Ex 3.3, 1 (Elimination) Solve the following pair of linear equations by the elimination method and the substitution method : (iv) π₯/2+2π¦/3=β1 πππ π₯βπ¦/3=3 Given x/2+2y/3=β1 (3(x) + 2(2y))/(2 Γ 3)=β1 (3π₯ + 4y)/6=β1 3x + 4y = β1 Γ 6 3x + 4y = β6 x β y/3=3 (3π₯ β π¦ )/3=3 3x β y = 3(3) 3x β y = 9 We use elimination method with equation (1) & (2) 5y = β 15 y=(β15)/( 5) y = β3 Putting y = β3 in equation (2) 3x β y = 9 3x β (β3) = 9 3x + 3 = 9 3x = 9 β 3 3x = 6 x = 6/3 x = 2 So, x = 2, y = β3 is the solution of the given equations Ex 3.3, 1 (Substitution) Solve the following pair of linear equations by the elimination method and the substitution method : (iv) π₯/2+2π¦/3=β1 πππ π₯βπ¦/3=3 Given x/2+2y/3=β1 (3(x) + 2(2y))/(2 Γ 3)=β1 (3π₯ + 4y)/6=β1 3x + 4y = β1 Γ 6 3x + 4y = β6 x β y/3=3 (3π₯ β π¦ )/3=3 3x β y = 3(3) 3x β y = 9 From (1) 3x + 4y = β6 3x = β6 β 4y Putting value of 3x in (2) 3x β y = 9 (β6 β 4y) β y = 9 β 4y β y = 9 + 6 β5y = 15 y = (β15)/( 5) y = β3 Putting y = β3 in equation (2) 3x β y = 9 3x β (β3) = 9 3x = 9 β 3 3x = 6 x = 6/3 x = 2 So, x = 2, y = β3 is the solution of the given equations