Ex 3.6, 1 (vii) and (viii) - Ex 3.6, 1 
Solve the following pairs of equations by reducing them to a pair of linear equations:
 - Mix questions - Equation given

  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: (vii) 10/(x + y) + 2/(x โˆ’ y) = 4 15/(x + y) - 5/(x โˆ’ y) = -2 10/(x + y) + 2/(๐‘ฅ โˆ’ y) = 4 15/(x + y) โ€“ 5/(๐‘ฅ โˆ’ 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 โˆ’ 2v)/10 Putting value of u in (4) 15u โ€“ 5v = โ€“ 2 15 ((4 โˆ’2v)/10) โ€“ 5v = โ€“ 2 3 ((4 โˆ’2v)/2) โ€“ 5v = โ€“ 2 Multiplying both sides by 2 2 ร— 3 ((4 โˆ’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 = (โˆ’16)/(โˆ’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 โˆ’๐‘ฆ) = 3/4 1/(2(3๐‘ฅ + ๐‘ฆ)) - 1/(2(3๐‘ฅ โˆ’๐‘ฆ)) = (โˆ’1)/8 1/(3๐‘ฅ + ๐‘ฆ) + 1/(3๐‘ฅ โˆ’๐‘ฆ) = 3/4 1/(2(3๐‘ฅ + ๐‘ฆ)) โ€“ 1/(2(3๐‘ฅ โˆ’๐‘ฆ)) = (โˆ’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 โˆ’ 4๐‘ฃ)/4 Putting (5) in (4) 4u โ€“ 4v = โ€“1 4((3 โˆ’ 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 = (โˆ’4)/(โˆ’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

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