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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.