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# 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 – 2y – 20 = 0

Solution:

11x +15y + 23 = 0 …………………………. (i)

7x – 2y – 20 = 0 …………………………….. (ii)

From (ii)

2y = 7x – 20

⇒ y = (7x −20)/2 ……………………………… (iii)

Putting y in (i) we get,

⇒ 11x + 15((7x−20) / 2) + 23 = 0

⇒ 11x + (105x − 300) / 2 + 23 = 0

Taking 2 as LCM

⇒ (22x + 105x – 300 + 46) = 0

⇒ 127x – 254 = 0

⇒ x = 2

Putting x in (iii)

⇒ y = (7(2) − 20)/2

⇒ y= -3

Therefore, x = 2 and y = -3

### Question 2. 3x – 7y + 10 = 0 and y – 2x – 3 = 0

Solution:

3x – 7y + 10 = 0 …………………………. (i)

y – 2x – 3 = 0 ……………………………….. (ii)

From (ii)

y – 2x – 3 = 0

y = 2x+3 ……………………………… (iii)

Substituting y in (i)

⇒ 3x – 7(2x+3) + 10 = 0

⇒ 3x – 14x – 21 + 10 = 0

⇒ -11x = 11

⇒ x = -1

Putting x in (iii)

⇒ y = 2(-1) + 3

⇒ y= 1

Therefore, x = -1 and y =1

### Question 3. 0.4x + 0.3y = 1.7 and 0.7x – 0.2y = 0.8

Solution:

0.4x + 0.3y = 1.7

0.7x – 0.2y = 0.8

Multiply LHS and RHS by 10

4x + 3y = 17 ……………………….. (i)

7x – 2y = 8 …………………………… (ii)

From (ii)

7x – 2y = 8

⇒ x = (8 + 2y) / 7……………………………… (iii)

Substituting x (i)

⇒ 4[(8 + 2y) / 7] + 3y = 17

Taking 7 as LCM

⇒ 32 + 8y + 21y = (17 × 7)

⇒ 29y = 87

⇒ y = 3

Putting y in (iii)

⇒ x = (8 + 2(3)) / 7

⇒ x = 14/7

⇒ 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

⇒ x + 2y = 1.6…… (i)

7/(x + y/2) = 10

⇒7 = 10(x + y/2)

⇒7 = 10x + 5y

Multiply (i) by 10

10x + 20y = 16 ……………………….. (ii)

10x + 5y = 7 …………………………… (iii)

(ii) – (iii)

⇒ 15y = 9

⇒ y = 3/5

Putting y in (ii)

x = [16 − 20(3/5)] / 10

⇒ (16 – 12) / 10 = 4/10

⇒ x = 2/5

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

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

Solution:

7(y + 3) – 2(x + 2) = 14…………………………. (i)

4(y – 2) + 3(x – 3) = 2……………………………….. (ii)

From (i)

⇒ 7y + 21 – 2x – 4 = 14

⇒ 7y = 14 + 4 – 21 + 2x

⇒ y = (2x – 3) / 7

From (ii)

⇒ 4y – 8 + 3x – 9 = 2

⇒ 4y + 3x – 17 – 2 = 0

⇒ 4y + 3x – 19 = 0 …………….. (iii)

Substituting y in (iii)

4[(2x − 3)/7] + 3x – 19=0

Taking 7 as LCM

⇒ 8x – 12 + 21x – (19 × 17) = 0

⇒ 29x = 145

⇒ x = 5

Putting x and in (iii)

⇒ 4y + 15 -19 = 0

⇒ 4y — 4 = 0

⇒ 4y = 4

⇒ y = 1

Therefore, x = 5 and y = 1

### Question 6. x/7 + y/3 = 5 and x/2 – y/9 = 6

Solution:

x/7 + y/3 = 5…………………………. (i)

x/2 – y/9 = 6………………………………..(ii)

From (i)

Taking 21 as LCM

x/7 + y/3 = 5

⇒3x + 7y = (5×21)

⇒ 3x =105 – 7y

⇒ x = (105 – 7y) / 3……. (iii)

From (ii)

x/2 – y/9 = 6

Taking 18 as LCM

⇒ 9x – 2y = 108 ……………………… (iv)

Substituting x in (iv)

9[(105 − 7y) / 3] – 2y = 108

Taking 3 as LCM

⇒ 945 – 63y – 6y = 324

⇒ 945 – 324 = 69y

⇒ 69y = 621

⇒ y = 9

Putting y in (iv)

x = (105 − 7(9))/3

⇒ x = (105 − 63)/3 = 42/3

⇒ x = 14

Therefore, x = 14 and y = 9

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

Solution:

x/3 + y/4 = 11…………………………. (i)

5x/6 − y/3 = −7……………………………….. (ii)

From (i)

x/3 + y/4 = 11

Taking 12 as LCM

⇒ 4x + 3y = (11×12)

⇒ 4x =132 – 3y

⇒ x = (132 – 3y)/4……. (iii)

From (ii)

5x/6 − y/3 = −7

Takin 6 as LCM

⇒ 5x – 2y = -42 ……………………… (iv)

Substituting x in equation (iv)

⇒ 5[(132 − 3y) / 4] – 2y = -42

Taking 4 as LCM

⇒ 660 – 15y – 8y = -42 x 4

⇒ 660 + 168 = 23y

⇒ 23y = 828

⇒ y = 36

Putting y in (iii)

x = (132 – 3(36))/4

⇒ x = (132 − 108)/4 = 24/4

⇒ x = 6

Therefore, the x = 6 and y = 36

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

Solution:

Taking 1/x = u

4u + 3y = 8…………………… (i)

6u – 4y = -5……………………. (ii)

From (i)

4u = 8 – 3y

⇒ u = (8 − 3y) / 4 …….. (iii)

Substituting u in (ii)

[6(8 − 3y) / 4] – 4y = -5

⇒ [3(8−3y)/2] − 4y = −5

Taking 2 as LCM

⇒ 24 − 9y −8y = −5 × 2

⇒ 24 – 17y = -10

⇒ -17y =- 34

⇒ y = 2

Putting y=2 in (iii)

u = (8 − 3(2)) / 4

⇒ u = (8 − 6)/4

⇒ u = 2/4 = 1/2

⇒ x = 1/u = 2

⇒  x = 2

Therefore, x = 2 and y = 2.

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

Solution:

x + y/2 = 4 ……………………. (i)

2y + x/3 = 5……………………. (ii)

From (i)

x + y/2 = 4

Taking 2 as LCM

⇒ 2x + y = 8

⇒ y = 8 – 2x …..(iii)

From (ii)

Taking 3 as LCM

x + 6y = 15 ……………… (iv)

Substituting y in (iii)

⇒ x + 6(8 – 2x) = 15

⇒ x + 48 – 12x = 15

⇒ -11x = 15 – 48

⇒ -11x = -33

⇒ x = 3

Putting x = 3 in (iii)

y = 8 – (2×3)

⇒  y = 8 – 6 = 2

Therefore, x = 3 and y = 2

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

Solution:

x + 2y = 3/2 …………………. (i)

2x + y = 3/2…………………… (ii)

Multiplying (i) by 4

⇒ 4x + 8y = 6 ………………………. (iii)

Multiplying (ii) by 2

4x + 2y = 3 ……………………………………………………. (iv)

Subtracting (iv) from (iii)

⇒ 6y = 3

⇒ y = 3/6

⇒ y = 1/2

Putting y = 1/2 in (iv)

⇒ 4x + 2(1/2) = 3

⇒ 4x + 1 = 3

⇒ 4x = 2

⇒ x = 2/4 = 1/2

Therefore, x = 1/2 and y = 1/2

### Question 11. √2x – √3y = 0 and √3x − √8y = 0

Solution:

√2x – √3y = 0……………………….. (i)

√3x − √8y = 0……………………….. (ii)

By substitution

√2x = √3y

By transposing

x = √(3/2)y ……………..(iii)

Substituting x in (ii)

√3x − √8y = 0

⇒ √3(√(3/2)y) − √8y = 0

⇒ (3/√2)y – √8y = 0

Taking √2 as the LCM

⇒ 3y – 4y = 0

⇒ -y = 0

⇒ y = 0

Putting the value of y in (iii)

⇒ x = 0

Therefore, x = 0 and y = 0

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

Solution:

3x – (y + 7)/11 + 2 = 10……………….. (i)

2y + (x + 11)/7 = 10…………………….. (ii)

By taking 11 as LCM in (i)

33x – y – 7 + 22 = (10 × 11)

⇒ 33x – y + 15 = 110

⇒ 33x + 15 – 110 = y

⇒ y = 33x – 95………. (iii)

By taking 7 as LCM in (ii)

14y + x + 11 = (10 × 7)

⇒ 14y + x + 11 = 70

⇒ 14y + x = 70 – 11

⇒ 14y + x = 59 …………………….. (iv)

Substituting (iii) in (iv)

14 (33x – 95) + x = 59

⇒ 462x – 1330 + x = 59

⇒ 463x = 1389

Transposing 463

⇒ x = 3

Putting x = 3 in (iii)

⇒ y = 33(3) – 95

⇒ y = 99 – 95

Therefore, y= 4

Therefore, x = 3 and y = 4

### Question 13. 2x – (3/y) = 9 and 3x + (7/y) = 2, y ≠ 0

Solution:

2x – (3/y) = 9……………………………. (i)

3x + (7/y) = 2…………………………… (ii)

Substituting 1/y = u

2x – 3u = 9 ………………………..(iii)

3x + 7u = 2………………………..(iv)

From (iii)

2x = 9 + 3u

⇒ x = (9+3u) / 2

Substituting the value of x in (iv)

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

Taking 2 as LCM

⇒ 27 + 9u + 14u = (2 x 2)

⇒ 27 + 23u = 4

Transposing 27

⇒ 23u = -23

Transposing 23

⇒ u = -1

y = 1/u = -1

Putting u = -1 in (iii)

⇒ x = (9 + 3(-1)) / 2

⇒ x = 6/2

⇒ x = 3

Therefore, x = 3 and y = -1

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