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Ex 3.6

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

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

Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (i) 1/2π₯ + 1/3π¦ = 2 1/3π₯ + 1/2π¦ = 13/6 1/2π₯ + 1/3π¦ = 2 1/3π₯ + 1/2π¦ = 13/6 Let 1/π₯ = u 1/π¦ = v So, our equations become 1/2 u + 1/3 v = 2 (3π’ + 2π£)/(2 Γ 3) = 2 3u + 2v = 12 1/3 u + 1/2 v = 13/6 (2π’ +3π£)/(2 Γ 3) = 13/6 2u + 3v = 13 Our equations are 3u + 2v = 12 β¦(3) 2u + 3v = 13 β¦(4) From (3) 3u + 2v = 12 3u = 12 β 2v u = (12 β 2π£)/3 Putting value of u in (4) 2u + 3v = 13 2 ((12 β2π£)/3) + 3v = 13 Multiplying both sides by 3 3 Γ 2((12 β 2π£)/3) + 3 Γ 3v = 3 Γ 13 2(12 β 2v) + 9v = 39 24 β 4v + 9v = 39 β 4v + 9v = 39 β 24 5v = 15 v = 15/5 v = 3 Putting v = 3 in (3) 3u + 2v = 12 3u + 2(3) = 12 3u + 6 = 12 3u = 12 β 6 3u = 6 u = 6/3 u = 2 Hence, v = 3, u = 2 But we have to find x & y We know that u = π/π 2 = 1/π₯ x = π/π v = π/π 3 = 1/π¦ y = π/π So, x = π/π , y = π/π 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: (ii) 2/βπ₯ + 3/βπ¦ = 2 4/βπ₯ β 9/βπ¦ = β1 2/βπ₯ + 3/βπ¦ = 2 4/βπ₯ β 9/βπ¦ = β1 So, our equations become 2u + 3v = 2 4u β 9v = β1 Our equations 2u + 3v = 2 β¦(3) 4u β 9v = β1 β¦(4) From (3) 2u + 3v = 2 2u = 2 β 3v u = (2 β 3π£)/2 Putting value of u in (4) 4u β 9v = β 1 4 ((2 β 3π£)/2) β 9v = β1 2(2 β 3v) β 9v = β1 4 β 6v β 9v = β1 β 6v β 9v = β1 β 4 β15v = β 5 v = (β5)/(β15) v = π/π Putting v = 1/3 in (3) 2u + 3v = 2 2u + 3 (1/3) = 2 2u + 1 = 2 2u = 2 β 1 u = π/π Hence, u = 1/2 & v = 1/3 But, we need to find x & y u = π/βπ 1/2 = 1/βπ₯ βπ₯ = 2 Squaring both sides (βπ₯)2 = (2)2 x = 4 v = π/βπ 1/3 = 1/βπ¦ βπ¦ = 3 Squaring both sides (βπ¦)2 = (3)2 y = 9 Therefore, x = 4, y = 9 is the solution of the given equation