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

Solve the following system of equations:

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

Solution:

0.5x + 0.7y = 0.74……………………… (i)

0.3x – 0.5y = 0.5 ………………………….. (ii)

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

50x +70y = 74 ……………………….. (iii)

30x + 50y = 50 …………………………… (iv)

From (iii)

50x = 74 – 70y

x = (74−70y) / 50 ……………………………… (v)

Substituting x in equation (iv)

30[(74−70y)/ 50] + 50y = 50

Taking 50 as LCM

⇒ 2220 – 2100y + 2500y = 2500

Dividing by 10

⇒ 222 – 210y + 250y = 250

⇒ -210y + 250y = -222 + 250

⇒ 40y = 28

Transposing 40

⇒ y = 0.7

Putting the value of y in (v)

⇒ x = [74 − 70(0.7)] / 50

⇒ x = (74 – 49) / 50 

⇒ 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) – 1/(3y) = 5

Solution:

1/(7x) + 1/(6y) = 3………………………….. (i)

1/(2x) – 1/(3y) = 5……………………………. (ii)

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

u/7 + v/6 = 3     ……………………………. (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)

⇒ 6u – 6u +7v +4v = 126 – 60

⇒ 11v = 66

Transposing 11

⇒ v = 66/11

⇒ v = 6

⇒ y = 1/v

⇒y = 1/6

Putting v in (vii)

⇒ 6u – 4(6) = 60

⇒ 6u = 60 + 24

⇒ 6u = 84

Transposing 6

⇒ u = 84/6

⇒ u = 14

⇒ 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 ………………(i)

u/3 + v/2 = 13/6 ……………(ii)

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

3u + 2v = 12…………… (iii)

2u + 3v = 13……………. (iv)

From (iii)

⇒ 3u = 12 – 2v

⇒ u = (12 – 2v) / 3    

Putting in (iv)

⇒ 2(12 – 2v) / 3 + 3v = 13

⇒ (24 – 4v) / 3 + 3v = 13

Taking 3 as LCM

⇒ 24 – 4v +9v = 39

⇒ 5v = 39 – 24

⇒ 5v = 15

⇒ v= 3

⇒ y = 1/v = 1/3

Putting the value of v in (iii)

⇒ 3u + 2(3) = 12

⇒ 3u + 6 = 12

⇒ 3u = 6

⇒ u = 2

⇒ 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 ………………………….. (i)

x + y = 36/5………………………. (ii)

From (i)

2y = 17 – 15x

⇒ y = (17 − 15x) / 2 …………………. (iii)

Substituting (iii) in equation (ii)

⇒ x + (17 − 15x) / 2 = 36/5

Taking 2 as LCM

⇒ 2x + 17 – 15x = (36 × 2)/ 5 

⇒ -13x = 72/5 – 17

Taking 5 as LCM

⇒ -13x = (72 – 85) / 5

⇒  -13x = -13/5

⇒ x = 1/5

⇒ u = 1/x = 5

Putting x = 1/5 in (ii)

1/5 + y = 36/5

⇒ y = 35/5

⇒ y = 7

⇒ v = 1/y = 1/7

Therefore, u = 5 and v = 1/7

Question 18. 3/x – 1/y = −9 and 2/x + 3/y = 5

Solution:

Let 1/x = u and 1/y = v

3u – v = -9…………………..(i)

2u + 3v = 5 ……………………….(ii)

Multiplying (i) by 3

9u – 3v = -27 ………………………….. (iii)

2u + 3v = 5 ……………………………… (iv)

Adding equation (iii) and (iv)

9u + 2u – 3v + 3v = -27 + 5

⇒ 11u = -22

⇒ u = -2

Putting u = -2 in (iv)

2(-2) + 3v = 5

⇒ -4 + 3v = 5

⇒ 3v = 9

⇒ v = 3

Therefore, x = 1/u = −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 = 1…………………..(i)

60u + 40v = 19 ……………………….(ii)

Multiplying equation by 8

16u + 40v = 8 ………………………….. (iii)

60u + 40v = 19 ……………………………… (iv)

Subtracting equation (iii) from (iv)

60u – 16u + 40v – 40v = 19 – 8

⇒ 44u = 11

⇒ u = 11/44

⇒ u = 1/4

Putting u = 1/4 in (iv)

60(1/4) + 40v = 19

⇒ 15 + 40v = 19

⇒ 40v = 4

⇒ v = 4/ 40 = 1/10

x = 1/u = 4 

y = 1/v = 10

Question 20. 1/(5x) + 1/(6y) = 12 and 1/(3x) – 3/(7y) = 8

Solution:

Let 1/x = u and 1/y = v

u/5 + v/6 = 12…………………..(i)

u/3 – 3v/7 = 8……………………….(ii)

Taking LCM for both equations

6u + 5v = 360………. (iii)

7u – 9v = 168……….. (iv)

Subtracting (iii) from (iv)

7u – 9v – (6u + 5v) = 168 – 360

⇒ u – 14v = -192

⇒ u = (14v – 192)………. (v)

Using (v) in equation (iii)

6(14v – 192) + 5v = 360

⇒ 84v -1152 + 5v = 360

⇒ 89v = 1512

⇒ v = 1512/89

⇒ y = 1/v = 89/1512

Now, substituting v in equation (v)

u = 14 x (1512/89) – 192

⇒ u = 21168/89 – 192

⇒ u = (21168 – 17088) / 89

⇒ u = 4080/89

⇒ x = 1/u = 89/ 4080

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

Question 21. 4/x + 3y = 14 and 3/x – 4y = 23

Solution:

Taking 1/x = u

4u + 3y = 14…………………….. (i)

3u – 4y = 23…………………….. (ii)

Adding (i) and (ii), we get

4u + 3y + 3u – 4y = 14 + 23

⇒ 7u – y = 37

⇒ y = 7u – 37……………………… (iii)

Putting (iii) in (i),

4u + 3(7u – 37) = 14

⇒ 4u + 21u – 111 = 14

⇒ 25u = 125

⇒ u = 5

⇒ x = 1/u = 1/5

Putting u= 5 in (iii)

y = 7(5) – 37

⇒ 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 = 7…………………….. (i)

3u + 4y = 5…………………….. (ii)

Multiplying (i) by 4

⇒ 16u + 20y = 28 …….. (iii)

Multiplying (ii) by 5

⇒ 15u + 20y = 25 …….. (iv)

(iii) – (iv)

⇒ 16u – 15u + 20y – 20y = 28–25

⇒ u = 3

⇒ x = 1/u = 1/3

Putting u in (i)

⇒ 12 + 5y = 7

⇒ 5y = -5

⇒ y = -1

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

Question 23. 2/x + 3/y = 13 and 5/x – 4/y = -2

Solution:

Let 1/x = u and 1/y = v

2u + 3v = 13………………….. (i)

5u – 4v = -2 ………………………. (ii)

Multiplying (i) by 4

⇒ 8u + 12v = 52 ……….. (iii)

Multiplying (ii) by 3

⇒ 15u – 12v = -6 ……….. (iv)

Adding (iii) and (iv)

⇒ 15u + 8u + 12v – 12v = 52 – 6

⇒ 23u = 46

⇒ u = 46/23

⇒ u = 2

Putting u = 2 in (i)

⇒ 4 + 3v = 13

⇒ 3v = 9

⇒ v = 3

⇒ x = 1/u = 1/2

⇒ y = 1/v = 1/3

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

Question 24. 2/√x + 3/√y = 2 and 4/√x – 9/√y = -1

Solution:

Let 1/x = u and 1/y = v

2u + 3v = 2………………….. (i)

4u – 9v = -1 ………………………. (ii)

Multiplying (i) by 3

⇒ 6u + 9v = 6 …….. (iii)

Adding (ii) and (iii)

6u + 9v + 4u – 9v = 6 – 1

⇒ 10u = 5

⇒ u = 1/2

Substituting u = 1/2 in (i)

2(1/2) + 3v = 2

⇒ 3v = 2 – 1

⇒ v = 1/3

1/√x = u 

⇒ x = 1/u²

⇒ x = 1/(1/2)² = 4

1/√y = v 

⇒ y = 1/v²

⇒ y = 1/(1/3)² = 9

Therefore, x = 4 and y = 9.

Question 25. (x + y)/xy = 2 and (x – y)/xy = 6

Solution:

(x + y)/xy = 2 

⇒ 1/y + 1/x = 2……. (i)

(x – y)/xy = 6 

⇒ 1/y – 1/x = 6………(ii)

Let 1/x = u and 1/y = v

v + u = 2……. (iii)

v – u = 6……..(iv)

Adding (iii) and (iv)

2v = 8

⇒ v = 4

⇒ y = 1/v = 1/4

Substituting v = 4 in (iii)

4 + u = 2

⇒ u = -2

⇒ 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 

⇒ 3x + 2y = 9………. (i)

(4y + 9x)/ xy = 21/xy 

⇒ 9x + 4y = 21………(ii)

Multiplying (i) by 2

⇒ 6x + 4y = 18 …….. (iii)

(ii) – (iii)

⇒ 9x – 6x + 4y – 4y = 21–18

⇒ 3x = 3

⇒ x = 1

Putting x = 1 in (i)

3(1) + 2y = 9

⇒ y = 6/2

⇒ y = 3

Therefore, x = 1 and y = 3