**NCERT Solutions for Class 9 Maths Chapter 4** , When preparing for the CBSE Class 9 Maths exams, pupils are believed to find linear equations in two variables to be of great assistance. Here, we provide in-depth solutions to the exercises from **NCERT Solutions for class 9 Maths Chapter 4** .

These questions from **NCERT Solutions for Class 9 Maths Chapter 4** have been compiled for students to review by the subject matter experts who prepared these NCERT Solutions. All of the questions from the NCERT books are answered precisely by us.

The most recent revision of the CBSE syllabus for 2023–2024 and its guidelines will be used to inform these **NCERT Solutions for Class 9 Maths Chapter 4** . These tasks will provide students enough practise, and they will also help them achieve top grades.

**NCERT Solutions for Class 9 Maths Chapter 4 **

**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.**

**(Take the cost of a notebook to be ₹ x and that of a pen to be ₹ y)**

Solution:

Let the cost of a notebook be = ₹ x

Let the cost of a pen be = ₹ y

According to the question,

The cost of a notebook is twice the cost of a pen.

i.e., cost of a notebook = 2×cost of a pen

x = 2×y

x = 2y

x-2y = 0

x-2y = 0 is the linear equation in two variables to represent the statement, ‘The cost of a notebook is twice the cost of a pen.’

**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.**

**(ii) x –(y/5)–10 = 0**

Solution:

The equation x –(y/5)-10 = 0 can be written as,

1x+(-1/5)y +(–10) = 0

Now comparing x+(-1/5)y+(–10) = 0 with ax+by+c = 0

We get,

a = 1

b = -(1/5)

c = -10

**(iii) –2x+3y = 6**

Solution:

–2x+3y = 6

Re -arranging the equation, we get,

–2x+3y–6 = 0

The equation –2x+3y–6 = 0 can be written as,

(–2)x+3y+(– 6) = 0

Now, comparing (–2)x+3y+(–6) = 0 with ax+by+c = 0

We get, a = –2

b = 3

c =-6

**(iv) x = 3y**

Solution:

x = 3y

Re -arranging the equation, we get,

x-3y = 0

The equation x-3y=0 can be written as,

1x+(-3)y+(0)c = 0

Now comparing 1x+(-3)y+(0)c = 0 with ax+by+c = 0

We get a = 1

b = -3

c =0

**(v) 2x = –5y**

Solution:

2x = –5y

Re -arranging the equation, we get,

2x+5y = 0

The equation 2x+5y = 0 can be written as,

2x+5y+0 = 0

Now, comparing 2x+5y+0= 0 with ax+by+c = 0

We get a = 2

b = 5

c = 0

**(vi) 3x+2 = 0**

Solution:

3x+2 = 0

The equation 3x+2 = 0 can be written as,

3x+0y+2 = 0

Now comparing 3x+0+2= 0 with ax+by+c = 0

We get a = 3

b = 0

c = 2

**(vii) y–2 = 0**

Solution:

y–2 = 0

The equation y–2 = 0 can be written as,

0x+1y+(–2) = 0

Now comparing 0x+1y+(–2) = 0with ax+by+c = 0

We get a = 0

b = 1

c = –2

**(viii) 5 = 2x**

Solution:

5 = 2x

Re-arranging the equation, we get,

2x = 5

i.e., 2x–5 = 0

The equation 2x–5 = 0 can be written as,

2x+0y–5 = 0

Now comparing 2x+0y–5 = 0 with ax+by+c = 0

We get a = 2

b = 0

c = -5

**NCERT Solutions for Class 9 Maths Chapter 4 **

**Exercise 4.2**

**Question 1. Which one of the following options is true, and why?**

**y = 3x+5 has**

**A unique solution****Only two solutions****Infinitely many solutions**

Solution:

Let us substitute different values for x in the linear equation y = 3x+5

From the table, it is clear that x can have infinite values, and for all the infinite values of x, there are infinite values of y as well.

Hence, (iii) infinitely many solutions is the only option true.

**2. Write four solutions for each of the following equations:**

**(i) 2x+y = 7**

Solution:

To find the four solutions of 2x+y =7, we substitute different values for x and y.

Let x = 0

Then,

2x+y = 7

(2×0)+y = 7

y = 7

(0,7)

Let x = 1

Then,

2x+y = 7

(2×1)+y = 7

2+y = 7

y = 7-2

y = 5

(1,5)

Let y = 1

Then,

2x+y = 7

(2x)+1 = 7

2x = 7-1

2x = 6

x = 6/2

x = 3

(3,1)

Let x = 2

Then,

2x+y = 7

(2×2)+y = 7

4+y = 7

y =7-4

y = 3

(2,3)

The solutions are (0, 7), (1,5), (3,1), (2,3)

**(ii) πx+y = 9**

Solution:

To find the four solutions of πx+y = 9, we substitute different values for x and y.

Let x = 0

Then,

πx+y = 9

(π×0)+y = 9

y = 9

(0,9)

Let x = 1

Then,

πx +y = 9

(π×1)+y = 9

π+y = 9

y = 9-π

(1, 9-π)

Let y = 0

Then,

πx+y = 9

πx+0 = 9

πx = 9

x = 9/π

(9/π,0)

Let x = -1

Then,

πx + y = 9

(π×-1) + y = 9

-π+y = 9

y = 9+π

(-1,9+π)

The solutions are (0,9), (1,9-π), (9/π,0), (-1,9+π)

**(iii) x = 4y**

Solution:

To find the four solutions of x = 4y, we substitute different values for x and y.

Let x = 0

Then,

x = 4y

0 = 4y

4y= 0

y = 0/4

y = 0

(0,0)

Let x = 1

Then,

x = 4y

1 = 4y

4y = 1

y = 1/4

(1,1/4)

Let y = 4

Then,

x = 4y

x= 4×4

x = 16

(16,4)

Let y = 1

Then,

x = 4y

x = 4×1

x = 4

(4,1)

The solutions are (0,0), (1,1/4), (16,4), (4,1)

**3. Check which of the following are solutions of the equation x–2y = 4 and which are not:**

**(i) (0, 2)**

**(ii) (2, 0)**

**(iii) (4, 0)**

**(iv) (√2, 4√2)**

**(v) (1, 1)**

Solutions:

**(i) (0, 2)**

(x,y) = (0,2)

Here, x=0 and y=2

Substituting the values of x and y in the equation x–2y = 4, we get,

x–2y = 4

⟹ 0 – (2×2) = 4

But, -4 ≠ 4

(0, 2) is **not** a solution of the equation x–2y = 4

**(ii) (2, 0)**

(x,y) = (2, 0)

Here, x = 2 and y = 0

Substituting the values of x and y in the equation x -2y = 4, we get,

x -2y = 4

⟹ 2-(2×0) = 4

⟹ 2 -0 = 4

But, 2 ≠ 4

(2, 0) is **not** a solution of the equation x-2y = 4

**(iii) (4, 0)**

Solution:

(x,y) = (4, 0)

Here, x= 4 and y=0

Substituting the values of x and y in the equation x -2y = 4, we get,

x–2y = 4

⟹ 4 – 2×0 = 4

⟹ 4-0 = 4

⟹ 4 = 4

(4, 0) is a solution of the equation x–2y = 4

**(iv) (√2,4√2)**

Solution:

(x,y) = (√2,4√2)

Here, x = √2 and y = 4√2

Substituting the values of x and y in the equation x–2y = 4, we get,

x –2y = 4

⟹ √2-(2×4√2) = 4

√2-8√2 = 4

But, -7√2 ≠ 4

(√2,4√2) is **not** a solution of the equation x–2y = 4

**(v) (1, 1)**

Solution:

(x,y) = (1, 1)

Here, x= 1 and y= 1

Substituting the values of x and y in the equation x–2y = 4, we get,

x –2y = 4

⟹ 1 -(2×1) = 4

⟹ 1-2 = 4

But, -1 ≠ 4

(1, 1) is **not** a solution of the equation x–2y = 4

**4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k.**

Solution:

The given equation is

2x+3y = k

According to the question, x = 2 and y = 1

Now, substituting the values of x and y in the equation 2x+3y = k,

We get,

(2×2)+(3×1) = k

⟹ 4+3 = k

⟹ 7 = k

k = 7

The value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k, is 7.