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Ex 3.6

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

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 March 29, 2023 by Teachoo

Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (iii) 4/π₯ + 3y = 14 3/π₯ β 4y = 23 4/π₯ + 3y = 14 3/π₯ β 4y = 23 So, our equations become 4u + 3y = 14 3u β 4y = 23 Now, our equations are 4u + 3y = 14 β¦(3) 3u β 4y = 23 β¦(4) From (3) 4u + 3y = 14 4u = 14 β 3y u = (14 β3π¦)/4 Putting value of u in (4) 3u β 4y = 23 3 ((14 β 3π¦)/4) β 4y = 23 Multiplying both sides by 4 4 Γ 3 ((14 β 3π¦)/4) β 4 Γ 4y = 4 Γ 23 3(14 β 3y) β 16y = 92 42 β 9y β 16y = 92 β 9y β 16y = 92 β 42 β25y = 50 y = 50/(β25) y = β 2 Putting y = β 2 in equation (3) 4u + 3y = 14 4u + 3(β2) = 14 4u β 6 = 14 4u = 14 + 6 u = 20/4 u = 5 But u = 1/π₯ 5 = 1/π₯ x = π/π Hence, x = π/π, y = β2 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/(π₯ β 1) + 1/(π¦ β 2) = 2 6/(π₯ β 1) β 3/(π¦ β 2) = 1 5/(π₯ β 1) + 1/(π¦ β 2) = 2 6/(π₯ β 1) β 3/(π¦ β 2) = 1 So, our equations become 5u + v = 2 6u β 3v = 1 Our equations are 5u + v = 2 β¦(3) 6u β 3v = 1 β¦(4) From (3) 5u + v = 2 v = 2 β 5u Putting value of v in (4) 6u β 3v = 1 6u β 3(2 β 5u) = 1 6u β 6 + 15u = 1 6u + 15u = 1 + 6 21u = 7 u = 7/21 u = π/π Putting u = 1/3 in (3) 5u + v = 2 5 (1/3) + v = 2 5/3 + v = 2 v = 2 β 5/3 v = (2(3) β 5)/3 v = π/π Hence, u = 1/3 & v = 1/3 We need to find x & y We know that u = π/(π β π) 1/3 = 1/(π₯ β 1) x β 1 = 3 x = 3 + 1 x = 4 v = π/(π β π) 1/3 = 1/(π¦ β2) y β 2 = 3 y = 3 + 2 y = 5 So, x = 4, y = 5 is the solution of our equations