1. Class 10
2. Important Questions for Exam - Class 10

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Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (v) (7๐ฅ โ 2๐ฆ)/๐ฅ๐ฆ = 5 (8๐ฅ +7๐ฆ)/๐ฅ๐ฆ = 15 Given (7๐ฅ โ 2๐ฆ)/๐ฅ๐ฆ = 5 (7๐ฅ )/๐ฅ๐ฆ - (2๐ฆ )/๐ฅ๐ฆ = 5 (7 )/๐ฆ โ(2 )/๐ฅ = 5 โ(2 )/๐ฅ +(7 )/๐ฆ = 5 Our equations are โ(2 )/๐ฅ +(7 )/๐ฆ = 5 โฆ(1) (7 )/๐ฅ +(8 )/๐ฆ = 15 ...(2) Hence, we solve โ2u + 7v = 5 โฆ(3) 7u + 8v = 15 โฆ(4) From (3) โ2u + 7v = 5 7v = 5 + 2u v = (5 + 2๐ข)/7 Putting value of v in (4) 7u + 8v = 15 7u + 8((5 + 2๐ข)/7) = 15 Multiplying 7 both sides 7 ร 7u + 7 ร 8((5+2๐ข)/7) = 7 ร 15 49u + 8(5 + 2u) = 105 49u + 40 + 16u = 105 49u + 16u = 105 โ 40 65u = 65 u = 65/65 u = 1 Putting value of u in (3) โ2u + 7v = 5 โ2(1) + 7v = 5 โ2 + 7v = 5 7v = 5 + 2 7v = 7 v = 7/7 v = 1 Hence u = 1 , v = 1 But we have to find x & y We know that u = 1/๐ฅ 1 = 1/๐ฅ x = 1 Hence, x = 1 , y = 1 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: (vi) 6x + 3y = 6xy 2x + 4y = 5xy Given 6x + 3y = 6xy Diving whole equation by xy (6๐ฅ+3๐ฆ)/๐ฅ๐ฆ = 6๐ฅ๐ฆ/๐ฅ๐ฆ 6๐ฅ/๐ฅ๐ฆ +3๐ฆ/๐ฅ๐ฆ = 6 6/๐ฆ +3/๐ฅ = 6 Hence, our equations are 6/๐ฆ +3/๐ฅ = 6 โฆ(1) 2/๐ฆ +4/๐ฅ = 5 โฆ(2) Now, we solve 6v + 3u = 6 โฆ(3) 2v + 4u = 5 โฆ(4) From (3) 6v + 3u = 6 6v = 6 โ 3u v = (6 โ 3๐ข)/6 Putting value of v in (4) 2v + 4u = 5 2((6 โ3๐ข)/6) + 4u = 5 ((6 โ3๐ข)/3) + 4u = 5 Multiplying both sides by 3 3 ร ((6 โ4๐ข)/3) + 3 ร 4u = 3 ร 5 (6 โ 3u) + 12u = 15 โ3u + 12u = 15 โ 6 9u = 9 u = 9/9 u = 1 Putting u = 1 in (3) 6v + 3u = 6 6v + 3(1) = 6 6v + 3 = 6 6v = 6 โ 3 6v = 3 v = 3/6 v = 1/2 Hence u = 1 , v = 1/2 But we have to find x & y Now, u = 1/๐ฅ 1 = 1/๐ฅ x = 1 Hence, x = 1 , y = 2 is the solution of the given equation

Class 10
Important Questions for Exam - Class 10