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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 part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1.

Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 4
Chapter Name Linear Equations in Two Variables
Exercise  Ex 4.1
Number of Questions Solved 2
Category NCERT Solutions

NCERT Solutions for Class 9 Maths 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)
Solution:
Let the cost of a notebook =Rs.x
and the cost of a pen = Rs.
y According to question,
Cost of a notebook = 2 (cost of a pen)
x = 2y ⇒ 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 = 9.3 \bar { 5 }
(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) Given linear equation, 2x + 3y= 9.35
∴ (2)x + (3)y + (- 9.3 \bar { 5 } ) = 0
On comparing with ax ‘+ bx + c = 0, we have
a = 2, b = 3 and c = – 9.3\bar { 5 }

(ii) Given linear equation x-\frac { y }{ 5 }-10 =0
⇒ x+(-\frac { y }{ 5 }) y+(-10)=0
On comparing with ax + by + c = 0, we get
a = 1, b = –\frac { 1 }{ 5 } and c = -10

(iii) Given linear equation -2x + 3y = 6
⇒ -2x + 3 y – 6 = 0
⇒ (-2)x + (3) y + (-6) = 0
On Comparing with ax + by + c – 0, we get
a = – 2, b = 3 and c = – 6

(iv) Given linear equation x = 3y
⇒ x-3y=0
⇒(1) x + (-3) y + 0 = 0
On comparing with ax + by + c = 0, we get
a = 1, b = -3 and c = 0.

(v) Given linear equation 2x = – 5y
⇒ 2x + 5y = 0
⇒ (2)x + (5)y + 0 = 0
On comparing with ax + by + c = 0, we get
a – 2, b = 5 and c = 0.

(vi) Given linear equation 3x + 2 = 0
⇒ 3x +2 + 0y = 0
⇒  (3)x + (0)y + (2) = 0
On comparing with ax + by + c = 0 , we get
a = 3, b = 0 and c = 2.

(vii) Given linear equation y – 2=0
⇒ (0)x + (l)y + (-2) = 0
On comparing with ax + by + c = 0, we get
a= 0, b-1 and c = – 2

(viii) Given linear equation
5= 2x
⇒ 5 – 2x = 0
⇒  -2x+0y + 5 = 0
⇒ (-2)x + (0)y + (5) = 0
On comparing with ax + by-c = 0, we get
a= – 2, b = 0 and c = 5.

 

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