# 10th cbse maths solution for exercise 3.6 part 1

This page 10th cbse maths solution for exercise 3.6 part 1is going to provide you solution for every problems that you find in the exercise no 3.6

## 10th CBSE maths solution for Exercise 3.6 part 1

(1)  Solve the following pairs of equations by reducing them to a pair of linear equations

(i) (1/2x) + (1/3y) = 2

(1/3x) + (1/2y) = 13/6

Solution: 3 a + 2 b = 12  -------- (1)

2 a + 3 b = 13   -------- (2)

(1) x 3 => 9 a + 6 b = 36

(2) x 2 => 4 a + 6 b = 26

(-)      (-)      (-)

--------------------

5 a = 10

a = 10/5

a = 2

Substitute a = 2 in the first equation to get the value of b

3 (2) + 3 b = 13

6 + 3 b = 13

3 b = 13 - 6

3 b = 7

b = 7/3

1/x = a                   1/y = b

1/x = 2                    1/y = 7/3

x = 1/2                      y = 3/7

(ii) (2/√x) + (3/√y) = 2

(4/√x) – (9/√y) = -1

Solution:

1/√x = a         1/√y = b

2 a + 3 b = 2  ------(1)

4 a – 9 b = -1  ------(2)

(1) x 3 = > 6 a + 9 b = 6

4 a – 9 b = - 1

-------------------

10 a = 5

a = 5/10

a = 1/2

1/√x = 1/2

2 = √x

x = 2²

x = 4

Substitute a = 1/2 in the first equation

2(1/2) + 3 b = 2

1 + 3 b = 2

3 b = 2 – 1

3 b = 1

b = 1/3

1/√y  = 1/3

3 = √y

y = 3²

y = 9

(iiii) (4/x)   + 3 y = 14

(3/x) – 4 y = 23

Solution:

Let 1/x = a and y = b

4 a + 3 b = 14  ------(1)

3 a – 4 b = 23  ------(2)

(1) x 4 => 16 a + 12 b = 56

(2) x 3 = > 9 a – 12 b = 69

---------------------

25 a = 125

a = 125/25

a = 5

substitute a = 5 in the first equation

4(5) + 3 b = 14

20 + 3 b = 14

3 b = 14 – 20

3 b = -6

b = -6/3

b = -2

1/x = a       y = b

1/x = 5      y = -2

x = 1/5         10th cbse maths solution for exercise 3.6 part 1

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