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# 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 - 2) = 1

(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 - 2) = 1

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 - 2) = 1 ⇒ 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

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 3

**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 - 2) = 1 (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

**Summary:**

The following pairs of equations are solved 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 - 2) = 1 (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 and the values are as follows: (i) x = 1/2 and y = 1/3, (ii) x = 4 and y = 9, (iii) x = 1/5 and y = - 2, (iv) x = 4 and y = 5, (v) x = 1 and y = 1, (vi) x = 1 and y = 2, (vii) x = 3 and y = 2, (viii) x = 1 and y = 1

**☛ Related Questions:**

- Formulate the following problems as a pair of equations, and hence find their solutions: (i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone. (iii) Roohi travels 300 km to her home partly by train and partly by She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
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