NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1.

- Linear Equations in Two Variables Class 9 Ex 4.1
- Linear Equations in Two Variables Class 9 Ex 4.2
- Linear Equations in Two Variables Class 9 Ex 4.3
- Linear Equations in Two Variables Class 9 Ex 4.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Linear Equations in Two Variables |

Exercise |
Ex 4.1 |

Number of Questions Solved |
2 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1

**Ex 4.1 Class 9 Maths Question 1.**

The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

(Take the cost of a notebook to be Rs. x and that of a pen to be Rs. y)

**Solution:**

Let the cost of a notebook =Rs.x

and the cost of a pen = Rs.

y According to question,

Cost of a notebook = 2 (cost of a pen)

x = 2y ⇒ x -2y = 0

**Ex 4.1 Class 9 Maths Question 2.**

**Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:**

(i) 2x + 3y = 9.3 \(\bar { 5 } \)

(ii) x- \(\frac { y }{ 5 }\) – 10 = 0

(iii) – 2x + 3y = 6

(iv) x = 3y

(v) 2x = – 5y

(vi) 3x+2=0

(vii) y-2=0

(viii) 5=2x

**Solution:**

**(i)** Given linear equation, 2x + 3y= 9.35

∴ (2)x + (3)y + (- 9.3 \(\bar { 5 } \)) = 0

On comparing with ax ‘+ bx + c = 0, we have

a = 2, b = 3 and c = – 9.3\(\bar { 5 } \)

**(ii)** Given linear equation x-\(\frac { y }{ 5 }\)-10 =0

⇒ x+(-\(\frac { y }{ 5 }\)) y+(-10)=0

On comparing with ax + by + c = 0, we get

a = 1, b = –\(\frac { 1 }{ 5 }\) and c = -10

**(iii)** Given linear equation -2x + 3y = 6

⇒ -2x + 3 y – 6 = 0

⇒ (-2)x + (3) y + (-6) = 0

On Comparing with ax + by + c – 0, we get

a = – 2, b = 3 and c = – 6

**(iv)** Given linear equation x = 3y

⇒ x-3y=0

⇒(1) x + (-3) y + 0 = 0

On comparing with ax + by + c = 0, we get

a = 1, b = -3 and c = 0.

**(v)** Given linear equation 2x = – 5y

⇒ 2x + 5y = 0

⇒ (2)x + (5)y + 0 = 0

On comparing with ax + by + c = 0, we get

a – 2, b = 5 and c = 0.

**(vi)** Given linear equation 3x + 2 = 0

⇒ 3x +2 + 0y = 0

⇒ (3)x + (0)y + (2) = 0

On comparing with ax + by + c = 0 , we get

a = 3, b = 0 and c = 2.

**(vii)** Given linear equation y – 2=0

⇒ (0)x + (l)y + (-2) = 0

On comparing with ax + by + c = 0, we get

a= 0, b-1 and c = – 2

**(viii)** Given linear equation

5= 2x

⇒ 5 – 2x = 0

⇒ -2x+0y + 5 = 0

⇒ (-2)x + (0)y + (5) = 0

On comparing with ax + by-c = 0, we get

a= – 2, b = 0 and c = 5.

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