RD Sharma Class 10 Ex 3.3 Solutions Chapter 3 Pair of Linear Equations in Two Variables

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TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 3
Chapter NamePair of Linear Equations in Two Variables
Exercise3.3
CategoryRD Sharma Solutions

RD Sharma Solutions for Class 10 Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Download PDF

Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.3

Question: 1

Solve the system of equations:

11x + 15y + 23 = 0 and 7x – 2y – 20 = 0

Solution:

The given system of equation is

11x + 15y + 23 = 0 ……. (i)

7x – 2y – 20 = 0 ….. (ii)

From (ii)

2y = 7x – 20

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

127x = 254

x = 2

Putting the value of x in the equation (iii)

y = – 3

The value of x and y are 2 and -3 respectively.

Question: 2

Solve the system of equations:

3x – 7y + 10 = 0, y – 2x – 3 = 0

Solution:

The given system of equation is

3x – 7y + 10 = 0 …. (i)

y – 2x – 3 = 0 ….. (ii)

From (ii)

y – 2x – 3 = 0

y = 2x + 3 …… (iii)

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

= 3x – 7(2x + 3) + 10 = 0

= 3x + 14x – 21 + 10 = 0

= -11x = 11

= x = -1

Putting the value of x in the equation (iii)

= y = 2(- 1) + 3

y = 1

The value of x and y are -1 and 1 respectively.

Question: 3

Solve the system of equations:

0.4x + 0.3y = 1.7, 0.7x – 0.2y = 0.8

Solution:

The given system of equation is

0.4x + 0.3y = 1.7

0.7x – 0.2y = 0.8

Multiplying both sides by 10

4x + 3y = 17 ….. (i)

7x – 2y = 8 …… (ii)

From (ii)

7x – 2y = 8

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

32 + 29y = 119

29y = 87

y = 3

Putting the value of y in the equation (iii)

x = 2

The value of x and y are 2 and 3 respectively.

Question: 4

Solution:

The given system of equation is

Therefore x + 2y = 1.6

x + 2y = 1.6

7 = 10x + 5y

Multiplying both sides by 10

10x + 20y = 16 ….. (i)

10x + 5y = 7 …… (ii)

Subtracting two equations we get,

15y = 9

The value of x and y are 2/5 and 3/5 respectively.

Question: 5

Solve the system of equations:

7(y + 3) – 2(x + 3) = 14

4(y – 2) + 3(x – 3) = 2

Solution:

The given system of equation is

7(y + 3) – 2(x + 3) = 14 ……. (i)

4(y – 2) + 3(x – 3) = 2 ….. (ii)

From (i)

7y + 21 – 2x – 4 = 14

7y = 14 + 4 – 21 + 2x

From (ii)

4y – 8 + 3x – 9 = 2

4y + 3x – 17 – 2 = 0

4y + 3x – 19 = 0 ….. (iii)

Substituting the value of y in equation (iii)

8x – 12 + 21x – 133 = 0

29x = 145

x = 5

Putting the value of x in the above equation

y = 1

The value of x and y are 5 and 1 respectively.

Question: 6

Solve the system of equations:

Solution:

The given system of equation is

From (i)

= 9x – 2y = 108 … (iii)

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

945 – 63y – 6y = 324

945 – 324 = 69y

69y = 621

y = 9

Putting the value of y in the above equation

y = 14

The value of x and y are 5 and 14 respectively.

Question: 7

Solve the system of equations:

Solution:

The given system of equation is

From (i)

4x + 3y = 132 … (iii)

From (ii)

5x – 2y = – 42 …… (iv)

Let us eliminate y from the given equations. The co efficient of y in the equation (iii) and (iv) are 3 and 2 respectively. The L.C.M of 3 and 2 is 6. So, we make the coefficient of y equal to 6 in the two equations.

Multiplying equation (iii) 2 and (iv) 3 we get

8x + 6y = 264 …. (v)

15x – 6y = -126 … (vi)

Adding equation (v) and (vi)

8x + 15x = 264 – 126

23x = 138

x = 6

Putting the value of x in the equation (iii)

24 + 3y = 132

3y = 108

y = 36

The value of x and y are 36 and 6 respectively.

Question: 8

Solve the system of equations:

6x – 4y = – 56x – 4y = –  5

Solution:

The new equation becomes

4u + 3y = 8 … (i)

6u – 4y = – 5 …. (ii)

From (i)

4u = 8 –  3y

u = (8-3y)/4

From (ii)

24 – 17y = – 10

– 17y = – 34

y = 2

x = 2

So the Solution of the given system of equation is x = 2 and y = 2

Question: 9

Solve the system of equations:

Solution:

The given system of equation is:

From (i) we get,

2x + y = 8

y = 8 –  2x

From (ii) we get,

x + 6y = 15   ……  (iii)

Substituting y = 8 – 2x in (iii), we get

x + 6(8 – 2x) = 15

x + 48 – 12x = 15

– 11x = 15 – 48

– 11x = – 33

x = 3

Putting x = 3 in y 8 – 2x, we get

y = 8 – (2×3)

y = 8 – 6

Y = 2

The Solution of the given system of equation are x = 3 and y = 2 respectively.    

Question: 10

Solve the system of equations:

Solution:

The given system of equation is

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)*1 and (ii)*2

4x + 2y = 3 ……. (iv)

Subtracting equation (iii) from (iv)

Putting x = 1/2in equation (iv)

2 + 2y = 3

y = 1/2

The Solution of the system of equation is x = 1/2 and y = ½

Question: 11

Solve the system of equations:

Solution:

From equation (i)

Substituting this value in equation (ii) we obtain

y = 0

Substituting the value of y in equation (iii) we obtain

x = 0

The value of x and y are 0 and 0 respectively.

Question: 12

Solve the system of equations:

Solution:

The given system of equation is:

From equation (i)

33x – y + 15 = 110

 33x + 15 – 110 = y

y = 33x – 95

From equation (ii)

14y + x + 11 = 70

14y + x = 70 – 11

14y + x = 59 ….. (iii)

Substituting y = 33x – 95 in (iii) we get,

14(33x – 95) + x = 59

462x – 1330 + x = 59

463x = 59 + 1330

463x = 1389

x = 1389/463

x = 3

Putting x = 3 in y = 33x – 95 we get,

y = 33(3) – 95

99 – 95 = 4

The Solution of the given system of equation is 3 and 4 respectively.

Question: 13

Solve the system of equations:

Solution:

Taking 1/y = u the given equation becomes,

2x – 3u = 9 ….. (iii)

3x + 7u = 2 ….. (iv)

From (iii)

2x = 9 + 3u

Substituting the valuein equation (iv) we get,

27 + 23u = 4

u = – 1

y = 1/u = – 1 

x = 3

The Solution of the given system of equation is 3 and -1 respectively.

Question: 14

Solve the system of equations:

0.5x + 0.7y = 0.74

0.3x + 0.5y = 0.5

Solution:

The given system of equation is

0.5x + 0.7y = 0.74 …… (i)

0.3x – 0.5y = 0.5 ….. (ii)

Multiplying both sides by 100

50x + 70y = 74 ….. (iii)

30x + 50y = 50 … (iv)

From (iii)

50x = 74 – 70y

Substituting the value of y in equation (iv) we get,

222 – 210y + 250y = 250

40y = 28

y = 0.7

Putting the value of y in the equation (iii)

x = 0.5

The value of x and y are 0.5 and 0.7 respectively.

Question: 15

Solve the system of equations:

Solution:

Multiplying (ii) by 1/2 we get,

Solving equation (i) and (iii)

Adding we get,

When, x = 1/14 we get,

Using equation (i)

The Solution of the given system of equation is x = 1/14 and y = 1/6 respectively.

Question: 16

Solve the system of equations:

Solution:

3u + 2v = 12 ….. (i)

 v = 3

1/u = x

x = 1/2

1/v = y

y = 1/3

Question: 17

Solve the system of equations:

Solution:

15x + 2y = 17 ….. (i)

x + y = 36/5  …. (ii)

From equation (i) we get,

2y = 17 – 15x

Substitutingin equation (ii) we get,

5(-13x + 17) = 72

– 65x = –13

x = 1/5

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

y = 7

The Solution of the given system of equation is 5 and 1/7 respectively.

Question: 18

Solve the system of equations:

Solution:

3u – v = – 9 …. (i)

2u + 3v = 5 …. (ii)

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

9u – 3v = -27 ….. (iii)

2u + 3v = 5 … (iv)

Adding equation (i) and equation (iv) we get,

9u + 2u – 3v + 3v = -27 + 5

u = -2

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

2(-2) + 3v = 5

3v = 9

v = 3

Question: 19

Solve the system of equations:

Solution:

Multiplying equation (i) adding equation (ii) we get,

2y + 3x = 9 ….. (iii)

4y + 9x = 21 …. (iv)

From (iii) we get,

3x = 9 – 2y

Substitutingin equation (iv) we get

4y + 3(9 – 2y) = 21

– 2y = 21 – 27

y = 3

 x = 1

Hence the Solutions of the system of equation are 1 and 3 respectively.

Question: 20

Solve the system of equations:

Solution:

6u + 5v = 360 …. (i)

7u – 9v = 168 …. (ii)

Let us eliminate v from the equation (i) and (ii) multiplying equation (i) by 9 and (ii) by 5

54u + 35u = 3240 + 840

89u = 4080

u = 4080/89

Putting u = 4080/89 in equation (i) we get,

So, the solution of the given system of equation is x = 89/4080, y = 89/1512

Question: 21

Solve the system of equations:

Solution:

Then, the given system of equation becomes,

6u = 7v + 3

6u – 7v = 3 ….. (i)

3u = 2v

3u – 2v = 0 … (ii)

Multiplying equation (ii) by 2 and (i) 1

6u – 7v = 3

6u – 4v = 0

Subtracting v = – 1 in equation (ii), we get

3u – 2(-1) = 0

3u + 2 = 0

3u = – 2

and v = –1

x – y = –1 … (vi)

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

Putting x = -2/3 in equation (vi)

Question: 22

Solve the system of equations:

Solution:

5xy = 6x + 6y …. (i) and

xy = 6(y – x)

xy = 6y – 6x … (ii)

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

6xy = 6y + 6y

6xy = 12y

x = 2

Putting x = 2 in equation (i) we get,

10y = 12 + 6y

10 – 6y = 12

4y = 12

y = 3

The Solution of the given system of equation is 2 and 3 respectively.

Question: 23

Solve the system of equations:

Solution:

Then the given system of equation becomes:

5u – 2v = -1 ….. (i)

15u + 7v = 10 …… (ii)

Multiplying equation (i) by 7 and (ii) by 2

35u – 14v = -7 …… (iii)

30u + 14v = 20 …… (iv)

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

– 2v = – 1 – 1

– 2v = – 2

v = 1

Now,

x + y = 5 ….. (v)

x – y = 1 ….. (vi)

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

2x = 6

x = 3

Putting the value of x in equation (v)

3 + y = 5

y = 2

The Solutions of the given system of equation are 3 and 2 respectively.

Question: 24

Solve the system of equations:

Solution:

Then the given system of equation becomes:

3u + 2v = 2 ….. (i)

9u + 4v = 1 …… (ii)

Multiplying equation (i) by 3 and (ii) by 1

6u + 4v = 4 …… (iii)

9u – 4v = 1 …… (iv)

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

45u = 5

u = 1/3

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

2v = 2 – 1

2v = 1

v = 1/2

Now,

x + y = 3 ….. (v)

x – y = 2 ….. (vi)

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

2x = 5

x = 2/5

Putting the value of x in equation (v)

5/2 + y = 11

y = 1/2

The Solutions of the given system of equation are 5/2 and 1/2 respectively.

Question: 25

Solve the system of equations:

Solution:

Then the given system of equation becomes:

3u + 10v = -9 ….. (i)

25u -12v = 61/3   …… (ii)

Multiplying equation (i) by 12 and (ii) by 10

36u + 120v = -108 …… (iii)

250u + 120v = 610/3   …… (iv)

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

36u + 250u = 610/3 – 108

286u =286/3

U = 1/3

Putting u = 61/3 in equation (i)

v = -1

Now,

3x – 2y = –1 ….. (vi)

Putting x = 1/2 in equation (v) we get,

The Solutions of the given system of equation are 1/2 and 5/4 respectively.

Question: 26

Solve the system of equations:

x + y = 5xy

3x + 2y = 13xy

Solution:

The given system of equations is:

x + y = 5xy ….. (i)

3x + 2y = 13xy …… (ii)

Multiplying equation (i) by 2 and equation (ii) 1 we get,

2x ++ 2y = 10xy …… (iii)

3x + 2y = 13xy ……. (iv)

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

3x – 2x = 13xy – 10xy

x = 3xy

x/3x = y

1/3 = y

Putting y = 1/3 = y in equation (i) we get,

Hence Solution of the given system of equation is 1/2 and 1/3

Question: 27

Solve the system of equations:

x + y = xy

Solution:

x + y = xy ….. (i)

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

2x = 2xy + 6xy

2x = 6xy

y = x + y = xy

y = 1/4

Putting= y = 1/4 in equation (i), we get,

Hence the Solution of the given system of equation is x = – 1/2 and y = 1/4 respectively.

Question: 28

Solve the system of equations:

2(3u – v) = 5uv

2(u + 3v) = 5uv

Solution:

2(3u – v) = 5uv

6u – 2v = 5uv …. (i)

2(u + 3v) = 5uv

2u + 6v = 5uv ….. (ii)

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

18u – 6v = 15uv ….. (iii)

2u + 6v = 5uv …….. (iv)

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

18u + 2u = 15uv + 5uv

v = 1

Putting v = 1 in equation (i) we get,

6u – 2 = 5u

u = 2

Hence the Solution of the given system of Solution of equation is 2 and 1 respectively.

Question: 29

Solve the system of equations:

Solution:

Then the given system of equation becomes:

5u – v = 2 …… (ii)

Multiplying equation (ii) by 3

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

13u = 13/5

u = 1/5

Putting u = 1/5 in equation (i)

v = 1

Now,

3x + 2y = 5  ….. (iv)

3x – 2y = 1 ….. (v)

Adding equation (iv) and (v) we get,

6x = 6

x =1

Putting the value of x in equation (v) we get,

3 + 2y = 5

y = 1

The Solutions of the given system of equation are 1 and 1 respectively.

Question: 30

Solve the system of equations:

Solution:

Then the given system of equation becomes:

44u + 30v = 10 ….. (i)

55u + 40v = 13 … (ii)

Multiplying equation (i) by 4 and (ii) by 3

176u + 120v = 40 …… (iii)

165u + 120v = 39 …… (iv)

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

176 – 165u = 40 – 39

u = 1/11

Putting the value of u in equation (i)

4 + 30v = 10

30v = 6

x + y = 11 ….. (v)

x – y = 5 ….. (vi)

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

2x = 16

x = 8

Putting the value of x in equation (v)

8 + y = 11

y = 3

The Solutions of the given system of equation are 8 and 3 respectively.

Question: 31

Solve the system of equations:

Solution:

Then the given system of equation becomes:

10p + 2q = 4 ….. (i)

15p – 5q = – 2 …… (ii)

Multiplying equation (i) by 4 and (ii) by 3

176u + 120v = 40 …… (iii)

165u + 120v = 39 …… (iv)

Using cross multiplication method we get,

x + y = 5 ….. 3

x – y = 1 …..4

Adding equation 3 and 4 we get,

x = 3

Substituting the value of x in equation 3 we get,

y = 2

The Solution of the given system of Solution is 3 and 2 respectively.

Question: 32

Solve the system of equations:

Solution:

Then the given system of equation becomes:

Can be written as 5p + q = 2 …… 3

6p – 3q = 1 ……. 4

Equation 3 and 4 from a pair of linear equation in the general form. Now, we can use any method to solve these equations.

We get

p = 1/3

q = 1/3

Substituting the 1/(x –1) for p, we have

x – 1 = 3

x = 4

y – 2 = 3

y = 5

The Solution of the required pair of equation is 4 and 5 respectively.

Question: 33

Solve the system of equations:

Solution:

The given equation s reduce to:

-2p + 7q = 5

-2p + 7q – 5 = 0 …… 3

7p + 8q = 15

7p + 8q – 15 = 0 …… 4

Using cross multiplication method we get,

p = 1/x

q = 1/y

x = 1 and y = 1

Question: 34

Solve the system of equations:

152x – 378y = – 74

– 378x + 152y = – 604

Solution:

152x – 378y = – 74 …. 1

-378x + 152y = – 604 …. 2

Adding the equations 1 and 2, we obtain

– 226x – 226y = -678

x + y = 3 ….. 3

Subtracting the equation 2 from equation 1, we obtain

530x + 530y = 530

x – y = 1 … 4

Adding equations 3 and 4 we obtain,

2x = 4

x = 2

Substituting the value of x in equation 3 we obtain y = 1

Question: 35

Solve the system of equations:

99x + 101y = 409

101x + 99y = 501

Solution:

The given system of equation are:

99x + 101y = 409 …. 1

101x + 99y = 501 ….. 2

Adding equation 1 and 2 we get,

99x + 101x + 101y + 99y = 49 + 501

200(x + y) = 1000

x + y = 5 ….. 3

Subtracting equation 1 from 2

101x – 99x + 99y – 101y = 501 – 499

2(x – y) = 2

x – y = 1 …. 4

Adding equation 3 and 4 we get,

2x = 6

x = 3

Putting x = 3 in equation 3 we get,

3 + y = 5

y = 2

The Solution of the given system of equation is 3 and 2 respectively.

Question: 36

Solve the system of equations:

23x – 29y = 98

29x – 23y = 110

Solution:

23x – 29y = 98 … 1

29x – 23y = 110 …… 2

Adding equation 1 and 2 we get,

6(x + y) = 12

x + y = 2 … 3

Subtracting equation 1 from 2 we get,

52(x-y) = 208

x – y = 4 …. 4

Adding equation 3 and 4 we get,

2x = 6

x = 3

Putting the value of x in equation 4

3 + y = 2

y = – 1

The Solution of the given system of equation is 3 and -1 respectively.

Question: 37

Solve the system of equations:

x – y + z = 4

x – 2y – 2z = 9

2x + y + 3z = 1

Solution:

x – y + z = 4 ….. 1

x – 2y – 2z = 9 …… 2

2x + y + 3z = 1 … 3

From equation 1

z = 4 – x + y

z = -x + y + 4

Subtracting the value of the z in equation 2 we get,

x – 2y – 2(- x + y + 4) = 9

x – 2y + 2x – 2y – 8 = 8

3x – 4y = 17 ….. 4

Subtracting the value of z in equation 3, we get,

2x + y + 3(-x + y + 4) = 1

2x + y + 3x +3y + 12 =1

– x + 4y = -11

Adding equation 4 and 5 we get,

3x – x – 4y + 4y = 17 – 11

2x = 6

x = 3

Putting x = 3 in equation 4, we get,

9 – 4y = 17

– 4y = 17 – 9

y = -2

Putting x = 3 and y = -2 in z = -x + y + 4, we get,

Z = -3 – 2 + 4

= -1

The Solution of the given system of equation are 3, – 2 and –1 respectively.

Question: 38

Solve the system of equations:

x – y + z = 4

x + y + z = 2

2x + y – 3z = 0

Solution:

x – y + z = 4 …… 1

x + y + z = 2 …. 2

2x + y – 3z = 0 …… 3

From equation 1

z = – x + y + 4

Substituting z = -x + y + 4 in equation 2, we get,

x + y + (-x + y + 4) = 2

x + y – x + y + 4 = 2

2y = 2

y = 1

Substituting the value of z in equation 3

2x + y – 3(-x + y + 4) = 0

2x + y + 3x – 3y -12 = 0

5x – 2y = 12 …… 4

Putting the y = – 1 in equation 4

5x – 2(-1) = 12

5x = 10

x = 2

Putting x = 2 and y = -1 in z = -x + y + 4

z = -2 – 1 + 4

= 1

The Solution of the given system of equations are 2, -1 and 1 respectively.

All Chapter RD Sharma Solutions For Class10 Maths

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