This page 10th cbse maths solution for exercise 3.6 part 3 is going to provide you solution for every problems that you find in the exercise no 3.6

(vii) 10/(x + y) + 2/(x –y) = 4

15/(x + y) – 5/(x – y) = -2

1/(x + y) = a 1/(x – y) = b

10 a + 2 b = 4 --------(1)

15 a – 5 b = -2 --------(2)

(1) x 5 => 50 a + 10 b = 20

(2) x 2 = > 30 a – 10 b = -4

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80 a = 16

a = 16/80

a = 1/5

Substitute a = 1/5 in the first equation

10(1/5) + 2 b = 4

2 + 2 b = 4

2 b = 4 – 2

2 b = 2

b = 2/2

b = 1

1/(x + y) = 1/5

5 = x + y

x + y = 5 -------(3)

1/(x – y) = 1

x – y = 1 -------(4)

x + y = 5

x – y = 1

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2 x = 6

x = 3

Substitute x = 3 in the third equation

3 + y = 5

y = 5- 3

y = 2

(viii) 1/(3 x + y) + 1/(3 x – y) = 3/4

1/2(3 x + y) - 1/2(3 x – y) = -1/8

**Solution:**

1/(3 x + y) = a and 1/(3 x – y) = b

a + b = 3/4

4(a + b) = 3

4 a + 4 b = 3 ------(1)

(a/2) –(b/2) = -1/8

4 a – 4 b = -1------(2)

4a + 4 b = 3

4 a – 4 b = 1

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8 a = 4

a = 1/2

Substitute a = 1/2 in the first equation

4(1/2) + 4 b = 3

2 + 4 b = 3

4 b = 3 - 2

4 b = 1

b = 1/4

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

2 = 3 x + y

3 x + y = 2-------(3)

1/(3 x – y) = 3/4

4 = 3 x – y

3 x – y = 4 --------(4)

3 x + y = 2

3 x – y = 4

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6 x = 6

x = 1

Substitute x = 1 in the third equation

3 (1) + y = 2

y = 2 – 3

y = -1

**Related pages**

**Go to exercise 3.6****Go to exercise 3.5****Go to exercise 3.4****Go to exercise 3.3****Go to exercise 3.2****Exercise 3.6 (part 1)****Exercise 3.6 (part 2)**

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