1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
2. Serial order wise

Transcript

Ex 3.6 , 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (i) 1/2π₯ + 1/3π¦ = 2 1/3π₯ + 1/2π¦ = 13/6 1/2π₯ + 1/3π¦ = 2 1/3π₯ + 1/2π¦ = 13/6 So, our equations become Our equations are 3u + 2v = 12 β¦(3) 2u + 3v = 13 β¦(4) From (3) 3u + 2v = 12 3u = 12 β 2v u = (12 β2π£)/3 Putting value of u in (4) 2u + 3v = 13 2((12 β2π£)/3) + 3v = 13 Multiplying both sides by 3 3 Γ 2((12 β2π£)/3) + 3 Γ 3v = 3 Γ 13 2(12 β 2v) + 9v = 39 24 β 4v + 9v = 39 β 4v + 9v = 39 β 24 5v = 15 v = 15/5 v = 3 Putting v = 3 in (3) 3u + 2v = 12 3u + 2(3) = 12 3u + 6 = 12 3u = 12 β 6 3u = 6 u = 6/3 u = 2 Hence v = 3, u = 2 But we have to find x & y We know that So, x = 1/2 , y = 1/3 is the solution of the given equation Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (ii) 2/βπ₯ + 3/βπ¦ = 2 4/βπ₯ β 9/βπ¦ = β1 2/βπ₯ + 3/βπ¦ = 2 4/βπ₯ β 9/βπ¦ = -1 Our equations 2u + 3v = 2 β¦(3) 4u β 9v = β1 β¦(4) From (3) 2u + 3v = 2 2u = 2 β 3v u = (2 β3π£)/2 Putting value of u in (4) 4u β 9v = β 1 4((2 β3π£)/2) β 9v = β 1 2(2 β 3v) β 9v = β 1 4 β 6v β 9v = β 1 β 6v β 9v = β 1 β 4 β 15v = β 5 v = (β5)/(β15) v = 1/3 Putting v = 1/3 in (3) 2u + 3v = 2 2u + 3(1/3) = 2 2u + 1 = 2 2u = 2 β 1 2u = 1 u = 1/2 Hence u = 1/2 & v = 1/3 But, we need to find x & y We need to find x & y Therefore, x = 4, y = 9 is the solution of the given equation