NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Linear Equations in Two Variables |

Exercise |
Ex 8.1 |

Number of Questions Solved |
2 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.1

**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 ₹ x and that of a pen to be ₹ y)**

**Solution:**

Let the cost of a notebook = ₹ x

and the cost of a pen = ₹ y

According to question,

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

∴ x = 2y

⇒ x- 2y = 0

**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.\(\overline { 35 }\)**

**(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)** 2x + 3y = 9.\(\overline { 35 }\)

⇒ 2x + 3y – 9..\(\overline { 35 }\) = 0

On comparing with ax+by+c = 0.

Then, the values of a = 2, b = 3 and c = 9..\(\overline { 35 }\)

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

On comparing with ax + by + c = 0, then the values of

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

**(iii)** – 2x+ 3y = 6

⇒ -2x + 3y – 6=0

On comparing with ax + by + c = 0, then the values of a = – 2, b = 3 and c = – 6

**(iv)** x = 3y ⇒ x-3y+0=0

On comparing with ax + by + c = 0, then the values of a = 1, b = – 3 and c = 0.

**(v)** 2x = -5y

⇒ 2x + 5y + 0=0

On comparing with ax + by + c = 0, then the values of a = 2, b = 5 and c = 0.

**(vi)** 3x + 2 = 0

⇒ 3x + 0y + 2 = 0

On comparing with ax + by + c = 0, then the values of a = 3, b = 0 and , c= 2.

**(vii)** y – 2 = 0

⇒0x+y-2 = 0

On comparing with ax + by + c = 0, then the values of a = 0, b = 1 and c = – 2.

**(viii)** 5 = 2x ⇒ 2x + 0y – 5 = 0

On comparing with ax + by + c = 0, then the values of a = 2, b = 0 and c = – 5.

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