# NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

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.

## 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 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:

(iii) Wehave -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.

## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2

### Ex 4.2 Class 9 Maths Question 1.

Which one of the following options is true, and why?

y = 3x + 5 has

(i) a unique solution,

(ii) only two solutions,

(iii) infinitely many solutions

**Solution:**

Option (iii) is true because for every value of x, we get a corresponding value of y and vice-versa in the given equation.

Hence, given linear equation has an infinitely many solutions.

### Ex 4.2 Class 9 Maths Question 2.

Write four solutions for each of the following equations:

(i) 2x + y = 7

(ii) πx + y = 9

(iii) x = 4y

**Solution:**

(i) 2x + y = 7

When x = 0, 2(0) + y = 7 ⇒ y = 7

∴ Solution is (0, 7)

When x =1, 2(1) + y = 7 ⇒ y = 7 - 2 ⇒ y = 5

∴ Solution is (1, 5)

When x = 2, 2(2) + y =7y = 7 - 4 ⇒ y = 3

∴ Solution is (2, 3)

When x = 3, 2(3) + y = 7y = 7 - 6 ⇒ y = 1

∴ Solution is (3, 1).

(ii) πx + y = 9

When x = 0, π(0) + y = 9 ⇒ y = 9 - 0 ⇒ y = 9

∴ Solution is (0, 9)

When x = 1, π(1) + y = 9 ⇒ y = 9 - π

∴ Solution is (1, (9 - π))

When x = 2, π(2) + y = 9 ⇒ y = 9 - 2π

∴ Solution is (2, (9 - 2π))

When x = -1,π(-1) + y = 9 ⇒ y = 9 + π

∴ Solution is (-1, (9 + π))

(iii) x = 4y

When x = 0, 4y = 1 ⇒ y = 0

∴ Solution is (0, 0)

When x = 1, 4y = 1 ⇒ y = 1/4

∴ Solution is (1,1/4)

When x = 4, 4y = 4 ⇒ y = 1

∴ Solution is (4, 1)

When x = 4, 4y = 4 ⇒ y = -1

∴ Solution is (-4, -1)

## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3

### Ex 4.3 Class 9 Maths Question 1.

Draw the graph of each of the following linear equations in two variables:

(i) x + y = 4

(ii) x – y = 2

(iii) y = 3x

(iv) 3 = 2x + y

**Solution:**

(i) x + y = 4 ? y = 4 – x

If we have x = 0, then y = 4 – 0 = 4

x = 1, then y =4 – 1 = 3

x = 2, then y = 4 – 2 = 2

∴ We get the following table:

Plot the ordered pairs (0, 4), (1,3) and (2,2) on the graph paper. Joining these points, we get a straight line AB as shown.

Thus, the line AB is the required graph of x + y = 4

(ii) x – y = 2 ? y = x – 2

If we have x = 0, then y = 0 – 2 = -2

x = 1, then y = 1 – 2 = -1

x = 2, then y = 2 – 2 = 0

∴ We get the following table:

Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown.

Thus, the ime is the required graph of x – y = 2

(iii) y = 3x

If we have x = 0,

then y = 3(0) ⇒ y = 0

x = 1, then y = 3(1) = 3

x= -1, then y = 3(-1) = -3

Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get a straight line LM as shown.

Thus, the line LM is the required graph of y = 3x.

(iv) 3 = 2x + y ⇒ y = 3 - 2x

If we have x = 0, then y = 3 - 2(0) = 3

x = 1,then y = 3 - 2(1) = 3 - 2 = 1

x = 2,then y = 3 - 2(2) = 3 - 4 = -1

∴ We get the following table:

Plot the ordered pairs (0, 3), (1, 1) and (2, – 1) on the graph paper. Joining these points, we get a straight line CD as shown.

Thus, the line CD is the required graph of 3 = 2x + y.

### Ex 4.3 Class 9 Maths Question 2.

Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

**Solution:**

(2, 14) means x = 2 and y = 14

Equations which have (2,14) as the solution are (i) x + y = 16, (ii) 7x – y = 0

There are infinite number of lines which passes through the point (2, 14), because infinite number of lines can be drawn through a point.

## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4

### Ex 4.4 Class 9 Maths Question 1.

Give the geometric representations of y = 3 as an equation

(i) in one variable

(ii) in two variables

**Solution:**

(i) y = 3

∴ y = 3 is an equation in one variable, i.e., y only.

∴ y = 3 is a unique solution on the number line as shown below:

(ii) y = 3

We can write y = 3 in two variables as 0.x + y = 3

Now, when x = 1, y = 3

x = 2, y = 3

x = -1, y = 3

∴ We get the following table:

Plotting the ordered pairs (1, 3), (2, 3) and (-1, 3) on a graph paper and joining them, we get aline AB as solution of 0. x + y = 3,

i.e. y = 3.