In this chapter, we provide RD Sharma Class 10 Ex 3.3 Solutions Chapter 3 Pair of Linear Equations in Two Variables for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 3.3 Solutions Chapter 3 Pair of Linear Equations in Two Variables pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 3 |

Chapter Name | Pair of Linear Equations in Two Variables |

Exercise | 3.3 |

Category | RD 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.

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