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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