# NCERT Solutions Class 9 Maths Chapter 4 Linear Equations In Two Variable Exercise 4.2

## Introduction:

In this exercise/article we will learn about Linear Equations In Two Variable. You have seen that every linear equation in one variable has a unique solution. What can you say about the solution of a linear equation involving two variables? As there are two variables in the equation, a solution means a pair of values, one for x and one for y which satisfy the given equation.

NCERT Class 9 Maths Chapter 4 Coordinate Geometry :

**Class 9 Maths Exercise 4.2 (Page-59)**

**Q1. **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 :**

(iii) infinitely many solutions

∴ It has infinitely many solutions because for same value of x there will be some value of y will exist .

**Q2. **Write four solutions for each of the following equations:

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

**Solution :**

We have 2x + y = 7

According to the question,

putting value of x = 0

Now, (i) 2 × 0 + y = 7

⇒ 0 + y = 7

⇒ y = 7

(ii) putting value of x = 1

⇒ 2 × 1 + y = 7

⇒ 2 + y = 7

⇒ y = 7 - 2

⇒ y = 5

(iii) putting value of x = 2

⇒ 2 × 2 + y = 7

⇒ 4 + y = 7

⇒ y = 7 - 4

⇒ y = 3

(iv) putting value of x = 3

⇒ 2 × 3 + y = 7

⇒ 6 + y = 7

⇒ y = 7 - 6

⇒ y = 1

∴ ( 0, 7 ) , ( 1, 5 ) , ( 2, 3 ) and ( 3, 1 ) are the four solutions of the equation 2x + y = 7 .

**(ii)** πx + y = 9

**Solution :**

We have πx + y = 9

According to the question,

putting value of x = 0

Now, (i) πx + y = 9

⇒ π × 0 + y = 9

⇒ 0 + y = 9

⇒ y = 9

(ii) putting value of x = 1

⇒ π × 1 + y = 9

⇒ π + y = 9

⇒ y = 9 - π

(iii) putting value of x = 2

⇒ π × 2 + y = 9

⇒ 2π + y = 9

⇒ y = 9 - 2π

(iv) putting value of x = 3

⇒ π × 3 + y = 9

⇒ 3π + y = 9

⇒ y = 9 - 3π

∴ ( 0, 9 ) , ( 1, 9 - π ) , ( 2, 9 - 2π ) and ( 3, 9 - 3π ) are the four solutions of the equation πx + y = 9 .

**(iii)** x = 4y

**Solution :**

We have x = 4y

According to the question,

putting value of y = 0

Now, (i) x = 4y

⇒ x = 4 × 0

⇒ x = 0

(ii) putting value of y = 1

⇒ x = 4 × 1

⇒ x = 4

(iii) putting value of y = 2

⇒ x = 4 × 2

⇒ x = 8

(iv) putting value of y = 3

⇒ x = 4 × 3

⇒ x = 12

∴ ( 0, 0 ) , ( 4, 1 ) , ( 8, 2 ) and ( 12, 3 ) are the four solutions of the equation x = 4y .

**Q3. **Check which of the following are solutions of the equation x - 2y = 4 and which are not:

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

**Solution :**

Given equation x - 2y = 4

Given values ⇒ x = 0, y = 2

According to the question,

Here, x - 2y = 4

put values of x and y

Now, 0 - 2 × 2 = 4

⇒ 0 - 4 = 4

⇒ - 4 = 4

LHS \(\displaystyle \ne \) RHS

∴ ( 0, 2 ) is a not solutions of the equation x - 2y = 4 .

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

**Solution :**

Given equation x - 2y = 4

Given values ⇒ x = 2, y = 0

According to the question,

Here, x - 2y = 4

put values of x and y

⇒ 2 - 2 × 0 = 4

⇒ 2 - 0 = 4

⇒ 2 = 4

LHS \(\displaystyle \ne \) RHS

∴ ( 2, 0 ) is a not solutions of the equation x - 2y = 4 .

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

**Solution :**

Given equation x - 2y = 4

Given values ⇒ x = 4, y = 0

According to the question,

Here, x - 2y = 4

put values of x and y

⇒ 4 - 2 × 0 = 4

⇒ 4 - 0 = 4

⇒ 4 = 4

LHS = RHS

∴ ( 4, 0 ) is a solutions of the equation x - 2y = 4 .

**(iv)** ( \(\displaystyle \sqrt{2}\), \(\displaystyle 4\sqrt{2}\) )

**Solution :**

Given equation x - 2y = 4

Given values ⇒ x = \(\displaystyle \sqrt{2}\), y = \(\displaystyle 4\sqrt{2}\)

According to the question,

Here, x - 2y = 4

put values of x and y

⇒ \(\displaystyle \sqrt{2}\) - 2 × \(\displaystyle 4\sqrt{2}\) = 4

⇒ \(\displaystyle \sqrt{2}\) - \(\displaystyle 8\sqrt{2}\) = 4

⇒ \(\displaystyle -7\sqrt{2}\) = 4

LHS \(\displaystyle \ne \) RHS

∴ ( \(\displaystyle \sqrt{2}\), \(\displaystyle 4\sqrt{2}\) ) is a not solutions of the equation x - 2y = 4 .

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

**Solution :**

Given equation x - 2y = 4

Given values ⇒ x = 1, y = 1

According to the question,

Here, x - 2y = 4

put values of x and y

⇒ 1 - 2 × 1 = 4

⇒ 1 - 2 = 4

⇒ - 1 = 4

LHS \(\displaystyle \ne \) RHS

∴ ( 1, 1 ) is a not solutions of the equation x - 2y = 4 .

**Q4. **Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

**Solution :**

Given equation 2x + 3y = k

Given value ⇒ x = 2, y = 1

According to the question

Here, 2x + 3y = k

put values of x and y

Now, 2 × 2 + 3 × 1 = k

⇒ 4 + 3 = k

⇒ 7 = k

∴ The value of k = 7 .