**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).

(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 the condition, we have

[Cost of a notebook] =2 x [Cost of a pen]

i. e„ (x) = 2 x (y) or, x = 2y

or, x – 2y = 0

Thus, the required linear equation is x – 2y = 0.

Solution:

Let the cost of a notebook = Rs. x

and the cost of a pen = Rs. y

According to the condition, we have

[Cost of a notebook] =2 x [Cost of a pen]

i. e„ (x) = 2 x (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 =

(ii)

(iii) – 2x + 3y = 6

(iv) x = 3y

(v) 2x = -5y

(vi) 3x + 2 = 0

(vii) y – 2 = 0

(viii) 5 = 2x

Solution:

(i) We have 2x + 3y =

or (2)x + (3)y + (

Comparing it with ax + by +c= 0, we geta = 2,

b = 3 and c= –

**(ii) We have
or x + (-
) y + (10) = 0
Comparing it with ax + by + c = 0, we get
a =1, b =-
and c= -10**

**(iii) We have -2x + 3y = 6 or (-2) x + (3)y + (-6) = 0
Comparing it with ax – 4 – by + c = 0,we get a = -2, b = 3 and c = -6.**

**(iv) We have x = 3y or (1) x + (-3) y + (0) = 0 Comparing it
with ax + by + c = 0, we get a = 1, b = -3 and c = 0.
(v) We have 2x = -5y or (2) x + (5) y + (0) = 0 Comparing it with ax + by + c =
0, we get a = 2, b = 5 and c = 0.
(vi) We have 3x + 2 = 0 or (3) x + (0) y + (2) = 0 Comparing it with ax + by +
c = 0, we get a = 3, b = 0 and c = 2.
(vii) We have y – 2 = 0 or (0) x + (1) y + (-2) = 0 Comparing it with ax + by +
c = 0, we get a = 0, b = 1 and c = -2.
(viii) We have 5 = 2x **

**⇒**

**5 – 2x = 0**

or -2x + 0y + 5 = 0

or (-2) x + (0) y + (5) = 0

Comparing it with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

or -2x + 0y + 5 = 0

or (-2) x + (0) y + (5) = 0

Comparing it with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

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