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

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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.