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Ex 3.6, 1 (iii) & (iv) : 4/x + 3y  = 14 , 3/x - 4y = 23 - Mix questions - Equation given

  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) 4/๐‘ฅ + 3y = 14 3/๐‘ฅ โ€“ 4y = 23 4/๐‘ฅ + 3y = 14 3/๐‘ฅ โ€“ 4y = 23 Now, our equations are 4u + 3y = 14 โ€ฆ(3) 3u โ€“ 4y = 23 โ€ฆ(4) From (3) 4u + 3y = 14 4u = 14 โ€“ 3y u = (14 โˆ’3๐‘ฆ)/4 Putting value of u in (4) 3u โ€“ 4y = 23 3((14 โˆ’3๐‘ฆ)/4) โ€“ 4y = 23 Multiplying both sides by 4 4 ร— 3((14 โˆ’3๐‘ฆ)/4) โ€“ 4 ร— 4y = 4 ร— 23 3(14 โ€“ 3y ) - 16y = 92 42 โ€“ 9y - 16y = 92 โ€“ 9y โ€“ 16y = 92 โ€“ 42 โ€“25y = 50 y = 50/(โˆ’25) = โ€“ 2 Putting y = โ€“ 2 in equation (3) 4u + 3y = 14 4u + 3(โ€“2) = 14 4u โ€“ 6 = 14 4u = 14 + 6 4u = 20 u = 20/4 u = 5 But u = 1/๐‘ฅ 5 = 1/๐‘ฅ x = 1/5 Hence, x = 1/5 , y = โ€“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/(๐‘ฅ โˆ’1) + 1/(๐‘ฆ โˆ’2) = 2 6/(๐‘ฅ โˆ’1) - 3/(๐‘ฆ โˆ’2) = 1 5/(๐‘ฅ โˆ’1) + 1/(๐‘ฆ โˆ’2) = 2 6/(๐‘ฅ โˆ’1) - 3/(๐‘ฆ โˆ’2) = 1 Our equations are 5u + v = 2 โ€ฆ(3) 6u โ€“ 3v = 1 โ€ฆ(4) From (3) 5u + v = 2 v = 2 โ€“ 5u Putting value of v in (4) 6u โ€“ 3v = 1 6u โ€“ 3(2 โ€“ 5u) = 1 6u โ€“ 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 โ€“ 5/3 v = (2(3) โˆ’ 5)/3 v = (6 โˆ’ 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

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