# Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.3 | Set 2

### Question 14. 0.5x + 0.7y = 0.74 and 0.3x + 0.5y = 0.5

Solution:

0.5x + 0.7y = 0.74â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (i)

0.3x â€“ 0.5y = 0.5 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

Multiply LHS and RHS by 100 in (i) and (ii)

50x +70y = 74 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (iii)

30x + 50y = 50 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iv)

From (iii)

50x = 74 â€“ 70y

x = (74âˆ’70y) / 50 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (v)

Substituting x in equation (iv)

30[(74âˆ’70y)/ 50] + 50y = 50

Taking 50 as LCM

â‡’ 2220 â€“ 2100y + 2500y = 2500

Dividing by 10

â‡’ 222 – 210y + 250y = 250

â‡’ -210y + 250y = -222 + 250

â‡’ 40y = 28

Transposing 40

â‡’ y = 0.7

Putting the value of y in (v)

â‡’ x = [74 âˆ’ 70(0.7)] / 50

â‡’ x = (74 – 49) / 50

â‡’ x = 25/ 50 = 1/2

Therefore, x = 0.5

Therefore, x = 0.5 and y = 0.7

### Question 15. 1/(7x) + 1/(6y) = 3 and 1/(2x) â€“ 1/(3y) = 5

Solution:

1/(7x) + 1/(6y) = 3â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

1/(2x) â€“ 1/(3y) = 5â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (ii)

Let 1/x = u and 1/y = v

u/7 + v/6 = 3     â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (iii)

u/2 – v/3 = 5      ……………………………….   (iv)

Taking LCM as 42 in (iii) and 6 in (iv)

6u + 7v = 126………………………………… (v)

3u – 2v = 30………………………………….  (vi)

Multiplying (vi) by 2

6u – 4v = 60…………………. (vii)

Subtracting (vii) from (v)

â‡’ 6u – 6u +7v +4v = 126 – 60

â‡’ 11v = 66

Transposing 11

â‡’ v = 66/11

â‡’ v = 6

â‡’ y = 1/v

â‡’y = 1/6

Putting v in (vii)

â‡’ 6u – 4(6) = 60

â‡’ 6u = 60 + 24

â‡’ 6u = 84

Transposing 6

â‡’ u = 84/6

â‡’ u = 14

â‡’ x = 1/u = 1/14

Therefore, x=1/14 and y=1/6 respectively.

### Question 16. 1/(2x) + 1/(3y) = 2 and 1/(3x) + 1/(2y) = 13/6

Solution:

Let 1/x = u and 1/y = v

u/2 + v/3 = 2 â€¦â€¦â€¦â€¦â€¦â€¦(i)

u/3 + v/2 = 13/6 â€¦â€¦â€¦â€¦â€¦(ii)

Taking 6 as LCM in (i) and (ii)

3u + 2v = 12…………… (iii)

2u + 3v = 13……………. (iv)

From (iii)

â‡’ 3u = 12 – 2v

â‡’ u = (12 – 2v) / 3

Putting in (iv)

â‡’ 2(12 – 2v) / 3 + 3v = 13

â‡’ (24 – 4v) / 3 + 3v = 13

Taking 3 as LCM

â‡’ 24 – 4v +9v = 39

â‡’ 5v = 39 – 24

â‡’ 5v = 15

â‡’ v= 3

â‡’ y = 1/v = 1/3

Putting the value of v in (iii)

â‡’ 3u + 2(3) = 12

â‡’ 3u + 6 = 12

â‡’ 3u = 6

â‡’ u = 2

â‡’ x = 1/u = 1/2

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

### Question 17. 15/u + 2/v = 17 and 1/u + 1/v = 36/5

Solution:

Let 1/u = x and 1/v = y

15x + 2y = 17 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

x + y = 36/5â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (ii)

From (i)

2y = 17 â€“ 15x

â‡’ y = (17 âˆ’ 15x) / 2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (iii)

Substituting (iii) in equation (ii)

â‡’ x + (17 âˆ’ 15x) / 2 = 36/5

Taking 2 as LCM

â‡’ 2x + 17 â€“ 15x = (36 Ã— 2)/ 5

â‡’ -13x = 72/5 â€“ 17

Taking 5 as LCM

â‡’ -13x = (72 – 85) / 5

â‡’  -13x = -13/5

â‡’ x = 1/5

â‡’ u = 1/x = 5

Putting x = 1/5 in (ii)

1/5 + y = 36/5

â‡’ y = 35/5

â‡’ y = 7

â‡’ v = 1/y = 1/7

Therefore, u = 5 and v = 1/7

### Question 18. 3/x â€“ 1/y = âˆ’9 and 2/x + 3/y = 5

Solution:

Let 1/x = u and 1/y = v

3u â€“ v = -9â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

2u + 3v = 5 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

Multiplying (i) by 3

9u â€“ 3v = -27 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (iii)

2u + 3v = 5 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iv)

9u + 2u â€“ 3v + 3v = -27 + 5

â‡’ 11u = -22

â‡’ u = -2

Putting u = -2 in (iv)

2(-2) + 3v = 5

â‡’ -4 + 3v = 5

â‡’ 3v = 9

â‡’ v = 3

Therefore, x = 1/u = âˆ’1/2, y = 1/v = 1/3

### Question 19. 2/x + 5/y = 1 and 60/x + 40/y = 19

Solution:

Let 1/x = u and 1/y = v

2u + 5v = 1â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

60u + 40v = 19 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

Multiplying equation by 8

16u + 40v = 8 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (iii)

60u + 40v = 19 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iv)

Subtracting equation (iii) from (iv)

60u â€“ 16u + 40v â€“ 40v = 19 â€“ 8

â‡’ 44u = 11

â‡’ u = 11/44

â‡’ u = 1/4

Putting u = 1/4 in (iv)

60(1/4) + 40v = 19

â‡’ 15 + 40v = 19

â‡’ 40v = 4

â‡’ v = 4/ 40 = 1/10

x = 1/u = 4

y = 1/v = 10

### Question 20. 1/(5x) + 1/(6y) = 12 and 1/(3x) â€“ 3/(7y) = 8

Solution:

Let 1/x = u and 1/y = v

u/5 + v/6 = 12â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

u/3 â€“ 3v/7 = 8â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

Taking LCM for both equations

6u + 5v = 360â€¦â€¦â€¦. (iii)

7u â€“ 9v = 168â€¦â€¦â€¦.. (iv)

Subtracting (iii) from (iv)

7u â€“ 9v â€“ (6u + 5v) = 168 â€“ 360

â‡’ u â€“ 14v = -192

â‡’ u = (14v â€“ 192)â€¦â€¦â€¦. (v)

Using (v) in equation (iii)

6(14v â€“ 192) + 5v = 360

â‡’ 84v -1152 + 5v = 360

â‡’ 89v = 1512

â‡’ v = 1512/89

â‡’ y = 1/v = 89/1512

Now, substituting v in equation (v)

u = 14 x (1512/89) â€“ 192

â‡’ u = 21168/89 – 192

â‡’ u = (21168 – 17088) / 89

â‡’ u = 4080/89

â‡’ x = 1/u = 89/ 4080

Therefore, x = 89/4080 and y = 89/ 1512

### Question 21. 4/x + 3y = 14 and 3/x â€“ 4y = 23

Solution:

Taking 1/x = u

4u + 3y = 14â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

3u â€“ 4y = 23â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

Adding (i) and (ii), we get

4u + 3y + 3u â€“ 4y = 14 + 23

â‡’ 7u â€“ y = 37

â‡’ y = 7u â€“ 37â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iii)

Putting (iii) in (i),

4u + 3(7u â€“ 37) = 14

â‡’ 4u + 21u â€“ 111 = 14

â‡’ 25u = 125

â‡’ u = 5

â‡’ x = 1/u = 1/5

Putting u= 5 in (iii)

y = 7(5) â€“ 37

â‡’ y = -2

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

### Question 22.4/x + 5y = 7 and 3/x + 4y = 5

Solution:

Taking 1/x = u

4u + 5y = 7â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

3u + 4y = 5â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

Multiplying (i) by 4

â‡’ 16u + 20y = 28 …….. (iii)

Multiplying (ii) by 5

â‡’ 15u + 20y = 25 …….. (iv)

(iii) – (iv)

â‡’ 16u – 15u + 20y – 20y = 28â€“25

â‡’ u = 3

â‡’ x = 1/u = 1/3

Putting u in (i)

â‡’ 12 + 5y = 7

â‡’ 5y = -5

â‡’ y = -1

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

### Question 23.2/x + 3/y = 13 and 5/x â€“ 4/y = -2

Solution:

Let 1/x = u and 1/y = v

2u + 3v = 13â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

5u â€“ 4v = -2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (ii)

Multiplying (i) by 4

â‡’ 8u + 12v = 52 ……….. (iii)

Multiplying (ii) by 3

â‡’ 15u – 12v = -6 ……….. (iv)

â‡’ 15u + 8u + 12v – 12v = 52 â€“ 6

â‡’ 23u = 46

â‡’ u = 46/23

â‡’ u = 2

Putting u = 2 in (i)

â‡’ 4 + 3v = 13

â‡’ 3v = 9

â‡’ v = 3

â‡’ x = 1/u = 1/2

â‡’ y = 1/v = 1/3

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

### Question 24. 2/âˆšx + 3/âˆšy = 2 and 4/âˆšx â€“ 9/âˆšy = -1

Solution:

Let 1/x = u and 1/y = v

2u + 3v = 2â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

4u â€“ 9v = -1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (ii)

Multiplying (i) by 3

â‡’ 6u + 9v = 6 …….. (iii)

6u + 9v + 4u â€“ 9v = 6 â€“ 1

â‡’ 10u = 5

â‡’ u = 1/2

Substituting u = 1/2 in (i)

2(1/2) + 3v = 2

â‡’ 3v = 2 â€“ 1

â‡’ v = 1/3

1/âˆšx = u

â‡’ x = 1/u2

â‡’ x = 1/(1/2)2 = 4

1/âˆšy = v

â‡’ y = 1/v2

â‡’ y = 1/(1/3)2 = 9

Therefore, x = 4 and y = 9.

### Question 25. (x + y)/xy = 2 and (x â€“ y)/xy = 6

Solution:

(x + y)/xy = 2

â‡’ 1/y + 1/x = 2â€¦â€¦. (i)

(x â€“ y)/xy = 6

â‡’ 1/y â€“ 1/x = 6â€¦â€¦â€¦(ii)

Let 1/x = u and 1/y = v

v + u = 2â€¦â€¦. (iii)

v â€“ u = 6â€¦â€¦..(iv)

2v = 8

â‡’ v = 4

â‡’ y = 1/v = 1/4

Substituting v = 4 in (iii)

4 + u = 2

â‡’ u = -2

â‡’ x = 1/u = -1/2

Therefore, x = -1/2 and y = 1/4

### Question 26. 2/x + 3/y = 9/xy and 4/x + 9/y = 21/xy

Solution:

Taking LCM as xy

(2y + 3x)/ xy = 9/xy

â‡’ 3x + 2y = 9â€¦â€¦â€¦. (i)

(4y + 9x)/ xy = 21/xy

â‡’ 9x + 4y = 21â€¦â€¦â€¦(ii)

Multiplying (i) by 2

â‡’ 6x + 4y = 18 …….. (iii)

(ii) – (iii)

â‡’ 9x – 6x + 4y – 4y = 21â€“18

â‡’ 3x = 3

â‡’ x = 1

Putting x = 1 in (i)

3(1) + 2y = 9

â‡’ y = 6/2

â‡’ y = 3

Therefore, x = 1 and y = 3

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