1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
2. Serial order wise
3. Ex 3.6

Transcript

Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (vii) 10/(x + y) + 2/(x โ y) = 4 15/(x + y) - 5/(x โ y) = -2 10/(x + y) + 2/(๐ฅ โ y) = 4 15/(x + y) โ 5/(๐ฅ โ y) = โ2 Now, we solve 10u + 2v = 4 โฆ(3) 15u โ 5v = โ2 โฆ(4) From (3) 10u + 2v = 4 10u = 4 โ 2v u = (4 โ 2v)/10 Putting value of u in (4) 15u โ 5v = โ 2 15 ((4 โ2v)/10) โ 5v = โ 2 3 ((4 โ2v)/2) โ 5v = โ 2 Multiplying both sides by 2 2 ร 3 ((4 โ2v)/2) โ 2 ร 5v = 2 ร โ 2 3(4 โ 2v) โ 10v = โ 4 12 โ 6v โ 10v = โ 4 โ6v โ 10v = โ 4 โ 12 โ16v = โ 16 v = (โ16)/(โ16) v = 1 Putting v = 1 in (3) 10u + 2v = 4 10u + 2(1) = 4 10u + 2 = 4 10u = 4 โ 2 10u = 2 u = 2/10 u = 1/5 Hence, u = 1/5 & v = 1 But, we need to find x & y So, our equations become x + y = 5 โฆ(5) x โ y = 1 โฆ(6) From (5) x + y = 5 y = 5 โ x Putting value of y in (6) x โ y = 1 x โ (5 โ x) = 1 x โ 5 + x = 1 x + x = 1 + 5 2x = 6 x = 6/2 x = 3 Putting x = 3 in equation (5) x + y = 5 3 + y = 5 y = 5 โ 3 y = 2 Therefore, x = 3, y = 2 is the solution of our equation Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (viii) 1/(3๐ฅ + ๐ฆ) + 1/(3๐ฅ โ๐ฆ) = 3/4 1/(2(3๐ฅ + ๐ฆ)) - 1/(2(3๐ฅ โ๐ฆ)) = (โ1)/8 1/(3๐ฅ + ๐ฆ) + 1/(3๐ฅ โ๐ฆ) = 3/4 1/(2(3๐ฅ + ๐ฆ)) โ 1/(2(3๐ฅ โ๐ฆ)) = (โ1)/8 So, our equations are 4u + 4v = 3 โฆ(3) 4u โ 4v = -1 โฆ(4) From (3) 4u + 4v = 3 4u = 3 โ 4v u = (3 โ 4๐ฃ)/4 Putting (5) in (4) 4u โ 4v = โ1 4((3 โ 4๐ฃ)/4) โ 4v = โ1 (3 โ 4v) โ 4v = โ 1 โ 4v โ 4v = โ1 โ 3 โ 8v = โ 4 v = (โ4)/(โ8) v = 1/2 Putting v = 1/2 in equation (3) 4u + 4v = 3 4u + 4(1/2) = 3 4u + 2 = 3 4u = 3 โ 2 4u = 1 u = 1/4 Hence u = 1/4 , v = 1/2 But we need to find x & y We know Hence, we solve 3x + y = 4 โฆ(5) 3x โ y = 2 โฆ(6) From (5) 3x + y = 4 y = 4 โ 3x Putting y = 4 โ 3x in (6) 3x โ (4 โ 3x ) = 2 3x โ 4 + 3x = 2 3x + 3x = 2 + 4 6x = 6 x = 6/6 x = 1 Putting x = 1 in (5) 3x + y = 4 3(1) + y = 4 3 + y = 4 y = 4 โ 3 y = 1 So, x = 1, y = 1 is the solution of the given equation

Ex 3.6