# 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

(ii) 2/√x + 3/√y = 2; 4/√x - 9/√y = -1

(iii) 4/x + 3y = 14; 3/x - 4y = 23

(iv) 5/(x - 1) + 1/(y - 2) = 2; 6/(x - 1) - 3/(y - 1) = 2

(v) (7x - 2 y)/xy = 5; (8x + 7y)/xy = 15

(vi) 6x + 3y = 6xy; 2x + 4 y = 5xy

(vii) 10/(x + y) + 2/(x - y) = 4; 15/(x + y) - 5/(x - y) = - 2

(viii) 1/(3x + y) + 1/(3x - y) = 3/4; 1/2(3x + y) - 1/2(3x - y) = -1/8

**Solution:**

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

Let 1/x = p and 1/y = q , then the equations change as follows:

1/2x + 1/3y = 2 ⇒ p/2 + q/3 = 2 ⇒ 3 p + 2q -12 = 0 ....(1)

1/3x + 1/2y = 13/6 ⇒ p/6 + q/3 = 13/6 ⇒ 2 p + 3q - 13= 0 ....(2)

Using cross-multiplication method, we obtain

p/-26 - (-36) = q/-24 - (-39) = 1/9 - 4

p/10 = q/15 = 1/5

p/10 = 1/5 and q/15 = 1/5

p = 2 and q = 3

Therefore, 1/x = 2 and 1/y = 3

Hence, x = 1/2 and y = 1/3

(ii) 2/√x + 3/√y = 2; 4/√x - 9/√y = -1

Substituting 1/√x = p and 1/√y = q in the given equations, we obtain

2/√x + 3/√y = 2 ⇒ 2 p +3q = 2 ....(1)

4/√x - 9/√y = - 14 ⇒ 4 p - 9q = - 14 ....(2)

Multiplying equation (1) by 3, we obtain 6 p + 9q = 6 ....(3)

Adding equation (2) and (3), we obtain

10 p = 5

p = 1/2

Putting p = 1/2 in equation (1), we obtain

2 × 1/2 + 3q = 2

3q = 2 - 1

q = 1/3

Therefore, p = 1/√x = 1/2

⇒ √x = 2

⇒ x = 4

And q = 1/√2 = 1/3

⇒ √y = 3

⇒ y = 9

Hence, x = 4 and y = 9

(iii) 4/x + 3y = 14; 3/x - 4 y = 23

Substituting 1/x = p in the given equations, we obtain

4 p + 3y = 14 ⇒ 4 p + 3y -14 = 0 ....(1)

3 p - 4 y = 23 ⇒ 3 p - 4 y - 23 = 0 ....(2)

By cross-multiplication, we obtain

p/(-69 - 56) = y/[-42 - (-92)] = 1/(-16 - 9)

p/-125 = y/50 = 1/-25

p/-125 = 1/-25 and y/50 = 1/-25

p = 5 and y = - 2

Therefore, p = 1/x = 5

⇒ x = 1/5

Hence, x = 1/5 and y = - 2

(iv) 5/(x - 1) + 1/(y - 2) = 2; 6/(x - 1) - 3/(y - 1) = 2

Putting 1/(x - 1) =p and 1/p/(y - 2) = q in the given equation, we obtain

5/(x - 1) + 1/(y - 2) = 2 ⇒ 5 p + q = 2 ....(1)

6/(x - 1) - 3/(y - 1) = 2 ⇒ 6 p - 3q = 1 ....(2)

Multiplying equation (1) by 3, we obtain

15 p + 3q = 6 ....(3)

Adding (2) and (3), we obtain

21p = 7

p = 1/3

Putting p = 1/3 in equation (1), we obtain

5 x 1/3 + q = 2

q = 2 - 5/3

q = 1/3

Therefore, p = 1/x - 1 = 1/3

⇒ x - 1 = 3

⇒ x = 4

and q = 1/y-2 = 1/3

⇒ y - 2 = 3

⇒ y = 5

Hence, x = 4 and y = 5

(v) (7x - 2 y)/xy = 5; (8x + 7 y)/xy = 15

(7x - 2 y)/xy = 5 ⇒ 7x/xy - 2 y/xy = 5 ⇒ 7/y - 2/x = 5 ....(1)

(8x + 7 y)/xy = 15 ⇒ 8x/xy + 7 y/xy = 15 ⇒ 8/y + 7/x = 15 ....(2)

Putting 1/x = p and 1/y = q in the equations (1) and (2), we obtain

7/x - 2/y = 5 ⇒ -2 p + 7q - 5 = 0

8/y + 7/x = 15 ⇒ 7 p + 8q - 15 = 0

By cross-multiplication method, we obtain

p/[-105 - (-40)] = q /(-35 - 30) = 1/(-16 - 49)

p/-65 = q/-65 = 1/-65

p/-65 = 1/-65 and q/-65 = 1/-65

p = 1 and q = 1

Therefore, p = 1/x = 1

⇒ x = 1

and, q = 1/y = 1

⇒ y = 1

Hence, x = 1 and y = 1

(vi) 6x + 3y = 6xy; 2x + 4 y = 5xy

By dividing both the given equations by (xy), we obtain

6x + 3y = 6xy ⇒ 6/y + 3/x = 6....(1)

2x + 4 y = 5xy ⇒ 2/y + 4/x = 5 ....(2)

Substituting 1/x = p and 1/y = q in the equations (1) and (2), we obtain

3 p + 6q - 6 = 0 ....(3)

4 p + 2q - 5 = 0 ....(4)

By cross-multiplication method, we obtain

p/-30 - (-12) = q/-24 - (-15) = 1/6 - 24

p/-18 = q/-9 = 1/-18

p/-18 = 1/-18 and q/-9 = 1/-18

p = 1 and q = 1/2

Therefore, p = 1/x = 1

⇒ x = 1

and, q = 1/y = 1/2

⇒ y = 2

Hence, x = 1 and y = 2

(vii) 10/(x + y) + 2/(x - y) = 4; 15/(x + y) - 5/(x - y) = - 2

Substituting 1/x + y = p and 1/x - y = q in the given equations, we obtain

10/(x + y) + 2/(x - y) = 4 ⇒ 10 p + 2q = 4 ⇒ 5 p + q - 2 = 0 ....(1)

15/(x + y) - 5/(x - y) = - 2 ⇒ 15 p - 5q = - 2 ⇒ 15 p - 5q + 2 = 0 ....(2)

Using cross-multiplication method, we obtain

p/(2 -10) = q/(-30 -10) = 1/(-25 - 15)

p/-8 = q/-40 = 1/-40

p/-8 = 1/-40 and q/-40 = 1/-40

p = 1/5 and q = 1

Therefore, p = 1/(x + y) = 1/5

⇒ x + y = 5 ....(3)

and q = 1/(x - y) = 1

⇒ x - y = 1 ....(4)

Adding equation (3) and (4), we obtain

2x = 6

x = 3

Substituting x = 3 in equation (3), we obtain

3 + y = 5

y = 2

Hence, x = 3 and y = 2

(viii) 1/(3x + y) + 1/(3x - y) = 4; 1/2(3x + y) - 1/2(3x - y) = 8

Substituting 1/(3x + y) = p and 1/(3x - y) = qin these equations, we obtain

1/(3x + y) + 1/(3x - y) = 3/4 ⇒ p + q = 3/4 ....(1)

1/2(3x + y) - 1/2(3x - y) = -1/8 ⇒ p/2 - q/2 = -1/8 ⇒ p - q = -1/4 ....(2)

Adding (1) and (2), we obtain

2p = 3/4 - 1/4

2p = 1/2

p = 1/4

Substituting p = 1/4 in (2), we obtain

1/4 - q = -1/4

q = 1/4 + 1/4

q = 1/2

Therefore, p = 1/3x + y = 1/4

⇒ 3x + y = 4 ....(3)

and, q = 1/3x - y = 1

⇒ 3x - y = 2 ....(4)

Adding equations (3) and (4), we obtain

6x = 6

x = 1

Substituting x =1 in (3), we obtain

3 × 1+ y = 4

y = 1

Hence, x = 1 and y = 1

**Video Solution:**

## 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 (ii) 2/√x + 3/√y = 2; 4/√x - 9/√y = -1 (iii) 4/x + 3y = 14; 3/x - 4y = 23 (iv) 5/(x - 1) + 1/(y - 2) = 2; 6/(x - 1) - 3/(y - 1) = 2 (v) (7x - 2 y)/xy = 5; (8x + 7 y)/xy = 15 (vi) 6x + 3y = 6xy; 2x + 4 y = 5xy (vii) 10/(x + y) + 2/(x - y) = 4; 15/(x + y) - 5/(x - y) = - 2 (viii) 1/(3x + y) + 1/(3x - y) = 3/4; 1/2(3x + y) - 1/2(3x - y) = -1/8

### NCERT Solutions for Class 10 Maths - Chapter 3 Exercise 3.6 Question 1:

(i) Hence, x = 1/2 and y = 1/3

(ii) Hence, x = 4 and y = 9

(iii) Hence, x = 1/5 and y = - 2

(iv) Hence, x = 4 and y = 5

(v) Hence, x = 1 and y = 1

(vi) Hence, x = 1 and y = 2

(vii) Hence, x = 3 and y = 2

(viii) Hence, x = 1 and y = 1