Skip to content
Related Articles

Related Articles

Class 9 RD Sharma Solutions – Chapter 13 Linear Equation in Two Variable – Exercise 13.1
  • Last Updated : 27 Nov, 2020
GeeksforGeeks - Summer Carnival Banner

Question 1: 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 = 12 

(ii) x – y/2 – 5 = 0 

(iii) 2x + 3y = 9.35

(iv) 3x = -7y 

(v) 2x + 3 = 0 



(vi) y – 5 = 0

(vii) 4 = 3x 

(viii) y = x/2

Solution:

(i) -2x + 3y = 12

Rearranging, 

– 2x + 3y – 12 = 0

On comparing with the given form of linear equation, ax + by + c = 0,

We get, 

a = – 2 

b = 3

c = -12

(ii) x – y/2 – 5 = 0

On comparing with the given form of linear equation, ax + by + c = 0,

We get,

a = 1

b = -1/2

c = -5



(iii) 2x + 3y = 9.35

Rearranging, 2x + 3y – 9.35 = 0

On comparing with the given form of linear equation, ax + by + c = 0,

We get,

a = 2 

b = 3 

c = -9.35

(iv) 3x = -7y

Rearranging, 3x + 7y + 0 = 0

On comparing with the given form of linear equation, ax + by + c = 0,

We get,

a = 3

b = 7

 c = 0

(v) 2x + 3 = 0

Rearranging, 2x + 0y + 3 = 0

On comparing with the given form of linear equation, ax + by + c = 0,

We get,

a = 2 

b = 0 

c = 3

(vi) y – 5 = 0

Rearranging, 0x + y – 5 = 0

On comparing with the given form of linear equation, ax + by + c = 0,

We get,

a = 0

b = 1 

c = -5

(vii) 4 = 3x

Rearranging, 3x + 0y – 4 = 0

On comparing with the given form of linear equation, ax + by + c = 0,

We get,

a = 3

b = 0

c = -4

(viii) y = x/2

Rearranging, x – 2y +0 = 0

On comparing with the given form of linear equation, ax + by + c = 0,

We get,

a = 1

b = -2

c = 0

Question 2: Write each of the following as an equation in two variables:

(i) 2x = -3 

(ii) y=3 

(iii) 5x = 7/ 2 

(iv) y = 3/2x

Solution:

(i) 2x = -3

Rearranging,

2x + 3 = 0

Now adding ‘y’ term,

2x + 0.y + 3 = 0

Required equation is,

2x + 0.y + 3 = 0

(ii) y = 3

Rearranging,

y – 3 = 0

Now adding ‘x’ term,

0.x + y – 3 = 0

Required equation is,

0.x + y – 3 = 0

(iii) 5x = 7/2

Rearranging,

10x = 7,

or 10x – 7 – 0;

Now adding ‘y’ term,

10x +0.y – 7 = 0

Required equation is,

10x + 0.y – 7 = 0

(iv) y = 3/2 x

Rearranging,

2y = 3x

or 3x – 2y = 0

Now adding the constant term,

3x – 2y + 0 = 0

Required equation is,

3x – 2y + 0 = 0

Question 3: The cost of ball pen is Rs 5 less than half of the cost of fountain pen. Write this statement as a linear equation in two variables.

Solution:

Let the cost of a ball pen and fountain pen be x and y respectively.

According to the question the following equation can be formed,

x = y/2 − 5

or x = (y – 10)/2

or 2x = y – 10

or 2x – y + 10 = 0

The required linear equation will be 2x – y + 10 = 0.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :