# Class 9 NCERT Solutions – Chapter 4 Linear Equations in two variables – Exercise 4.1

**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.

**Solution:**

Let’s take the cost of a notebook as Rs. x and the cost of a pen as Rs. y.

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free classeswhich will definitely help them in making a wise career choice in the future.Given that cost of a notebook (x) is twice the cost of a pen(y).

So, x = 2y.

x – 2y = 0

x – 2y = 0 is the linear equation in two variables that represent the statement.

**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 = 9.35 **

**Solution:**

2x + 3y – 9.35 = 0 (Transposing 9.35 to LHS)

(2)x + (3)y + (-9.35) = 0 (converting it in the form ax + by + c = 0)

On comparing equation with ax + by + c = 0,

We get

a = 2, b = 3, c = -9.35

**(ii) x – y/5 -10 = 0**

**Solution:**

(5x – y – 50)/5 = 0 (Multiply and divide the whole equation by 5)

5x – y – 50 = 0

(5)x + (-1)y + (-50) = 0 (converting it in the form ax + by + c = 0)

On comparing equation with ax + by +c =0,

We get

a = 5, b = -1, c = -50

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

**Solution:**

-2x + 3y – 6 = 0 (Transposing 6 to LHS)

(-2)x + (3)y+ (-6) = 0 (converting it in the form ax + by + c = 0)

On comparing equation with ax + by +c =0,

We get

a = -2, b = 3, c = -6

**(iv) x = 3y**

**Solution:**

x – 3y = 0 (Transposing 3y to LHS)

(1)x + (-3)y + 0 = 0 (converting it in the form ax + by + c = 0)

On comparing equation with ax + by + c = 0,

We get

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

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

**Solution:**

2x + 5y = 0 (Transposing 5y to LHS)

(2)x + (5)y + 0 = 0 (converting it in the form ax + by + c = 0)

On comparing equation with ax + by +c =0,

We get

a = 2, b = 5, c = 0

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

**Solution:**

(1)x + (0)y + 2 = 0 (converting it in the form ax + by + c = 0)

On comparing equation with ax + by +c =0,

We get

a = 1, b = 0, c = 2

**(vii) y – 2 = 0**

**Solution:**

(0)x + (1)y + (-2) = 0 (converting it in the form ax + by + c = 0)

On comparing equation with ax + by +c =0,

We get

a = 0, b = 1, c = -2

**(viii) 5 = 2x**

**Solution:**

5 – 2x = 0 (Transposing 2x to LHS)

(-2)x + (0)y + 5 = 0 (converting it in the form ax + by + c = 0)

On comparing equation with ax + by + c = 0,

We get

a = -2, b = 0, c = 5