# NCERT Solutions Class 9 Maths Chapter 4 Linear Equations In Two Variable Exercise 4.1

## Introduction:

In this exercise/article we will learn about Linear Equations In Two Variable. Can you write down a linear equations in one variable? We may say that x + 1 = 0, x + √2 = 0 and √2y + √3 = 0 are examples of linear equations in one variable. You also know that such equations have a unique (i.e., one and only one) solution. You may also remember how to represent the solution on a number line. You shall also use the concepts you studied in chapter 3 to answer these questions.

NCERT Class 9 Maths Chapter 4 Coordinate Geometry :

- NCERT Class 9 Maths Chapter 4 Linear Equations In Two Variable Exercise 4.1
- NCERT Class 9 Maths Chapter 4 Linear Equations In Two Variable Exercise 4.2

**Class 9 Maths Exercise 4.1 (Page-57)**

**Q1. **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 = x

Let the cost of a pen = y

According to the question,

The cost of a notebook is twice the cost of a pen

⇒ The cost of a notebook = 2 × the cost of a pen

⇒ x = 2 × y

⇒ x = 2y

⇒ x - 2y = 0

∴ The required linear equation = x - 2y = 0 .

**Q2. **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\(\displaystyle \overline{5}\)

**Solution :**

We have 2x + 3y = 9.3\(\displaystyle \overline{5}\)

Re-arranging the equation, we get

2x + 3y - c9.3\(\displaystyle \overline{5}\) = 0

Now, comparing with ax + by + c = 0

we get,

⇒ a2 + b3 - c9.3\(\displaystyle \overline{5}\) = 0

⇒ a = 2 , b = 3 and c = -9.3\(\displaystyle \overline{5}\)

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

**Solution :**

We have x - \(\displaystyle \frac{y}{5}\) - 10 = 0

Now, comparing with ax + by + c = 0

we get,

⇒ ax - b\(\displaystyle \frac{y}{5}\) - c10 = 0

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

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

**Solution :**

We have -2x + 3y = 6

Re-arranging the equation, we get

-2x + 3y - 6 = 0

Now, comparing with ax + by + c = 0

we get,

⇒ -a2 + b3 - 6 = 0

⇒ a = -2 , b = 3 and c = -6

**(iv)** x = 3y

**Solution :**

We have x = 3y

Re-arranging the equation, we get

x - 3y + 0 = 0

Now, comparing with ax + by + c = 0

we get,

⇒ ax - b3 + c0 = 0

⇒ a = 1 , b = -3 and c = 0

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

**Solution :**

We have 2x = -5y

Re-arranging the equation, we get

2x + 5y + 0 = 0

Now, comparing with ax + by + c = 0

we get,

⇒ a2 + b5 + c0 = 0

⇒ a = 2 , b = 5 and c = 0

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

**Solution :**

We have 3x + 2 = 0

Re-arranging the equation, we get

3x + 0 + 2 = 0

Now, comparing with ax + by + c = 0

we get,

⇒ a3 + b0 + c2 = 0

⇒ a = 3 , b = 0 and c = 2

**(vii)** y - 2 = 0

**Solution :**

We have y - 2 = 0

Re-arranging the equation, we get

0 + y - 2 = 0

Now, comparing with ax + by + c = 0

we get,

⇒ a0 + b1 - c2 = 0

⇒ a = 0 , b = 1 and c = -2

**(viii)** 5 = 2x

**Solution :**

We have 5 = 2x

Re-arranging the equation, we get

-2x + 0 + 5 = 0

Now, comparing with ax + by + c = 0

we get,

⇒ -a2 + b0 + c5 = 0

⇒ a = -2 , b = 0 and c = 5