RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3 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 4.3.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Linear Equations in Two Variables |

Exercise |
Ex 4.3 |

Number of Questions Solved |
7 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

In each of the following system of linear equation (RBSESolutions.com) determine whether the system of equations has a unique solution or no solution or infinitely many solutions. In case there is a unique solution, find it.

Question 1.

2x + y = 35

3x + 4y = 65

Solution

The given system of equations are

2x + y = 35

3x + 4y = 65

Hence, the given system of equations is (RBSESolutions.com) consistent and have unique solution.

2x + y = 35 …(i)

3x + 4y = 65 …(ii)

Multiplying equation (i) by 4 and subtracting (ii) from it, we get

5x = 75

⇒ x = 15

Putting x = 15 in equation (i), we get y = 5

Hence, x = 15 and y = 5 is the required unique solution.

Question 2.

2x – y = 6

x – y = 2

Solution

The given system (RBSESolutions.com) of equations are

2x – y = 6

x – y = 2

Hence the given system of equations (RBSESolutions.com) is consistent and have unique solution.

2x – y = 6 …(i)

x – y = 2 …(ii)

Subtracting (ii) from (i), we get x = 4

Putting the value of x in (ii), we get y = 2

Hence x = 4 and y = 2 is the required unique solution.

Question 3.

3x + 2y + 25 = 0

2x + y + 10 = 0

Solution

The given system of equations are

3x + 2y + 25 = 0

2x + y + 10 = 0

Hence the given system of equations (RBSESolutions.com) is consistent and hence it will have unique solution

3x + 2y + 25 = 0 …(i)

2x + y + 10 = 0 …(ii)

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

-x + 5 = 0

x = 5

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

y = – 20

Hence, x = 5 and y = – 20 is the required unique solution.

Question 4.

x + 2y + 1 = 0

2x – 3y – 12 = 0

Solution

The given system (RBSESolutions.com) of equations are

x + 2y + 1 = 0

2x – 3y – 12 = 0

Hence, the given system of equations is consistent and it has unique solution.

x + 2y + 1 = 0 …(i)

2x – 3y – 12 = 0 …(ii)

Multiplying equation (i) by 2 and subtracting (ii) from it, we get

7y + 14 = 0

⇒ y = – 2

Putting the value of y in (RBSESolutions.com) equation (i), we get x = 3.

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

Question 5.

Find the value of ‘k’ for which the system

(i) 2x + ky = 1, 3x – 5y = 7

(ii) kx + 2y = 5, 3x + y = 1 has no solution.

Solution.

(i) The system of equations are

2x + ky – 1 = 0 and 3x – 5y – 1 = 0

∴ For no (RBSESolutions.com) solution, condition is

Question 6.

Solve the following (RBSESolutions.com) system of equations

mx – ny = m² + n²

x + y = 2m

Solution.

The given equations are

mx – ny = m² + n² …(i)

and x + y = 2m …(ii)

Multiplying equation (ii) by n and then adding, we get

(m + n)x = m² + n² + 2 mn

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

(m + n) + y = 2m

⇒ y = 2m – m – n

⇒ y = m – n

Hence, required solution is m + n, m – n.

Question 7.

Find the value of ‘λ’ for which the system

3x + λy + 1 = 0

2x + y – 9 = 0 will have

(i) unique solution

(ii) no solution

Solution.

The given system (RBSESolutions.com) of equations are

3x + λy + 1 = 0 …(i)

2x + y – 9 = 0 …(ii)

Here we have

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