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

NCERT Solutions for Class
9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 are the part of
NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1.

**NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1****NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2****NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3****NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4**

**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 be Rs. X and the cost of a pen be Rs. Y.

According to the question, we have

[The cost of a notebook] = 2 × [The cost of a pen]

x = 2 × y

Or, x = 2y

Or, x – 2y
= 0

Thus, the
required linear equation is 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 = 9.35¯

**(ii)**x – 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)** The
given linear equation is, 2x + 3y = 9.35¯

∴ (2)x + (3)y + (-9.35¯) = 0

On
comparing the above equation with ax
+ bx + c =
0, we have

a =
2, b =
3 and c = –9.35¯

**(ii)** The given linear equation is, x – y/5 – 10 = 0

⇒ x + (-1/5)y + (-10) = 0

On comparing the above equation with ax + by + c = 0, we have

a
= 1, b = –1/5 and
c = -10

**(iii)** The given linear equation is, -2x +
3y = 6

⇒ -2x + 3y – 6 = 0

⇒ (-2)x + (3)y + (-6) = 0

On comparing the above equation with ax + by + c – 0, we have

a = –2, b = 3 and c = –6

**(iv)** The given linear equation is, x =
3y

⇒ x – 3y = 0

⇒ (1)x + (-3)y + 0 = 0

On comparing the above equation with ax + by + c = 0, we have

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

**(v)** The given linear equation is, 2x =
–5y

⇒ 2x + 5y = 0

⇒ (2)x + (5)y + 0 = 0

On comparing the above equation with ax + by + c = 0, we have

a = 2, b = 5 and c = 0.

**(vi)** The given linear equation is, 3x +
2 = 0

⇒ 3x + 2 + 0y = 0

⇒ (3)x + (0)y + (2) = 0

On comparing the above equation with ax + by + c = 0 , we have

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

**(vii)** The given linear equation is, y – 2
= 0

⇒ (0)x + (1)y + (-2) = 0

On comparing the above equation with ax + by + c = 0, we have

a = 0, b = 1 and c = –2.

**(viii)** The given linear equation is, 5 =
2x

⇒ 5 – 2x = 0

⇒ -2x + 0y + 5 = 0

⇒ (-2)x + (0)y + (5) = 0

On comparing the above equation with ax + by + c = 0, we have

a = –2, b = 0 and c = 5.

**NCERT Solutions for Maths Class 10**