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

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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.