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Ex 3.6

Ex 3.6, 1 (i) and (ii)
Important
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Ex 3.6, 1 (iii) and (iv) Deleted for CBSE Board 2023 Exams

Ex 3.6, 1 (v) and (vi) Important Deleted for CBSE Board 2023 Exams

Ex 3.6, 1 (vii) and (viii) Important Deleted for CBSE Board 2023 Exams You are here

Ex 3.6, 2 (i) Important

Ex 3.6, 2 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 3.6, 2 (iii) Important Deleted for CBSE Board 2023 Exams

Last updated at Dec. 18, 2020 by Teachoo

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Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (vii) 10/(π₯ + π¦) + 2/(π₯ β π¦) = 4 15/(π₯ + π¦) β 5/(π₯ β π¦) = β2 10/(π₯ + π¦) + 2/(π₯ β π¦) = 4 15/(π₯ + π¦) β 5/(π₯ β π¦) = β2 So, our equations become 10u + 2v = 4 15u β 5v = β2 Now, we solve 10u + 2v = 4 β¦(3) 15u β 5v = β2 β¦(4) From (3) 10u + 2v = 4 10u = 4 β 2v u = (4 β 2π£)/10 Putting value of u in (4) 15u β 5v = β2 15 ((4 β 2π£)/10) β 5v = β2 3 ((4 β 2π£)/2) β 5v = β2 Multiplying both sides by 2 2 Γ 3 ((4 β 2π£)/2) β 2 Γ 5v = 2 Γ β2 3(4 β 2v) β 10v = β4 12 β 6v β 10v = β4 β6v β 10v = β4 β 12 β16v = β16 v = (β16)/(β16) v = 1 Putting v = 1 in (3) 10u + 2v = 4 10u + 2(1) = 4 10u + 2 = 4 10u = 4 β 2 10u = 2 u = 2/10 u = π/π Hence, u = 1/5 & v = 1 But, we need to find x & y u = π/(π + π) 1/5 = 1/(π₯ + π¦) x + y = 5 v = π/(π β π) 1 = 1/(π₯ β π¦) x β y = 1 So, our equations become x + y = 5 β¦(5) x β y = 1 β¦(6) Adding (5) and (6) (x + y) + (x β y) = 5 + 1 2x = 6 x = 6/2 x = 3 Putting value of y in (5) x + y = 5 3 + y = 5 y = 5 β 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/(3π₯ + π¦) + 1/(3π₯ β π¦) = 3/4 1/(2(3π₯ + π¦)) β 1/(2(3π₯ β π¦)) = (β1)/8 1/(3π₯ + π¦) + 1/(3π₯ β π¦) = 3/4 1/(2(3π₯ + π¦)) β 1/(2(3π₯ β π¦)) = (β1)/8 So, our equations become u + v = π/π 4(u + v) = 3 4u + 4v = 3 π/π u β π/π v = (βπ)/π (π’ β π£ )/2 = (β1)/8 8 Γ (π’ β π£ )/2 = β1 4(u β v) = β1 4u β 4v = β1 So, our equations are 4u + 4v = 3 β¦(3) 4u β 4v = β1 β¦(4) Adding (3) and (4) (4u + 4v) + (4u β 4v) = 3 + (β1) 8u = 2 u = 2/8 u = π/π Putting u = 1/4 in (3) 4u + 4v = 3 4 Γ 1/4 + 4v = 3 1 + 4v = 3 4v = 3 β 1 4v = 2 v = 2/4 v = π/π Hence u = 1/4 , v = 1/2 But we need to find x & y We know u = π/(ππ + π) 1/4 = 1/(3π₯ + π¦) 3x + y = 4 v = π/(ππ β π) 1/2 = 1/(3π₯ β π¦) 3x β y = 2 Hence, we solve 3x + y = 4 β¦(5) 3x β y = 2 β¦(6) Adding (5) and (6) (3x + y) + (3x β y) = 4 + 2 6x = 6 x = 6/6 x = 1 Putting x = 1 in (5) 3x + y = 4 3(1) + y = 4 3 + y = 4 y = 4 β 3 y = 1 So, x = 1, y = 1 is the solution of the given equation