RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Apart from the graphical methods of solving simultaneous linear equations in two variables, we also have the algebraic methods too. Under the algebraic methods of solving simultaneous linear equations in two variables, we have three methods. Method of elimination by substitution and method of elimination by equating the coefficients are the prominent concepts covered in this exercise. Students can make use of the RD Sharma Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3 PDF given below for any clarifications related to these exercise problems.

RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

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Access RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Solve the following system of equations:

1. 11x + 15y + 23 = 0  

7x – 2y – 20 = 0

Solution:

The given pair of equations are

11x +15y + 23 = 0 …………………………. (i)

7x – 2y – 20 = 0 …………………………….. (ii)

From (ii),

2y = 7x – 20

⇒ y = (7x −20)/2 ……………………………… (iii)

Now, substituting y in equation (i), we get

⇒ 11x + 15((7x−20)/2) + 23 = 0

⇒ 11x + (105x − 300)/2 + 23 = 0

⇒ (22x + 105x – 300 + 46) = 0

⇒ 127x – 254 = 0

⇒ x = 2

Next, putting the value of x in equation (iii), we get,

⇒ y = (7(2) − 20)/2

∴ y= -3

Thus, the value of x and y is found to be 2 and -3, respectively.

2.  3x – 7y + 10 = 0

y – 2x – 3 = 0

Solution:

The given pair of equations are

3x – 7y + 10 = 0 …………………………. (i)

y – 2x – 3 = 0 ……………………………….. (ii)

From (ii),

y – 2x – 3 = 0

y = 2x+3 ……………………………… (iii)

Now, substituting y in equation (i), we get

⇒ 3x – 7(2x+3) + 10 = 0

⇒ 3x – 14x – 21 + 10 = 0

⇒ -11x = 11

⇒ x = -1

Next, putting the value of x in equation (iii), we get

⇒ y = 2(-1) + 3

∴ y= 1

Thus, the value of x and y is found to be -1 and 1, respectively.

3. 0.4x + 0.3y = 1.7

0.7x – 0.2y = 0.8

Solution:

The given pair of equations are

0.4x + 0.3y = 1.7

0.7x – 0.2y = 0.8

Let’s, multiply LHS and RHS by 10 to make the coefficients an integer.

4x + 3y = 17 ……………………….. (i)

7x – 2y = 8 …………………………… (ii)

From (ii),

7x – 2y = 8

x = (8 + 2y)/7……………………………… (iii)

Now, substituting x in equation (i), we get

⇒ 4[(8 + 2y)/7] + 3y = 17

⇒ 32 + 8y + 21y = (17 x 7)

⇒ 29y = 87

⇒ y = 3

Next, putting the value of y in equation (iii), we get

⇒ x = (8 + 2(3))/ 7

⇒ x = 14/7

∴ x = 2

Thus, the value of x and y is found to be 2 and 3, respectively.

4. x/2 + y = 0.8

7/(x+ y/2) = 10

Solution:

The given pair of equations are

x/2 + y = 0.8 ⇒ x + 2y = 1.6…… (a)

7/(x + y/2) = 10 ⇒7 = 10(x + y/2) ⇒7 = 10x + 5y

Let’s, multiply the LHS and RHS of equation (a) by 10 for easy calculation.

So, we finally get

10x + 20y = 16 ……………………….. (i) And,

10x + 5y = 7 …………………………… (ii)

Now, subtracting two equations, we get

⇒ (i) – (ii)

15y = 9

⇒ y = 3/5

Next, putting the value of y in the equation (i), we get

x = [16 − 20(3/5)]/10

⇒  (16 – 12)/10 = 4/10

∴ x = 2/5

Thus, the value of x and y obtained are 2/5 and 3/5, respectively.

5. 7(y + 3) – 2(x + 2) = 14

4(y – 2) + 3(x – 3) = 2

Solution:

The given pair of equations are

7(y+3) – 2(x+2) = 14…………………………. (i)

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

From (i), we get

7y + 21 – 2x – 4 = 14

7y = 14 + 4 – 21 + 2x

⇒ y = (2x – 3)/7

From (ii), we get

4y – 8 + 3x – 9 = 2

4y + 3x – 17 – 2 = 0

⇒ 4y + 3x – 19 = 0 …………….. (iii)

Now, substituting y in equation (iii),

4[(2x − 3)/7] + 3x – 19=0

8x – 12 + 21x – (19 x 17) = 0 [after taking LCM]

29x = 145

⇒ x = 5

Now, putting the value of x and in equation (ii),

4(y-2) + 3(5-3) = 2

⇒ 4y -8 + 6 = 2

⇒ 4y = 4

∴ y = 1

Thus, the value of x and y obtained are 5 and 1, respectively.

6. x/7 + y/3 = 5

x/2 – y/9 = 6

Solution:

The given pair of equations are

x/7 + y/3 = 5…………………………. (i)

x/2 – y/9 = 6………………………………..(ii)

From (i), we get

x/7 + y/3 = 5

⇒3x + 7y = (5×21) [After taking LCM]

⇒ 3x =105 – 7y

⇒ x = (105 – 7y)/3……. (iv)

From (ii), we get

x/2 – y/9 = 6

⇒ 9x – 2y = 108 ……………………… (iii) [After taking LCM]

Now, substituting x in equation (iii), we get

9[(105 − 7y)/3] – 2y = 108

⇒ 945 – 63y – 6y = 324 [After taking LCM]

⇒ 945 – 324 = 69y

⇒ 69y = 621

⇒ y = 9

Now, putting the value of y in equation (iv),

x = (105 − 7(9))/3

⇒ x = (105 − 63)/3 = 42/3

∴ x = 14

Thus, the value of x and y obtained are 14 and 9, respectively.

7. x/3 + y/4 = 11

5x/6 − y/3 = −7

Solution:

The given pair of equations are

x/3 + y/4 = 11…………………………. (i)

5x/6 − y/3 = −7……………………………….. (ii)

From (i), we get

x/3 + y/4 = 11

⇒4x + 3y = (11×12) [After taking LCM]

⇒ 4x =132 – 3y

⇒ x = (132 – 3y)/4……. (iv)

From (ii), we get

5x/6 − y/3 = −7

⇒ 5x – 2y = -42 ……………………… (iii) [After taking LCM]

Now, substituting x in equation (iii), we get

5[(132 − 3y)/4] – 2y = -42

⇒ 660 – 15y – 8y = -42 x 4 [After taking LCM]

⇒ 660 + 168 = 23y

⇒ 23y = 828

⇒ y = 36

Now, putting the value of y in equation (iv),

x = (132 – 3(36))/4

⇒ x = (132 − 108)/4 = 24/4

∴ x = 6

Thus, the value of x and y obtained are 6 and 36, respectively.

8. 4/x + 3y = 8

6/x −4y = −5

Solution:

Taking 1/x = u

Then, the two equations become

4u + 3y = 8…………………… (i)

6u – 4y = -5……………………. (ii)

From (i), we get

4u = 8 – 3y

⇒ u = (8 − 3y)/4 …….. (iii)

Substituting u in (ii),

⇒  [3(8−3y)/2] − 4y = −5

⇒ 24 − 9y −8y = −5 x 2 [After taking LCM]

⇒ 24 – 17y = -10

⇒ -17y =- 34

⇒ y = 2

Putting y=2 in (iii), we get

u = (8 − 3(2))/4

⇒ u = (8 − 6)/4

⇒ u = 2/4 = 1/2

⇒ x = 1/u = 2

∴ x = 2

So, the solution of the pair of equations given is x=2 and y =2.

9. x + y/2 = 4

2y + x/3 = 5

Solution:

The given pair of equations are

x + y/2 = 4 ……………………. (i)

2y + x/3 = 5……………………. (ii)

From (i), we get

x + y/2 = 4

⇒ 2x + y = 8 [After taking LCM]

⇒ y = 8 – 2x …..(iv)

From (ii), we get

x + 6y = 15 ……………… (iii) [After taking LCM]

Substituting y in (iii), we get

x + 6(8 – 2x) = 15

⇒ x + 48 – 12x = 15

⇒ -11x = 15 – 48

⇒ -11x = -33

⇒ x = 3

Putting x = 3 in (iv), we get

y = 8 – (2×3)

∴ y = 8 – 6 = 2

Hence, the solutions of the given system of the equation are x = 3 and y = 2, respectively.

10. x + 2y = 3/2

2x + y = 3/2

Solution:

The given pair of equations are

x + 2y = 3/2 …………………. (i)

2x + y = 3/2…………………… (ii)

Let us eliminate y from the given equations. The coefficients of y in the given equations are 2 and 1, respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal to 2 in the two equations.

Multiplying equation (i)x1 and (ii)x2 ⇒

x + 2y = 3/2 ………………………. (iii)

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

Subtracting equation (iii) from (iv),

(4x – x) + (2y-2y) = 3x = 3 – (3/2)

⇒ 3x = 3/2

⇒ x = 1/2

Putting x = 1/2 in equation (iv),

4(1/2) + 2y = 3

⇒ 2 + 2y = 3

∴ y= 1/2

The solution of the system of equations is x = 1/2 and y = 1/2

11. √2x – √3y = 0

√3x − √8y = 0

Solution:

The given pair of equations are

√2x – √3y = 0……………………….. (i)

√3x − √8y = 0……………………….. (ii)

From equation (i),

x = √(3/2)y ……………..(iii)

Substituting this value in equation (ii), we obtain

√3x − √8y = 0

⇒ √3(√(3/2)y)  − √8y = 0

⇒ (3/√2)y – √8y = 0

⇒ 3y – 4y = 0

⇒ y = 0

Now, substituting y in equation (iii), we obtain

⇒ x=0

Thus, the value of x and y obtained are 0 and 0, respectively.

12. 3x – (y + 7)/11 + 2 = 10

2y + (x + 11)/7 = 10

Solution:

The given pair of equations are

3x – (y + 7)/11 + 2 = 10……………….. (i)

2y + (x + 11)/7 = 10…………………….. (ii)

From equation (i),

33x – y – 7 + 22 = (10 x 11) [After taking LCM]

⇒ 33x – y + 15 = 110

⇒ 33x + 15 – 110 = y

⇒ y = 33x – 95………. (iv)

From equation (ii),

14 + x + 11 = (10 x 7) [After taking LCM]

⇒ 14y + x + 11 = 70

⇒ 14y + x = 70 – 11

⇒ 14y + x = 59 …………………….. (iii)

Substituting (iv) in (iii), we get

14 (33x – 95) + x = 59

⇒ 462x – 1330 + x = 59

⇒ 463x = 1389

⇒ x = 3

Putting x = 3 in (iii), we get

⇒ y = 33(3) – 95

∴ y= 4

The solution for the given pair of equations is x = 3 and y = 4, respectively.

13. 2x – (3/y) = 9

3x + (7/y) = 2 ,y ≠ 0

Solution:

The given pair of equations are

2x – (3/y) = 9……………………………. (i)

3x + (7/y) = 2…………………………… (ii)

Substituting 1/y = u, the above equations become

2x – 3u = 9 ………………………..(iii)

3x + 7u = 2………………………..(iv)

From (iii)

2x = 9 + 3u

⇒ x = (9+3u)/2

Substituting the value of x from above in equation (iv), we get

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

⇒ 27 + 9u + 14u = (2 x 2)

⇒ 27 + 23u = 4

⇒ 23u = -23

⇒ u = -1

So, y = 1/u = -1

And putting u = -1 in x = (9 + 3u)/2, we get

⇒ x = [9 + 3(−1)]/2 = 6/2

∴ x = 3

The solutions of the pair of equations given are y = 3 and x = -1, respectively.

14. 0.5x + 0.7y = 0.74

0.3x + 0.5y = 0.5

Solution:

The given pair of equations are

0.5x + 0.7y = 0.74……………………… (i)

0.3x – 0.5y = 0.5 ………………………….. (ii)

Now, let’s multiply LHS and RHS by 100 for both (i) and (ii) to make integral coefficients and constants.

(i) x100 ⇒

50x +70y = 74 ……………………….. (iii)

(ii) x100 ⇒

30x + 50y = 50 …………………………… (iv)

From (iii),

50x = 74 – 70y

x = (74−70y)/ 50 ……………………………… (v)

Now, substituting x in equation (iv), we get

30[(74−70y)/ 50] + 50y = 50

⇒ 222 – 210y + 250y = 250 [After taking LCM]

⇒ 40y = 28

⇒ y = 0.7

Now, by putting the value of y in the equation (v), we get

⇒ x = [74 − 70(0.7)]/ 50=0

⇒ x =25/ 50 = 1/2

∴ x = 0.5

Thus, the values of x and y so obtained are 0.5 and 0.7, respectively.

15. 1/(7x) + 1/(6y) = 3

1/(2x) – 1/(3y) = 5

Solution:

The given pair of equations are

1/(7x) + 1/(6y) = 3………………………….. (i)

1/(2x) – 1/(3y) = 5……………………………. (ii)

Multiplying (ii) by 1/2, we get

1/(4x) – 1/(6y) = 52……………………………. (iii)

Now, solving equations (i) and (iii),

1/(7x) + 1/(6y) = 3………………………….. (i)

1/(4x) – 1/(6y) = 5/2……………………………. (iii)

Adding (i) + (iii), we get

1/x(1/7 + 1/4 ) = 3 + 5/2

⇒ 1/x(11/28) = 11/2

⇒ x = 1/14

Using x =1/ 14, we get, in (i)

1/[7(1/14)] + 1/(6y) = 3

⇒ 2 + 1/(6y)=3

⇒ 1/(6y) = 1

⇒ y = 1/6

The solution for the given pair of equations is x=1/14 and y=1/6, respectively.

16. 1/(2x) + 1/(3y) = 2

  1/(3x) + 1/(2y) = 13/6

Solution:

Let 1/x = u and 1/y = v,

So, the given equations become

u/2 + v/3 = 2 ………………(i)

u/3 + v/2 = 13/6 ……………(ii)

From (i), we get

u/2 + v/3 = 2

⇒ 3u + 2v = 12

⇒ u = (12 – 2v)/3 ………….(iii)

Using (iii) in (ii),

⇒ (12 – 2v)/9 + v/2 = 13/6

⇒ 24 – 4v + 9v = (13/6) x 18 [after taking LCM]

⇒ 24 + 5v = 39

⇒ 5v = 15

⇒ v = 3

Substituting v in (iii),

u = (12 – 2(3))/3

⇒ u = 2

Thus, x = 1/u ⇒ x = 1/2 and

y = 1/v ⇒ y = 1/3

The solution for the given pair of equations is x = 1/2 and y = 1/3, respectively.

17. 15/u + 2/v = 17

1/u + 1/v = 36/5

Solution:

Let 1/x = u and 1/y = v

So, the given equations become

15x + 2y = 17 ………………………….. (i)

x + y = 36/5………………………. (ii)

From equation (i), we get

2y = 17 – 15x

=y = (17 − 15x)/ 2 …………………. (iii)

Substituting (iii) in equation (ii), we get

= x + (17 − 15x)/ 2 = 36/5

2x + 17 – 15x = (36 x 2)/ 5 [after taking LCM]

-13x = 72/5 – 17

= -13x = -13/5

⇒ x = 1/5

⇒ u = 1/x = 5

Putting x = 1/5in equation (ii), we get

1/5 + y = 36/5

⇒ y = 7

⇒ v = 1/y = 1/7

The solutions of the pair of equations given are u = 5 and v = 1/7, respectively.

18. 3/x – 1/y = −9

2/x + 3/y = 5

Solution:

Let 1/x = u and 1/y = v

So, the given equations become

3u – v = -9…………………..(i)

2u + 3v = 5 ……………………….(ii)

Multiplying equation (i) x 3 and (ii) x 1, we get

9u – 3v = -27 ………………………….. (iii)

2u + 3v = 5 ……………………………… (iv)

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

9u + 2u – 3v + 3v = -27 + 5

⇒ 11u = -22

⇒ u = -2

Now putting u =-2 in equation (iv), we get

2(-2) + 3v = 5

⇒ 3v = 9

⇒ v = 3

Hence, x = 1/u = −1/2 and,

y = 1/v = 1/3

19. 2/x + 5/y = 1

60/x + 40/y = 19

Solution:

Let 1/x = u and 1/y = v

So, the given equations become

2u + 5v = 1…………………..(i)

60u + 40v = 19 ……………………….(ii)

Multiplying equation (i) x 8 and (ii) x 1, we get

16u + 40v = 8 ………………………….. (iii)

60u + 40v = 19 ……………………………… (iv)

Subtracting equation (iii) from (iv), we get

60u – 16u + 40v – 40v = 19 – 8

⇒ 44u = 11

⇒ u = 1/4

Now putting u = 1/4 in equation (iv), we get

60(1/4) + 40v = 19

⇒ 15 + 40v = 19

⇒ v = 4/ 40 = 1/10

Hence, x = 1/u = 4 and

y = 1/v = 10

20. 1/(5x) + 1/(6y) = 12

1/(3x) – 3/(7y) = 8

Solution:

Let 1/x = u and 1/y = v

So, the given equations become

u/5 + v/6 = 12…………………..(i)

u/3 – 3v/7 = 8……………………….(ii)

Taking LCM for both equations, we get

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), we get

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), we find u.

u = 14 x (1512/89) – 192

⇒ u = 4080/89

⇒ x = 1/u = 89/ 4080

Hence, the solution for the given system of equations is x = 89/4080 and y = 89/ 1512.

21. 4/x + 3y = 14

3/x – 4y = 23

Solution:

Taking 1/x = u, the given equations become

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)

Using (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), we find y

y = 7(5) – 37

⇒ y = -2

Hence, the solution for the given system of equations is x = 1/5 and y = -2.

22. 4/x + 5y = 7

3/x + 4y = 5

Solution:

Taking 1/x = u, the given equations become

4u + 5y = 7…………………….. (i)

3u + 4y = 5…………………….. (ii)

Subtracting (ii) from (i), we get

4u + 5y – (3u + 4y) = 7 – 5

⇒ u + y = 2

⇒ u = 2 – y……………………… (iii)

Using (iii) in (i),

4(2 – y) + 5y = 7

⇒ 8 – 4y + 5y = 7

⇒ y = -1

Putting y = -1 in (iii), we find u.

u = 2 – (-1)

⇒ u = 3

⇒ x = 1/u = 1/3

Hence, the solution for the given system of equations is x = 1/3 and y = -1.

23. 2/x + 3/y = 13

5/x – 4/y = -2

Solution:

Let 1/x = u and 1/y = v

So, the given equations become

2u + 3v = 13………………….. (i)

5u – 4v = -2 ………………………. (ii)

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

2u + 3v + 5u – 4v = 13 – 2

⇒ 7u – v = 11

⇒ v = 7u – 11…….. (iii)

Using (iii) in (i), we get

2u + 3(7u – 11) = 13

⇒ 2u + 21u – 33 = 13

⇒ 23u = 46

⇒ u = 2

Substituting u = 2 in (iii), we find v.

v = 7(2) – 11

⇒ v = 3

Hence, x = 1/u = 1/2 and,

y = 1/v = 1/3

24. 2/x + 3/y = 2

4/x – 9/y = -1

Solution:

Let 1/√x = u and 1/√y = v,

So, the given equations become

2u + 3v = 2………………….. (i)

4u – 9v = -1 ………………………. (ii)

Multiplying (ii) by 3 and

Adding equations (i) and (ii)x3, we get

6u + 9v + 4u – 9v = 6 – 1

⇒ 10u = 5

⇒ u = 1/2

Substituting u = 1/2 in (i), we find v

2(1/2) + 3v = 2

⇒ 3v = 2 – 1

⇒ v = 1/3

Since, 1/√x = u we get x = 1/u²

⇒ x = 1/(1/2)² = 4

And,

1/√y = v y = 1/v²

⇒ y = 1/(1/3)² = 9

Hence, the solution is x = 4 and y = 9.

25. (x + y)/xy = 2

(x – y)/xy = 6

Solution:

The given pair of equations are

(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, so the equations (i) and (ii) become

v + u = 2……. (iii)

v – u = 6……..(iv)

Adding (iii) and (iv), we get

2v = 8

⇒ v = 4

⇒ y = 1/v = 1/4

Substituting v = 4 in (iii) to find x,

4 + u = 2

⇒ u = -2

⇒ x = 1/u = -1/2

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

26. 2/x + 3/y = 9/xy

4/x + 9/y = 21/xy

Solution:

Taking LCM for both the given equations, we have

(2y + 3x)/ xy = 9/xy ⇒ 3x + 2y = 9………. (i)

(4y + 9x)/ xy = 21/xy ⇒ 9x + 4y = 21………(ii)

Performing (ii) – (i)x2⇒

9x + 4y – 2(3x + 2y) = 21 – (9×2)

⇒ 3x = 3

⇒ x = 1

Using x = 1 in (i), we find y

3(1) + 2y = 9

⇒ y = 6/2

⇒ y = 3

Thus, the solution for the given set of equations is x = 1 and y = 3.