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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 You are here

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: (v) (7π₯ β 2π¦)/π₯π¦ = 5 (8π₯ + 7π¦)/π₯π¦ = 15 Given (7π₯ β 2π¦)/π₯π¦ = 5 (7π₯ )/π₯π¦ β (2π¦ )/π₯π¦ = 5 (7 )/π¦ β(2 )/π₯ = 5 (βπ )/π +(π )/π = 5 (8π₯ + 7π¦)/π₯π¦ = 15 (8π₯ )/π₯π¦ + (7π¦ )/π₯π¦ = 15 (8 )/π¦ +(7 )/π₯ = 15 (π )/π +(π )/π = 15 Our equations are (β2 )/π₯ +(7 )/π¦ = 5 β¦(1) (7 )/π₯ +(8 )/π¦ = 15 ...(2) So, our equations become β2u + 7v = 5 7u + 8v = 15 Hence, we solve β2u + 7v = 5 β¦(3) 7u + 8v = 15 β¦(4) From (3) β2u + 7v = 5 7v = 5 + 2u v = (5 + 2π’)/7 Putting value of v in (4) 7u + 8v = 15 7u + 8((5 + 2π’)/7) = 15 Multiplying 7 both sides 7 Γ 7u + 7 Γ 8 ((5 + 2π’)/7) = 7 Γ 15 49u + 8(5 + 2u) = 105 49u + 40 + 16u = 105 49u + 16u = 105 β 40 65u = 65 u = 65/65 u = 1 Putting value of u in (3) β2u + 7v = 5 β2(1) + 7v = 5 β2 + 7v = 5 7v = 5 + 2 7v = 7 v = 7/7 v = 1 Hence, u = 1, v = 1 But we have to find x & y We know that u = π/π 1 = 1/π₯ x = 1 v = π/π 1 = 1/π¦ y = 1 Hence, x = 1 , y = 1 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: (vi) 6x + 3y = 6xy 2x + 4y = 5xy Given 6x + 3y = 6xy Diving whole equation by xy (6π₯ + 3π¦)/π₯π¦ = 6π₯π¦/π₯π¦ 6π₯/π₯π¦ +3π¦/π₯π¦ = 6 π/π +π/π = 6 2x + 4y = 5xy Diving whole equation by xy (2π₯ + 4π¦)/π₯π¦ = 5π₯π¦/π₯π¦ 2π₯/π₯π¦ +4π¦/π₯π¦ = 5 π/π +π/π = 5 Hence, our equations are 6/π¦ +3/π₯ = 6 β¦(1) 2/π¦ +4/π₯ = 5 β¦(2) So, our equations become 6v + 3u = 6 2v + 4u = 5 Now, we solve 6v + 3u = 6 β¦(3) 2v + 4u = 5 β¦(4) From (3) 6v + 3u = 6 6v = 6 β 3u v = (6 β 3π’)/6 Putting value of v in (4) 2v + 4u = 5 2((6 β 3π’)/6) + 4u = 5 ((6 β 3π’)/3) + 4u = 5 Multiplying both sides by 3 3 Γ ((6 β3π’)/3) + 3 Γ 4u = 3 Γ 5 (6 β 3u) + 12u = 15 β3u + 12u = 15 β 6 9u = 9 u = 9/9 u = 1 Putting u = 1 in (3) 6v + 3u = 6 6v + 3(1) = 6 6v + 3 = 6 6v = 6 β 3 6v = 3 v = 3/6 v = π/π Hence, u = 1 , v = 1/2 But we have to find x & y Now, u = π/π 1 = 1/π₯ x = 1 v = π/π 1/2 = 1/π¦ y = 2 Hence, x = 1 , y = 2 is the solution of the given equation