# RS Aggarwal Solutions Chapter 3 Linear Equations in Two Variables Exercise - 3B Class 10 Maths

 Chapter Name RS Aggarwal Chapter 3 Linear Equations in Two Variables Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 3AExercise 3CExercise 3DExercise 3EExercise 3F Related Study NCERT Solutions for Class 10 Maths

### Exercise 3B Solutions

1. Solve for x and y:

x + y = 3, 4x – 3y = 26

Solution

The given system of equation is:

x + y = 3 ….(i)

4x – 3y = 26 …(ii)

On multiplying (i) by 3, we get:

3x + 3y = 9 ….(iii)

On adding (ii) and (iii), we get:

7x = 35

⇒ x = 5

On substituting the value of x = 5 in (i), we get:

5 + y = 3

⇒ y = (3 – 5) = -2

Hence, the solution is x = 5 and y = -2

2. Solve for x and y:

x – y = 3, x/3 + y/2 = 6

Solution

The given system of equations is

x – y = 3  ...(i)

x/3 + y/2 = …(ii)

From (i), write y in terms of x to get

y = x – 3

Substituting y = x – 3 in (ii), we get

x/3 + (x – 3)/2 =

⇒ 2x + 3(x – 3) = 36

⇒2x + 3x – 9 = 36

⇒ x = 45/5 =

Now, substituting x = 9 in (i), we have

9 – y = 3

⇒ y = 9 – 3 = 6

Hence, x = 9 and y = 6.

3. Solve for x and y:

2x + 3y = 0, 3x + 4y = 5

Solution

The given system of equation is:

2x + 3y = 0 …(i)

3x + 4y = 5 …(ii)

On multiplying (i) by 4 and (ii) by 3, we get:

8x + 12y = 0 …(iii)

9x + 12y = 15 ….(iv)

On subtracting (iii) from (iv) we get:

x = 15

On substituting the value of x = 15 in (i), we get:

30 + 3y = 0

⇒ 3y = -30

⇒ y = -10

Hence, the solution is x = 15 and y = -10.

4. Solve for x and y:

2x - 3y = 13, 7x - 2y = 20

Solution

The given system of equation is:

2x - 3y = 13  …(i)

7x - 2y = 20 …(ii)

On multiplying (i) by 2 and (ii) by 3, we get:

4x - 6y = 26 …(iii)

21x - 6y = 60 ….(iv)

On subtracting (iii) from (iv) we get:

17x = (60 – 26) = 34

⇒ x = 2

On substituting the value of x = 2 in (i), we get:

4 – 3y = 13

⇒ 3y = (4 – 13) = -9

⇒ y = -3

Hence, the solution is x = 2 and y = -3.

5. Solve for x and y:

3x - 5y - 19 = 0, -7x + 3y + 1 = 0

Solution

The given system of equation is:

3x - 5y - 19 = 0  …(i)

-7x + 3y + 1 = 0 …(ii)

On multiplying (i) by 3 and (ii) by 5, we get:

9x - 15y = 57 …(iii)

-35x + 15y = -5 ….(iv)

On subtracting (iii) from (iv) we get:

-26x = (57 – 5) = 52

⇒ x = -2

On substituting the value of x = -2 in (i), we get:

-6 – 5y – 19 = 0

⇒ 5y = (– 6 – 19) = -25

⇒ y = -5

Hence, the solution is x = -2 and y = -5.

6. Solve for x and y:

2x – y + 3 = 0, 3x – 7y + 10 = 0

Solution

The given system of equation is:

2x – y + 3 = 0  ….(i)

3x – 7y + 10 = 0 …(ii)

From (i), write y in terms of x to get

y = 2x + 3

Substituting y = 2x + 3 in (ii), we get

3x – 7(2x + 3) + 10 = 0

⇒ 3x – 14x – 21 + 10 = 0

⇒ -7x = 21 – 10 = 11

x = –11/7

Now substituting x = -11/7 in (i), we have

-22/7 – y + 3 = 0

y = 3 – 22/7 = -1/7

Hence, x = -11/7 and y = -1/77.

7. Solve for x and y:

9x - 2y = 108, 3x + 7y = 105

Solution

The given system of equation can be written as:

9x - 2y = 108 …(i)

3x + 7y = 105 …(ii)

On multiplying (i) by 7 and (ii) by 2, we get:

63x + 6x = 108 × 7 + 105 × 2

⇒ 69x = 966

⇒ x = 966/69 = 14

Now, substituting x = 14 in (i), we get:

9 × 14 – 2y = 108

⇒ 2y = 126 – 108

⇒ y = 18/2 = 9

Hence, x = 14 and y = 9.

8. Solve for x and y:

x/3 + y/4= 11, 5x/6 – y/3 + 7 = 0

Solution

The given equations are:

x/3 + y/4 = 11

⇒ 4x + 3y = 132 ….(i)

and 5x/6 – y/3 + 7 =

⇒ 5x – 2y = -42 ...(ii)

On multiplying (i) by 2 and (ii) by 3, we get:

8x + 6y = 264 ...(iii)

15x – 6y = -126 ….(iv)

On adding (iii) and (iv), we get:

23x = 138

⇒ x = 6

On substituting x = 6 in (i), we get:

24 + 3y = 132

⇒ 3y = (132 – 24) = 108

⇒ y = 36

Hence, the solution is x = 6 and y = 36.

9. Solve for x and y:

4x - 3y = 8, 6x - y = 29/3

Solution

The given system of equation is:

4x - 3y = 8 …(i)

6x - y = 29/3 …(ii)

On multiplying (ii) by 3, we get:

18x – 3y = 29  ….(iii)

On subtracting (iii) from (i) we get:

-14x = -21

x = 21/14 = 3/2

Now, substituting the value of x = 3/2 in (i), we get:

4 × 3/2 – 3y = 8

⇒ 6 – 3y = 8

⇒ 3y = 6 – 8 = -2

y = -2/3

Hence, the solution x = 3/2 and y = -2/3.

10. Solve for x and y:

2x – 3y/4 = 3, 5x = 2y + 7

Solution

The given equations are:

2x – 3y/4 = 3 ...(i)

5x = 2y + 7 ...(ii)

On multiplying (i) by 2 and (ii) by 3/4, we get:

4x – 3/2y = 6 ...(iii)

15/4x = 3/2y + 21/4 ...(iv)

On subtracting (iii) and (iv), we get:

-1/4x = -3/4

⇒ x = 3

On substituting x = 3 in (i), we get:

2 × 3 – 3y/4 = 3

⇒ 3y/4 = (6 – 3) = 3

⇒ y = 3 × 4 = 4

Hence, the solution is x = 3 and y = 4.

11. Solve for x and y:

2x + 5y = 8/3, 3x – 2y =

Solution

The given equations are:

2x – 5y = 8/3 ...(i)

3x – 2y = 5/6 ...(ii)

On multiplying (i) by 2 and (ii) by 5, we get:

4x - 10y = 16/3 ...(iii)

15x – 10y = 25/6 …(iv)

On adding (iii) and (iv), we get:

19x = 57/6

⇒ x = 57/(6 × 19) = 3/6 = 1/2

On substituting x = 1/2 in (i), we get:

2 × 1/2 + 5y = 8

⇒ 5y = (8/3− 1) = 5/3

⇒ y = 5/(3 × 5) = 13

Hence, the solution is x = 1/2 and y = 1/3.

12. Solve for x and y:

2x + 3y + 1 = 0

(7 – 4x)/3 = y

Solution

The given equations are:

(7 – 4x)/3 = y

⇒ 4x + 3y = 7 …..(i)

and 2x + 3y + 1 = 0

⇒ 2x + 3y = -1 ….(ii)

On subtracting (ii) from (i), we get:

2x = 8

⇒ x = 4

On substituting x = 4 in (i), we get:

16x + 3y = 7

⇒ 3y = (7 – 16) = -9

⇒ y = -3

Hence, the solution is x = 4 and y = -3.

13. Solve for x and y:

0.4x + 0.3y = 1.7, 0.7x – 0.2y = 0.8.

Solution

The given system of equations is

0.4x + 0.3y = 1.7 ….(i)

0.7x – 0.2y = 0.8 ….(ii)

Multiplying (i) by 0.2 and (ii) by 0.3 and adding them, we get

0.8x + 2.1x = 3.4 + 2.4

⇒ 2.9x = 5.8

⇒ x = (5.8)/(2.9) = 2

Now, substituting x = 2 in (i), we have

0.4 × 2 + 0.3y = 1.7

⇒ 0.3y = 1.7 – 0.8

⇒ y = 0.9/0.3 = 3

Hence, x = 2 and y = 3.

14. Solve for x and y:

0.3x + 0.5y = 0.5, 0.5x + 0.7y = 0.74

Solution

The given system of equations is

0.3x + 0.5y = 0.5 ….(i)

0.5x + 0.7y = 0.74 ….(ii)

Multiplying (i) by 5 and (ii) by 3 and subtracting (ii) from (i), we get

2.5y - 2.1y = 2.5 - 2.2

⇒ 0.4y = 0.28

⇒ y = 0.28/0.4 = 0.7

Now, substituting y = 0.7 in (i), we have

0.3x + 0.5 × 0.7 = 0.5

⇒ 0.3x = 0.50 – 0.35 = 0.15

⇒ x = 0.15/0.3 = 0.5

Hence, x = 0.5 and y = 0.7.

15. Solve for x and y:

7(y + 3) – 2(x + 2) = 14, 4(y – 2) + 3(x – 3) = 2

Solution

The given equations are:

7(y + 3) – 2(x + 2) = 14

⇒ 7y + 21 – 2x – 4 = 14

⇒ -2x + 7y = -3 …..(i)

and 4(y – 2) + 3(x – 3) = 2

⇒ 4y – 8 + 3x – 9 = 2

⇒ 3x + 4y = 19 ….(ii)

On multiplying (i) by 4 and (ii) by 7, we get:

-8x + 28y = -12 …(iii)

21x + 28y = 133 …(iv)

On subtracting (iii) from (iv), we get:

29x = 145

⇒ x = 5

On substituting x = 5 in (i), we get:

-10 + 7y = -3

⇒ 7y = (-3 + 10) = 7

⇒ y = 1

Hence, the solution is x = 5 and y = 1.

16. Solve for x and y:

6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)

Solution

The given equations are:

6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)

⇒ 6x + 5y = 2(x + 6y – 1)

⇒ 6x + 5y = 2x + 12y – 2

⇒ 6x – 2x + 5y – 12y = -2

⇒ 4x – 7y = -2 ….(i)

and 7x + 3y + 1 = 2(x + 6y – 1)

⇒ 7x + 3y + 1 = 2x + 12y – 2

⇒ 7x – 2x + 3y – 12y = -2 – 1

⇒ 5x – 9y = -3 ….(ii)

On multiplying (i) by 9 and (ii) by 7, we get:

36x - 63y = -18 …(iii)

35x - 63y = -21 …(iv)

On subtracting (iv) from (iii), we get:

x = (-18 + 21) = 3

On substituting x = 3 in (i), we get:

12 - 7y = -2

⇒ 7y = (2 + 12) = 14

⇒ y = 2

Hence, the solution is x = 3 and y = 2.

17. Solve for x and y:

(x + y – 8) = (x + 2y – 14)/3 = (3x + y – 12)/11

Solution

The given equations are:

(x + y – 8)/2 = (x + 2y – 14)/3 = (3x + y – 12)/11

i.e., (x + y – 8)/2 = (3x + y – 12)/11

By cross multiplication, we get:

11x + 11y – 88 = 6x + 2y – 24

⇒ 11x – 6x + 11y – 2y = -24 + 88

⇒ 5x + 9y = 64 ….(i)

And (x + 2y – 14)/3 = (3x + y – 12)/11

⇒ 11x + 22y – 154 = 9x + 3y – 36

⇒ 11x – 9x + 22y – 3y = -36 + 154

⇒ 2x + 19y = 118 ….(ii)

On multiplying (i) by 19 and (ii) by 9, we get:

95x + 171y = 1216 …(iii)

18x + 171y = 1062 …(iv)

On subtracting (iv) from (iii), we get:

77x = 154

⇒ x = 2

On substituting x = 2 in (i), we get:

10 + 9y = 64

⇒ 9y = (64 – 10) = 54

⇒ y = 6

Hence, the solution is x = 2 and y = 6.

18. Solve for x and y:

5/x + 6y = 13, 3/x + 4y = 7

Solution

The given equations are:

5/x + 6y = 13 …..(i)

3/x + 4y = 7 …..(ii)

Putting 1/x = u, we get:

5u + 6y = 13 ….(iii)

3u + 4y = 7 …(iv)

On multiplying (iii) by 4 and (iv) by 6, we get:

20u + 24y = 52 …..(v)

18u + 24y = 42 …..(vi)

On subtracting (vi) from (v), we get:

2u = 10 ⇒ u = 5

⇒ 1/x = 5 ⇒ x = 1/5

On substituting x = 1/5 in (i), we get:

5/(1/3) + 6y = 13

⇒ 25 + 6y = 13

⇒ 6y = (13 – 25) = -12

⇒ y = -2

Hence, the required solution is x = 1/5and y = -2.

19. Solve for x and y:

x + 6/y = 6, 3x – 8/y = 5

Solution

The given equations are:

x + 6/y = 6 …..(i)

3x – 8/y = 5 ...(ii)

Putting 1/y = v, we get:

x + 6v = 6 ….(iii)

3x – 8v = 5 …(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:

4x + 24v = 24 …..(v)

9x – 24v = 15 …..(vi)

On adding (v) from (vi), we get:

13x = 39 ⇒ x = 3

On substituting x = 3 in (i), we get:

3 + 6/y = 6

⇒ 6/y = (6 – 3) = 3

⇒ 3y = 6

⇒ y = 2

Hence, the required solution is x = 3 and y = 2.

20. Solve for x and y:2x - 3/y = 9, 3x + 7/y = 2

Solution

The given equations are:

2x -3/y = 9 ...(i)

3x + 7/y = 2 ...(ii)

Putting 1/y = v, we get:

2x - 3v = 6 ….(iii)

3x + 7v = 2 …(iv)

On multiplying (iii) by 7 and (iv) by 3, we get:

14x - 21v = 63  ...(v)

9x + 21v = 6 ...(vi)

On adding (v) from (vi), we get:

23x = 69

⇒ x = 3

On substituting x = 3 in (i), we get:

2 × 3 – 3/y = 9

⇒ 6 – 3/y = 9

⇒ 3/y = -3

⇒ y = -1

Hence, the required solution is x = 3 and y = -1.

21. Solve for x and y:

3/x – 1/y + 9 = 0, 2/x + 3/y = 5

Solution

3/x – 1/y + 9 = 0

The given equations are:

⇒ 3/x – 1/y = -9 ...(i)

⇒ 2/x – 3/y = 5 …..(ii)

Putting 1/x = u and 1/y = v, we get:

3u – v = -9 ….(iii)

2u + 3v = 5 …(iv)

On multiplying (iii) by 3, we get:

9u – 3v = -27 …..(v)

On adding (iv) and (v), we get:

11u = -22 ⇒ u = -2

⇒ 1/x = -2 ⇒ x = -1/2

On substituting x = −1/2 in (i), we get:

3/(-1/2) – 1/y = - 9

⇒ -6 – 1/y = -9

⇒ 1/y = (-6 + 9) = 3

⇒ y = 1/3

Hence, the required solution is x = −1/2and y = 1/3.

22. Solve for x and y:

9/x – 4/y = 8, 13/x + 7/y = 101

Solution

The given equations are:

9/x – 4/y …..(i)

13/x + 7/y = 101 …..(ii)

Putting 1/x = u and 1/y = v, we get:

9u - 4v = 8 ….(iii)

13u + 7v = 101 …(iv)

On multiplying (iii) by 7 and (iv) by 4, we get:

63u - 28v = 56 …..(v)

52u + 28v = 404 …..(vi)

On adding (v) from (vi), we get:

115u = 460 ⇒ u = 4

⇒ 1/x = 4 ⇒ x = 1/4

On substituting x = 1/4 in (i), we get:

⇒ 36 – 4/y = 8

⇒ 4/y = (36 – 8) = 28

⇒ y = 4/28 = 1/7

Hence, the required solution is x = 1/4and y = 1/7.

23. Solve for x and y:

5/x – 3/y = 1, 3/2x + 2/3y = 5

Solution

The given equations are:

5/x – 3/y …..(i)

3/2x + 2/3y = 5 …..(ii)

Putting 1/x = u and 1/y = v, we get:

5u - 3v = 1 ….(iii)

⇒ 3/2u + 2/3v = 5

⇒ (9u + 4v)/6 = 5

⇒ 9u + 4v = 30 ….(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:

20u - 12v = 4 …..(v)

27u + 12v = 90 ...(vi)

On adding (iv) and (v), we get:

47u = 94 ⇒ u = 2

⇒ 1/x = 2 ⇒ x = 1/2

On substituting x = 1/2 in (i), we get:

⇒ 10 – 3/y = 1

⇒ 3/y = (10 – 1) = 9

⇒ y = 3/9 = 1/3

Hence, the required solution is x = 1/2 and y = 1/3.

24. Solve for x and y:

3/x + 2/y = 12, 2/x + 3/y = 13

Solution

The given equations are:

3/x + 2/y = 12 …..(i)

2/x + 3/y = 13 …..(ii)

Multiplying (i) by 3 and (ii) by 2 and subtracting (ii) from (i), we get:

9/x – 4/x = 36 – 26

⇒ 5/x = 10

⇒ x = 5/10 = 1/2

Now, substituting x = 1/2 in (i), we have

6 + 2/y = 12

⇒ 2/y = 6

⇒ y = 1/3

Hence, x = 1/2 and y = 1/3.

25. Solve for x and y:

4x + 6y = 3xy, 8x + 9y = 5xy

Solution

The given equations are:

4x + 6y = 3xy  ...(i)

8x + 9y = 5xy …(ii)

From equation (i), we have:

(4x + 6y)/xy = 3

⇒ 4/y + 6/x = 3  …(iii)

For equation (ii), we have:

(8x + 9y)/xy = 5

⇒ 8/y + 9/x = 5 …(iv)

On substituting 1/y = v and 1/x = u, we get:

4v + 6u = 3 …(v)

8v + 9u = 5 ….(vi)

On multiplying (v) by 9 and (vi) by 6, we get:

36v + 54u = 27 ….(vii)

48v + 54u = 30 ….(viii)

On subtracting (vii) from (viii), we get:

12v = 3 ⇒ v = 3/12 = 1/4

⇒ 1/y = 1/4 ⇒ y = 4

On substituting y = 4 in (iii), we get:

4/4 + 6/x = 3

⇒ 1 + 6/x = 3

⇒ 6/x = (3 – 1) = 2

⇒ 2x = 6

⇒ x = 6/2 = 3

Hence, the required solution is x = 3 and y = 4.

26. Solve for x and y:

x + y = 5xy, 3x + 2y = 13xy

Solution

The given equations are:

x + y = 5xy  …..(i)

3x + 2y = 13xy  …(ii)

From equation (i), we have:

(x + y)/xy = 5

⇒ 1/y + 1/x = 5 …(iii)

For equation (ii), we have:

(3x + 2y)/xy = 13

⇒ 3/y + 2/x = 13 …(iv)

On substituting 1/y = v and 1/x = u, we get:

v + u = 5 …(v)

3v + 2u = 13 ….(vi)

On multiplying (v) by 2, we get:

2v + 2u = 10 ….(vii)

On subtracting (vii) from (vi), we get:

v = 3

⇒ 1/y = 3 ⇒ y = 1/3

On substituting y = 1/3 in (iii), we get:

1/(1/3) + 1/x = 5

⇒ 3 + 1/x = 5

⇒ 1/x = 2

⇒ x = 1/2

Hence, the required solution is x = 1/2 and y = 1/3 or x = 0 and y = 0.

27. Solve for x and y:

5/(x + y) – 2/(x – y) = -1, 15/(x + y) – 7/(x – y) = 10

Solution
The given equations are

5/(x + y) – 2/(x – y) = -1 .....(i)

15/(x + y) – 7/(x – y) = 10  …(ii)

Substituting 1/(x + y) = u and 1/(x – y) = v in (i) and (ii), we get

5u – 2v = -1 …..(iii)

15u + 7v = 10 ….(iv)

Multiplying (iii) by 3 and subtracting it from (iv), we get 7v + 6v = 10 + 3

⇒ 13v = 13

⇒ v = 1

⇒ x – y = 1  (∵ 1/(x – y) = v) ...(v)

Now, substituting v = 1 in (iii), we get

5u – 2 = -1

⇒ 5u = 1

⇒ u = 1/5

x + y = 5 ….(vi)

Adding (v) and (vi), we get

2x = 6 ⇒ x = 3

Substituting x = 3 in (vi), we have

3 + y = 5

⇒ y = 5 – 3 = 2

Hence, x = 3 and y = 2.

28. Solve for x and y:

3/(x + y) + 2/(x – y) = 2, 3/(x + y) + 2/(x – y) = 2

Solution
The given equations are

3/(x + y) + 2/(x – y) …(i)

9/(x + y) – 4/(x – y) …(ii)

Substituting 1/(x + y) = u and 1/(x – y) = v, we get:

3u + 2v = 2 …..(iii)

9u - 4v = 1 ….(iv)

On multiplying (iii) by 2, we get:

6u + 4v = 4 ...(v)

On adding (iv) and (v), we get:

15u = 5

⇒ u = 5/15 = 1/3

⇒ 1/(x + y) = 1/3

⇒ x + y = 3  ...(vi)

On substituting u = 1/3 in (iii), we get

1 + 2v = 2

⇒ 2v = 1

⇒ v = 1/2

⇒ 1/(x – y) = 1/2

⇒ x – y = 2 ….(vii)

On adding (vi) and (vii), we get 2x = 5

2x = 5

⇒ x = 5/2

On substituting x = 5/2 in (vi), we have

5/2 + y = 3

⇒ y = (3 – 5/2) = 1/2

Hence, the required solution is x = 5/2 and y = 1/2.

29. Solve for x and y:

5/(x + 1) + 2/(y – 1) = 1/2, 10/(x + 1) – 2/(y – 1) = 5/2, where x ≠ 1, y ≠ 1.

Solution

The given equations are

5/(x + 1) + 2/(y – 1) = 1/2  …(i)

10/(x + 1) – 2/(y – 1) = 5/2  …(ii)

Substituting 1/(x + 1) = u and 1/(y – 1) = v, we get:

5u - 2v = 1/2 …..(iii)

10u + 2v = 5/2 ….(iv)

On adding (iii) and (iv), we get:

15u = 3

⇒ u = 3/15 = 1/5

⇒ 1/(x + 1) = 1/5

⇒ x + 1 = 5

⇒ x = 4

On substituting u = 1/5 in (iii), we get

5 × 1/5 - 2v = 1/2

⇒ 1 – 2v = 1/2

⇒ 2v = 1/2

⇒ v = 1/4

⇒ 1/(y – 1) = 1/4

⇒ y – 1 = 4

⇒ y = 5

Hence, the required solution is x = 4 and y = 5.

30. Solve for x and y:

44/(x + y) + 30/(x – y) = 10, 55/(x + y) – 40/(x – y) = 13.

Solution

The given equations are

44/(x + y) + 30/(x – y) = 10 …(i)

55/(x + y) – 40/(x – y) = 13 .....(ii)

Putting 1/(x + y) u and 1/(x – y) = v, we get:

44u + 30v = 10 ...(iii)

55u + 40v = 13 ….(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:

176u + 120v = 40  …..(v)

165u + 120v = 39 ...(vi)

On subtracting (vi) and (v), we get:

11u = 1

⇒ u = 1/11

⇒ 1/(x + y) = 1/11

⇒ x + y = 11 ….(vii)

On substituting u = 1/11 in (iii), we get:

4 + 30v = 10

⇒ 30v = 6

⇒ v = 6/30 = 1/5

⇒ 1/(x – y) = 1/5

⇒ x – y = 5 ….(viii)

On adding (vii) and (viii), we get

2x = 16

⇒ x = 8

On substituting x =8 in (vii), we get:

8 + y = 11

⇒ y = 11 – 8 = 3

Hence, the required solution is x = 8 and y =3.

31. Solve for x and y:

10/(x + y) + 2/(x – y) = 4, 15/(x + y) – 9/(x – y) = - 2, where x ≠ y, x ≠ -y.

Solution

The given equations are

10/(x + y) + 2/(x – y) = 4 …(i)

15/(x + y) – 9/(x – y) = -2 …(ii)

Substituting 1/(x + y) = u and 1/(x – y) = v in (i) and (ii), we get:

10u + 2v = 4 ...(iii)

15u - 9v = -2 ...(iv)

Multiplying (iii) by 9 and (iv) by 2 and adding, we get: 90u + 30u = 36 – 4

⇒ 120u = 32

⇒ u = 32/120 = 4/15

⇒ x + y = 15/4     (∵ 1/(x + y) = u)  ...(v)

On substituting u = 4/15 in (iii), we get:

10 × 4/15 + 2v = 4

8/3 + 2v = 4

⇒ 2v = 4 - 8/3 = 4/3

⇒ v = 2/3

⇒ x – y = 3/2  (∵ 1/(x – y) = v) ...(vi)

Adding (v) and (vi), we get

2x = 15/4 + 3/2

⇒ 2x = 21/4

⇒ x = 218

Substituting x = 21/8 in (v), we have

21/8 + y = 15/4

⇒ y = 15/4 – 21/8 = 9/8

Hence, x = 21/8 and y = 9/8

32. Solve for x and y:

71x + 37y = 253, 37x + 71y = 287

Solution

The given equations are:

71x + 37y = 253 ...(i)

37x + 71y = 287 …(ii)

On adding (i) and (ii), we get:

108x + 108y = 540

⇒ 108(x + y) = 540

⇒ (x + y) = 5 …(iii)

On subtracting (ii) from (i), we get:

34x – 34y = -34

⇒ 34(x – y) = -34

⇒ (x – y) = -1 …(iv)

On adding (iii) from (i), we get:

2x = 5 – 1 = 4

⇒ x = 2

On subtracting (iv) from (iii), we get:

2y = 5 + 1 = 6

⇒ y = 3

Hence, the required solution is x = 2 and y = 3.

33. Solve for x and y:

217x + 131y = 913, 131x + 217y = 827

Solution

The given equations are:

217x + 131y = 913 ...(i)

131x + 217y = 827 …(ii)

On adding (i) and (ii), we get:

348x + 348y = 1740

⇒ 348(x + y) = 1740

⇒ x + y = 5 …(iii)

On subtracting (ii) from (i), we get:

86x – 86y = 86

⇒ 86(x – y) = 86

⇒ x – y = 1 …(iv)

On adding (iii) from (i), we get:

2x = 6

⇒ x = 3

On substituting x = 3 in (iii), we get:

3 + y = 5

⇒ y = 5 – 3 = 2

Hence, the required solution is x = 3 and y = 2.

34. Solve for x and y:

23x - 29y = 98, 29x - 23y = 110

Solution

The given equations are:

23x - 29y = 98 ...(i)

29x - 23y = 110 …(ii)

Adding (i) and (ii), we get:

52x – 52y = 208

⇒ x – y = 4 …(iii)

Subtracting (i) from (ii), we get:

6x + 6y = 12

⇒ x + y = 2 …(iv)

Now, adding equation (iii) and (iv), we get:

2x = 6

⇒ x = 3

On substituting x = 3 in (iv), we have:

3 + y = 2

⇒ y = 2 – 3 = -1

Hence, x = 3 and y = -1.

35. Solve for x and y:

Solution

The given equations can be written as

5/x + 2/y = 6 ….(i)

-5/x + 4/y = -3 …(ii)

Adding (i) and (ii), we get

6/y = 3

⇒ y = 2

Substituting y = 2 in (i), we have

5/x + 2/2 = 6

⇒ x = 1

Hence, x = 1 and y = 2.

36. Solve for x and y: 1/(3x + y) + 1/(3x – y) = 3/4, 1/2(3x + y) – 1/2(3x – y) = - 1/8

Solution
The given equations are

1/(3x + y) + 1/(3x – y) = 3/4  …(i)

⇒ 1/2(3x + y) – 1/2(3x – y) = - 1/8

⇒ 1/(3x + y) + 1/(3x – y) = -1/4   (Multiplying by 2) …(ii)

Substituting 1/(3x + y) = u and 1/(3x – y) = v in (i) and (ii), we get:

u + v = 3/4 …..(iii)

u – v = -1/4 ….(iv)

Adding (iii) and (iv), we get:

2u = 1/2

⇒ u = 1/4

⇒ 3x + y = 4  (∵ 1/(3x + y) = u)…..(v)

Now, substituting u = 1/4 in (iii), we get:

1/4 + v = 3/4

v = 3/4 - 1/4

⇒ v = 1/2

⇒ 3x – y = 2  (∵ 1/(3x – y) = v) ...(vi)

Adding (v) and (vi), we get

6x = 6

⇒ x = 1

Substituting x = 1 in (v), we have

3 + y = 4

⇒ y = 1

Hence, x = 1 and y = 1.

37. Solve for x and y:

1/2(x + 2y) + 5/(3(3x – 2y) = -(3/2), 1/{4(x + 2y)} – 3/{5(3x – 2y)} = 61/60 where x + 2y ≠ 0 and 3x – 2y ≠ 0.

Solution
The given equations are

1/{2(x + 2y) + 5/{3(3x – 2y)} = -(3/2)  …(i)

1/{4(x + 2y) – 3/{5(3x – 2y) = 61/60 …(ii)

Putting 1/(x + 2y) = u and 1/(3x – 2y) = v, we get:

1/2.u + 5/3v = -3/2 ...(iii)

5/4.u – 3/5.v = 61/60 ….(iv)

On multiplying (iii) by 6 and (iv) by 20, we get:

3u + 10v = -9 …..(v)

25u – 12v = 61/3 ...(vi)

On multiplying (v) by 6 and (vi) by 5, we get

18u + 60v = -54 ….(vii)

125u – 60v = 305/3  ...(viii)

On adding (vii) and (viii), we get:

143u = 305/3 – 54 = (305 – 162)/3 = 143/3

⇒ u = 1/3 = 1/(x + 2y)

⇒ x + 2y = 3  ….(ix)

On substituting u = 1/3 in (v), we get:

1 + 10v = -9

⇒ 10v = -10

⇒ v = -1

⇒ 1/(3x – 2y) = - 1

⇒ 3x – 2y = -1 ….(x)

On adding (ix) and (x), we get:

4x = 2

⇒ x = 1/2

On substituting x = 1/2 in (x), we get:

3/2 – 2y = - 1

⇒ 2y = (3/2 + 1) = 5/2

⇒ y = 5/4

Hence, the required solution is x = 1/2 and y = 5/4.

38. Solve for x and y:

2/(3x + 2y) + 3/(3x – 2y) = 17/5, 5/(3x + 2y) + 1/(3x – 2y) = 2

Solution

The given equations are

2/(3x + 2y) + 3/(3x – 2y) = 17/5 …(i)

5/(3x + 2y) + 1/(3x – 2y) = 2  …(ii)

Substituting 1/(3x + 2y) = u and 1/(3x – 2y) = v, in (i) and (ii), we get:

2u + 3v = 17/5 ...(iii)

5u + v = 2 ...(iv)

Multiplying (iv) by 3 and subtracting from (iii), we get:

2u – 15u = 17/5 – 6

⇒ –13u = -13/5 ⇒ u = 1/5

⇒ 3x + 2y = 5  (∵ 1/(3x + 2y) = u) ...(v)

Now, substituting u = 1/5 in (iv), we get

1 + v = 2

⇒ v = 1

⇒ 3x – 2y = 1 (∵ 1/(3x – 2y) = v) ....(vi)

Adding (v) and (vi), we get:

⇒ 6x = 6 ⇒ x = 1

Substituting x = 1 in (v), we get:

3 + 2y = 5

⇒ y = 1

Hence, x = 1 and y = 1.

39. Solve for x and y:

3/x + 6/y = 7, 9/x + 3/y = 11

Solution

The given equations can be written as

3/x + 6/y = 7 ….(i)

9/x + 3/y = 11 …(ii)

Multiplying (i) by 3 and subtracting (ii) from it, we get

18/y - 3/y = 21 – 11

⇒ 15/y = 10

⇒ y = 15/10 = 3/2

Substituting y = 3/2 in (i), we have

⇒ 3/x + (6 × 2)/3 = 7

⇒ 3/x = 7 – 4 = 3

Hence, x = 1 and y = 3/2

40. Solve for x and y:

x + y = a + b, ax – by = a2 – b2

Solution

The given equations are

x + y = a + b ….(i)

ax – by = a2 – b2 ...(ii)

Multiplying (i) by b and adding it with (ii), we get bx + ax = ab + b2 + a2 – b2

⇒ x = a

Substituting x = a in (i), we have

a + y = a + b

⇒ y = b

Hence, x = a and y = b.

41. Solve for x and y:

x/a + y/b = 2, ax – by = (a2 – b2

Solution

The given equations are:

x/a + y/b = 2

⇒ (bx + ay)/ab = 2 [Taking LCM]

⇒ bx + ay = 2ab ….(i)

Again, ax – by = (a2– b2) ...(ii)

On multiplying (i) by b and (ii) by a, we get:

b2x + bay = 2ab2 ...(iii)

a2x – bay = a(a2– b2….(iv)

On adding (iii) from (iv), we get:

(b2 + a2)x = 2a2b + a(a2– b2

⇒ (b2 + a2)x = 2ab2 + a3– ab2

⇒ (b2 + a2)x = ab2 + a3

⇒ (b2 + a2)x = a(b2 + a2

⇒ x = a(b2 + a2)/(b2 + a2) = a

On substituting x = a in (i), we get:

ba + ay = 2ab

⇒ ay = ab

⇒ y = b

Hence, the solution is x = a and y = b.

42. Solve for x and y:

px + qy = p – q,

qx – py = p + q

Solution

The given equations are

px + qy = p – q …(i)

qx – py = p + q …(ii)

Multiplying (i) by p and (ii) by q and adding them, we get

p2x + q2x = p2 – pq + pq + q2

x = (p2 + q2)/(p2 + q2) = 1

Substituting x = 1 in (i), we have

p + qy = p – q

⇒ qy = -p

⇒ y = -1

Hence, x = 1 and y = -1.

43. Solve for x and y:

x/a – y/b = 0, ax + by = a2 + b2

Solution

The given equations can be written as

x/a – y/b = 0 ….(i)

ax + by = a2 + b2 …(ii)

From (i),

y = bx/a

Substituting y = bx/a in (ii), we get

ax + (b × bx)/a = a2 + b2

⇒ x = {(a2 + b2) × a}/(a2 + b2) = a

Now, substitute x = a in (ii) to get

a2 + by = a2 + b2

⇒ by = b2

⇒ y = b

Hence, x = a and y = b.

44. Solve for x and y:

6(ax + by) = 3a + 2b,

6(bx – ay) = 3b – 2a

Solution

The given equations are

6(ax + by) = 3a + 2b

⇒ 6ax + 6by = 3a + 2b  …(i)

and 6(bx – ay) = 3b – 2a

⇒ 6bx – 6ay = 3b – 2a  …(ii)

On multiplying (i) by a and (ii) by b, we get

6a2x + 6aby = 3a2 + 2ab ….(iii)

6b2x - 6aby = 3b2- 2ab  ….(iv)

On adding (iii) and (iv), we get

6(a2 + b2)x = 3(a2 + b2

x = 3(a2 + b2)/6(a2 + b2) = 1/2

On substituting x = 1/2 in (i), we get:

6a × 1/2 + 6by = 3a + 2b

⇒ 6by = 2b

⇒ y = 2b/6b = 1/3

Hence, the required solution is x = 1/2 and y = 1/3.

45. Solve for x and y:

ax - by = a2 + b2, x + y = 2a

Solution

The given equations are

ax - by = a2 + b2…(i)

x + y = 2a …(ii)

From (ii)

y = 2a - x

Substituting y = 2a – x in (i), we get

ax – b(2a – x) = a2 + b2

⇒ ax – 2ab + bx = a2 + b2

⇒ x = (a2 + b2 + 2ab)/(a + b) = (a + b)2/(a + b) = a + b

Now, substitute x = a + b in (ii) to get

a + b + y = 2a

⇒ y = a – b

Hence, x = a + b and y = a – b.

46. Solve for x and y:

a + b = 0, bx – ay + 2ab = 0

Solution

The given equations are:

bx/a – ay/b + a + b = 0

By taking LCM, we get:

b2x – a2y = -a2b – b2a ...(i)

and bx – ay + 2ab = 0

bx – ay = -2ab ….(ii)

On multiplying (ii) by a, we get:

abx – a2y = -2a2b …(iii)

On subtracting (i) from (iii), we get:

abx – b2x = 2a2b + a2b + b2a = -a2b + b2a

⇒ x(ab – b2) = -ab(a – b)

⇒ x(a – b)b = -ab(a – b)

∴ x = -ab(a – b)(a – b)b = -a

On substituting x = -a in (i), we get:

b2(-a) – a2y = -a2b – b2

⇒ -b2a – a2y = -a2b – b2

⇒ -a2y = -a2

⇒ y = b

Hence, the solution is x = -a and y = b.

47. Solve for x and y:

Solution

The given equations are:

bx/a + ay/b = a2 + b2

By taking LCM, we get:

(b2x + a2y)/ab = a2 + b2

⇒ b2x + a2y = (ab)a2 + b2

⇒ b2x + a2y = a3b +ab3….(i)

Also, x + y = 2ab ….(ii)

On multiplying (ii) by a2, we get:

a2x + a2y = 2a3b …(iii)

On subtracting (iii) from (i), we get:

(b2 – a2)x = a3b +ab3 – 2a3

⇒ (b2 – a2)x = -a3b + ab3

⇒ (b2 – a2)x = ab(b2 – a2

⇒ (b2 – a2)x = ab(b2 – a2

∴ x = ab(b2 – a2)/(b2 – a2) = ab

On substituting x = ab in (i), we get:

b2(ab) + a2y = a3b + ab3

⇒ a2y = a3b

⇒ ab

Hence, the solution is x = ab and y = ab.

48. Solve for x and y:

x + y = a + b, ax - by = a2- b2

Solution

The given equations are

x + y = a + b …(i)

ax - by = a2- b2 …(ii)

From (i)

y = a + b - x

Substituting y = a + b - x in (ii), we get

ax – b(a + b - x) = a2 - b2

⇒ ax – ab - b2 + bx = a2 - b2

⇒ x = (a2 + ab)/(a + b) = a

Now, substitute x = a in (i) to get

a + y = a + b

⇒ y = b

Hence, x = a and y = b.

49. Solve for x and y:

a2x + b2y = c2, b2x + a2y = d2

Solution

The given equations are

a2x + b2y = c2  …(i)

b2x + a2y = d2 …(ii)

Multiplying (i) by a2and (ii) by b2and subtracting, we get

a4x – b4x = a2c2 - b2d2

⇒ x = (a2c2 – b2d2)/(a4 – b4)

Now, multiplying (i) by b2 and (ii) by a2 and subtracting, we get

b4y – a4y = b2c2 - a2d2

⇒ y = (b2c2 – a2d2)/(b4 – a4)

Hence, x = (a2c2 – b2d2)/(a4 – b4) and y = (b2c2 – a2d2)/(b4 – a4)

50. Solve for x and y:

x/a + y/b = a + b, x/a2 + y/b2 = 2

Solution

The given equations are

x/a + y/b = a + b ....(i)

x/a2 + y/b2 = 2 ...(ii)

Multiplying (i) by b and (ii) by b2 and subtracting, we get

bx/a – b2x/a2 = ab + b2 – 2b2

⇒ (ab - b2)/a2.x = ab – b2

⇒ x = (ab – b2)a2/(ab – b2) = a2

Now, substituting x = a2  in (i) we get

a2/a + y/b = a + b

⇒ = a + b – a = b

⇒ y = b2

Hence, x = a2and y = b2