Ex 3.6

Chapter 3 Class 10 Pair of Linear Equations in Two Variables
Serial order wise

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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 (βπ )/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 (β2 )/π₯ +(7 )/π¦ = 5 β¦(1) (7 )/π₯ +(8 )/π¦ = 15 ...(2) So, our equations become β2u + 7v = 5 7u + 8v = 15 Hence, we solve β2u + 7v = 5 β¦(3) 7u + 8v = 15 β¦(4) From (3) β2u + 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) β2u + 7v = 5 β2(1) + 7v = 5 β2 + 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 β3π’)/3) + 3 Γ 4u = 3 Γ 5 (6 β 3u) + 12u = 15 β3u + 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

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.