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

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 3A Exercise 3C Exercise 3D Exercise 3E Exercise 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 = 6  …(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 = 6 

⇒ 2x + 3(x – 3) = 36 

⇒2x + 3x – 9 = 36 

⇒ x = 45/5 = 9 

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 = 0 

⇒ 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 = ⁸3 

⇒ 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 – b²a 

⇒ -b²a – a²y = -a²b – b²a 

⇒ -a²y = -a²b 

⇒ 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 – 2a³b 

⇒ (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