# Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.3 | Set 1

### Question 1. 11x + 15y + 23 = 0 and 7x â€“ 2y â€“ 20 = 0

Solution:

11x +15y + 23 = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

7x â€“ 2y â€“ 20 = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

From (ii)

2y = 7x â€“ 20

â‡’ y = (7x âˆ’20)/2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iii)

Putting y in (i) we get,

â‡’ 11x + 15((7xâˆ’20) / 2) + 23 = 0

â‡’ 11x + (105x âˆ’ 300) / 2 + 23 = 0

Taking 2 as LCM

â‡’ (22x + 105x â€“ 300 + 46) = 0

â‡’ 127x â€“ 254 = 0

â‡’ x = 2

Putting x in (iii)

â‡’ y = (7(2) âˆ’ 20)/2

â‡’ y= -3

Therefore, x = 2 and y = -3

### Question 2. 3x â€“ 7y + 10 = 0 and y â€“ 2x â€“ 3 = 0

Solution:

3x â€“ 7y + 10 = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

y â€“ 2x â€“ 3 = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

From (ii)

y â€“ 2x â€“ 3 = 0

y = 2x+3 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iii)

Substituting y in (i)

â‡’ 3x â€“ 7(2x+3) + 10 = 0

â‡’ 3x â€“ 14x â€“ 21 + 10 = 0

â‡’ -11x = 11

â‡’ x = -1

Putting x in (iii)

â‡’ y = 2(-1) + 3

â‡’ y= 1

Therefore, x = -1 and y =1

### Question 3. 0.4x + 0.3y = 1.7 and 0.7x â€“ 0.2y = 0.8

Solution:

0.4x + 0.3y = 1.7

0.7x â€“ 0.2y = 0.8

Multiply LHS and RHS by 10

4x + 3y = 17 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

7x â€“ 2y = 8 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (ii)

From (ii)

7x â€“ 2y = 8

â‡’ x = (8 + 2y) / 7â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iii)

Substituting x (i)

â‡’ 4[(8 + 2y) / 7] + 3y = 17

Taking 7 as LCM

â‡’ 32 + 8y + 21y = (17 Ã— 7)

â‡’ 29y = 87

â‡’ y = 3

Putting y in (iii)

â‡’ x = (8 + 2(3)) / 7

â‡’ x = 14/7

â‡’ x = 2

Therefore, x = 2 and y = 3

### Question 4. x/2 + y = 0.8 and 7/(x + y/2) = 10

Solution:

x/2 + y = 0.8

Taking 2 as LCM

â‡’ x + 2y = 1.6â€¦â€¦ (i)

7/(x + y/2) = 10

â‡’7 = 10(x + y/2)

â‡’7 = 10x + 5y

Multiply (i) by 10

10x + 20y = 16 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

10x + 5y = 7 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iii)

(ii) – (iii)

â‡’ 15y = 9

â‡’ y = 3/5

Putting y in (ii)

x = [16 âˆ’ 20(3/5)] / 10

â‡’ (16 â€“ 12) / 10 = 4/10

â‡’ x = 2/5

Therefore, x = 2/5 and y = 3/5

### Question 5. 7(y + 3) â€“ 2(x + 2) = 14 and 4(y â€“ 2) + 3(x â€“ 3) = 2

Solution:

7(y + 3) â€“ 2(x + 2) = 14â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

4(y – 2) + 3(x – 3) = 2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

From (i)

â‡’ 7y + 21 â€“ 2x â€“ 4 = 14

â‡’ 7y = 14 + 4 â€“ 21 + 2x

â‡’ y = (2x â€“ 3) / 7

From (ii)

â‡’ 4y â€“ 8 + 3x â€“ 9 = 2

â‡’ 4y + 3x â€“ 17 â€“ 2 = 0

â‡’ 4y + 3x â€“ 19 = 0 â€¦â€¦â€¦â€¦â€¦.. (iii)

Substituting y in (iii)

4[(2x âˆ’ 3)/7] + 3x â€“ 19=0

Taking 7 as LCM

â‡’ 8x â€“ 12 + 21x â€“ (19 Ã— 17) = 0

â‡’ 29x = 145

â‡’ x = 5

Putting x and in (iii)

â‡’ 4y + 15 -19 = 0

â‡’ 4y â€” 4 = 0

â‡’ 4y = 4

â‡’ y = 1

Therefore, x = 5 and y = 1

### Question 6. x/7 + y/3 = 5 and x/2 â€“ y/9 = 6

Solution:

x/7 + y/3 = 5â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

x/2 â€“ y/9 = 6â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

From (i)

Taking 21 as LCM

x/7 + y/3 = 5

â‡’3x + 7y = (5Ã—21)

â‡’ 3x =105 â€“ 7y

â‡’ x = (105 â€“ 7y) / 3â€¦â€¦. (iii)

From (ii)

x/2 â€“ y/9 = 6

Taking 18 as LCM

â‡’ 9x â€“ 2y = 108 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iv)

Substituting x in (iv)

9[(105 âˆ’ 7y) / 3] â€“ 2y = 108

Taking 3 as LCM

â‡’ 945 â€“ 63y â€“ 6y = 324

â‡’ 945 â€“ 324 = 69y

â‡’ 69y = 621

â‡’ y = 9

Putting y in (iv)

x = (105 âˆ’ 7(9))/3

â‡’ x = (105 âˆ’ 63)/3 = 42/3

â‡’ x = 14

Therefore, x = 14 and y = 9

### Question 7. x/3 + y/4 = 11 and 5x/6 âˆ’ y/3 = âˆ’7

Solution:

x/3 + y/4 = 11â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

5x/6 âˆ’ y/3 = âˆ’7â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

From (i)

x/3 + y/4 = 11

Taking 12 as LCM

â‡’ 4x + 3y = (11Ã—12)

â‡’ 4x =132 â€“ 3y

â‡’ x = (132 â€“ 3y)/4â€¦â€¦. (iii)

From (ii)

5x/6 âˆ’ y/3 = âˆ’7

Takin 6 as LCM

â‡’ 5x â€“ 2y = -42 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iv)

Substituting x in equation (iv)

â‡’ 5[(132 âˆ’ 3y) / 4] â€“ 2y = -42

Taking 4 as LCM

â‡’ 660 â€“ 15y â€“ 8y = -42 x 4

â‡’ 660 + 168 = 23y

â‡’ 23y = 828

â‡’ y = 36

Putting y in (iii)

x = (132 â€“ 3(36))/4

â‡’ x = (132 âˆ’ 108)/4 = 24/4

â‡’ x = 6

Therefore, the x = 6 and y = 36

### Question 8. 4/x + 3y = 8 and 6/x âˆ’ 4y = âˆ’5

Solution:

Taking 1/x = u

4u + 3y = 8â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (i)

6u â€“ 4y = -5â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (ii)

From (i)

4u = 8 â€“ 3y

â‡’ u = (8 âˆ’ 3y) / 4 â€¦â€¦.. (iii)

Substituting u in (ii)

[6(8 âˆ’ 3y) / 4] â€“ 4y = -5

â‡’ [3(8âˆ’3y)/2] âˆ’ 4y = âˆ’5

Taking 2 as LCM

â‡’ 24 âˆ’ 9y âˆ’8y = âˆ’5 Ã— 2

â‡’ 24 â€“ 17y = -10

â‡’ -17y =- 34

â‡’ y = 2

Putting y=2 in (iii)

u = (8 âˆ’ 3(2)) / 4

â‡’ u = (8 âˆ’ 6)/4

â‡’ u = 2/4 = 1/2

â‡’ x = 1/u = 2

â‡’  x = 2

Therefore, x = 2 and y = 2.

### Question 9. x + y/2 = 4 and 2y + x/3 = 5

Solution:

x + y/2 = 4 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

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

From (i)

x + y/2 = 4

Taking 2 as LCM

â‡’ 2x + y = 8

â‡’ y = 8 â€“ 2x â€¦..(iii)

From (ii)

Taking 3 as LCM

x + 6y = 15 â€¦â€¦â€¦â€¦â€¦â€¦ (iv)

Substituting y in (iii)

â‡’ x + 6(8 â€“ 2x) = 15

â‡’ x + 48 â€“ 12x = 15

â‡’ -11x = 15 â€“ 48

â‡’ -11x = -33

â‡’ x = 3

Putting x = 3 in (iii)

y = 8 â€“ (2Ã—3)

â‡’  y = 8 â€“ 6 = 2

Therefore, x = 3 and y = 2

### Question 10. x + 2y = 3/2 and 2x + y = 3/2

Solution:

x + 2y = 3/2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

2x + y = 3/2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (ii)

Multiplying (i) by 4

â‡’ 4x + 8y = 6 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (iii)

Multiplying (ii) by 2

4x + 2y = 3 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (iv)

Subtracting (iv) from (iii)

â‡’ 6y = 3

â‡’ y = 3/6

â‡’ y = 1/2

Putting y = 1/2 in (iv)

â‡’ 4x + 2(1/2) = 3

â‡’ 4x + 1 = 3

â‡’ 4x = 2

â‡’ x = 2/4 = 1/2

Therefore, x = 1/2 and y = 1/2

### Question 11. âˆš2x â€“ âˆš3y = 0 and âˆš3x âˆ’ âˆš8y = 0

Solution:

âˆš2x â€“ âˆš3y = 0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

âˆš3x âˆ’ âˆš8y = 0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

By substitution

âˆš2x = âˆš3y

By transposing

x = âˆš(3/2)y â€¦â€¦â€¦â€¦â€¦..(iii)

Substituting x in (ii)

âˆš3x âˆ’ âˆš8y = 0

â‡’ âˆš3(âˆš(3/2)y) âˆ’ âˆš8y = 0

â‡’ (3/âˆš2)y â€“ âˆš8y = 0

Taking âˆš2 as the LCM

â‡’ 3y â€“ 4y = 0

â‡’ -y = 0

â‡’ y = 0

Putting the value of y in (iii)

â‡’ x = 0

Therefore, x = 0 and y = 0

### Question 12. 3x â€“ (y + 7)/11 + 2 = 10 and 2y + (x + 11)/7 = 10

Solution:

3x â€“ (y + 7)/11 + 2 = 10â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

2y + (x + 11)/7 = 10â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

By taking 11 as LCM in (i)

33x â€“ y â€“ 7 + 22 = (10 Ã— 11)

â‡’ 33x â€“ y + 15 = 110

â‡’ 33x + 15 â€“ 110 = y

â‡’ y = 33x â€“ 95â€¦â€¦â€¦. (iii)

By taking 7 as LCM in (ii)

14y + x + 11 = (10 Ã— 7)

â‡’ 14y + x + 11 = 70

â‡’ 14y + x = 70 â€“ 11

â‡’ 14y + x = 59 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (iv)

Substituting (iii) in (iv)

14 (33x â€“ 95) + x = 59

â‡’ 462x â€“ 1330 + x = 59

â‡’ 463x = 1389

Transposing 463

â‡’ x = 3

Putting x = 3 in (iii)

â‡’ y = 33(3) â€“ 95

â‡’ y = 99 – 95

Therefore, y= 4

Therefore, x = 3 and y = 4

### Question 13. 2x â€“ (3/y) = 9 and 3x + (7/y) = 2, y â‰  0

Solution:

2x â€“ (3/y) = 9â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

3x + (7/y) = 2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (ii)

Substituting 1/y = u

2x â€“ 3u = 9 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

3x + 7u = 2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iv)

From (iii)

2x = 9 + 3u

â‡’ x = (9+3u) / 2

Substituting the value of x in (iv)

3[(9 + 3u)/2] + 7u = 2

Taking 2 as LCM

â‡’ 27 + 9u + 14u = (2 x 2)

â‡’ 27 + 23u = 4

Transposing 27

â‡’ 23u = -23

Transposing 23

â‡’ u = -1

y = 1/u = -1

Putting u = -1 in (iii)

â‡’ x = (9 + 3(-1)) / 2

â‡’ x = 6/2

â‡’ x = 3

Therefore, x = 3 and y = -1

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