RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Linear Equations in Two Variables |

Exercise |
Ex 4.2 |

Number of Questions Solved |
15 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2

Solve each of the following systems of simultaneous linear (RBSESolutions.com) equations by the method of substitution. [Q1. to Q6.]

Question 1.

2x + 3y = 9

3x + 4y = 5

Solution.

The given equations are

2x + 3y = 9 ….(i)

and 3x + 4y = 5 ….(ii)

From (i) 2x = 9 – 3y

Hence, the (RBSESolutions.com) required solution is x = – 21 and y = 17.

Question 2.

x + 2y = – 1

2x – 3y = 12

Solution

The given equations are

x + 2y = – 1 ..(i)

and 2x – 3y = 12 …(ii)

From (i) x = – 1 – 2y …(iii)

Substituting this (RBSESolutions.com) value of x in (ii), we get

2( – 1 – 2y) – 3y = 2

⇒ – 2 – 4y – 3y = 12

⇒ – 7y = 14

⇒ y = -2

Putting y = – 2 in (iii), we get

x = -1-2×(-2) = -1+4 = 3

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

Question 3.

3x + 2y = 11

2x + 3y = 4

Solution

The given equations are

3x + 27 = 11 ….(i)

and 2x + 3y = 4 …(ii)

From (i) 2y = 11 – 3x

Hence the required (RBSESolutions.com) solution is x = 5 and y = – 2

Question 4.

8x + 5y = 9

3x + 2y = 4

Solution

The given equations are

8x + 5y = 9 …(i)

and 3x + 2y = 4 …(ii)

From (ii) 2y = 4 – 3x

Hence, the required (RBSESolutions.com) solution is x = – 2 and y = 5.

Question 5.

4x – 5y = 39

2x – 7y = 51

Solution

The given equations are

4x – 5y = 39 …(i)

and 2x – 7y = 51 …(ii)

From equation (ii),

2x – 51 = 7y

Hence the required (RBSESolutions.com) solution is x = 1, y = -7.

Question 6.

5x – 2y = 19

3x + y = 18

Solution

The given equations are

5x – 2y = 19 …(i)

and 3x + y = 18 …(ii)

From (ii), y = 18 – 3x …(iii)

Substituting y from (iii) in (i), we get

5x – 2(18 – 3x) = 19

⇒ 5x – 36 + 6x = 19

⇒ 11x = 55

⇒ x = 5

Putting x = 5 in (iii), we get

y = 18 – 3 × 5

y = 18 – 15 = 3

Hence the required (RBSESolutions.com) solution is x = 5 and y = 3.

Solve each of the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.

Question 7.

2x + y = 13

5x – 3y = 16

Solution

The given equations are

2x + y = 13 …(i)

and 5x – 3y= 16 …(ii)

Multiplying equation (i) by 3 and then adding in (ii), we get

11x = 55

Substituting (RBSESolutions.com) the value of x in (i), we get

2 × 5 + y = 13

⇒ 10 + y = 13

⇒ y = 13 – 10 = 3

Hence the required solution is x = 5, y = 3.

Question 8.

0.4x + 0.3y = 1.7

0.7x – 0.2y = 0.8

Solution

The given equations are

0.4x + 0.3y = 1.7 …(i)

0.7x – 0.2y = 0.8 …(ii)

Multiplying equation (i) by 2 and (ii) by 3 and then (RBSESolutions.com) adding, we get

2.9x = 5.8

⇒ x = 2

Substituting the value of x in (i), we get

0.4 x 2 + 0.3y = 1.7

⇒ 0.3y = 1.7 – 0.8

⇒ 0.3y = 0.9

⇒ y = 3

Question 9.

Solution

The given (RBSESolutions.com) equations are

Hence the required (RBSESolutions.com) solution is x = 14 and y = 9

Question 10.

11x + 15y = – 23

7x – 2y = 20

Solution

The given equations are

11x + 15y = – 23 …(i)

and 7x -2y = 20 …(ii)

Multiplying (RBSESolutions.com) equation (i) by 2 and (ii) by 15 and then adding, we get

127x = 254

⇒ x = 2

Putting x = 2 in (ii), we get

7 × 2 – 2y = 20

⇒ 14 – 20 = 2y

⇒ 2y = – 6

⇒ y = – 3

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

Question 11.

3x – 7y + 10 = 0

y – 2x = 3

Solution

The given (RBSESolutions.com) equations are

3x – 7y + 10 = 0 …(i)

and y – 2x = 3 …(ii)

Multiplying equation (i) by 2 and (ii) by 3 and then adding, we get

– 11y = – 11

⇒ y = 1

Putting y = 1 in (ii), we get

1 – 2x = 3 ⇒ 1 – 3 = 2x

⇒ 2x = -2

⇒ x = -1

Hence the required solution is x = – 1 and y = 1

Question 12.

x + 2y =

2x + y =

Solution

The given (RBSESolutions.com) equations are

Solve each of the (RBSESolutions.com) following (Q13. to Q15.)

Question 13.

8v – 3u = 5uv

6v – 5u = – 2uv

Solution

The given equations are

8v – 3u = 5uv …(i)

and 6v – 5u = – 2uv …(ii)

Dividing equation (i) and (ii) both side by uv, we get

Question 14.

Solution

The given equations are

Question 15.

Solution

The given equations are

Adding (v) and (vi), we get

2x = 6

⇒ x = 3 and 3 + y = 5

⇒ y = 2

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

We hope the given RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2, drop a comment below and we will get back to you at the earliest.

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