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

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Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (iii) 4/π₯ + 3y = 14 3/π₯ β 4y = 23 4/π₯ + 3y = 14 3/π₯ β 4y = 23 Now, our equations are 4u + 3y = 14 β¦(3) 3u β 4y = 23 β¦(4) From (3) 4u + 3y = 14 4u = 14 β 3y u = (14 β3π¦)/4 Putting value of u in (4) 3u β 4y = 23 3((14 β3π¦)/4) β 4y = 23 Multiplying both sides by 4 4 Γ 3((14 β3π¦)/4) β 4 Γ 4y = 4 Γ 23 3(14 β 3y ) - 16y = 92 42 β 9y - 16y = 92 β 9y β 16y = 92 β 42 β25y = 50 y = 50/(β25) = β 2 Putting y = β 2 in equation (3) 4u + 3y = 14 4u + 3(β2) = 14 4u β 6 = 14 4u = 14 + 6 4u = 20 u = 20/4 u = 5 But u = 1/π₯ 5 = 1/π₯ x = 1/5 Hence, x = 1/5 , y = β2 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: (iv) 5/(π₯ β1) + 1/(π¦ β2) = 2 6/(π₯ β1) - 3/(π¦ β2) = 1 5/(π₯ β1) + 1/(π¦ β2) = 2 6/(π₯ β1) - 3/(π¦ β2) = 1 Our equations are 5u + v = 2 β¦(3) 6u β 3v = 1 β¦(4) From (3) 5u + v = 2 v = 2 β 5u Putting value of v in (4) 6u β 3v = 1 6u β 3(2 β 5u) = 1 6u β 6 + 15u = 1 6u + 15u = 1 + 6 21u = 7 u = 7/21 u = 1/3 Putting u = 1/3 in (3) 5u + v = 2 5(1/3) + v = 2 5/3 + v = 2 v = 2 β 5/3 v = (2(3) β 5)/3 v = (6 β 5)/3 v = 1/3 Hence, u = 1/3 & v = 1/3 Hence, u = 1/3 & v = 1/3 We need to find x & y We know that So, x = 4, y = 5 is the solution of our equations