**Download NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.2**

**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

(ii) only two solutions,

(iii) infinitely many solutions

**Solution :**

Given Equation : y = 3x + 5

As it is an equation in two variables.

Therefore, for every value of x, there will be a value of y satisfying the above equation and vice-versa.

Thus, we conclude that the given has infinitely many solutions.

Hence (iii) is the correct answer.

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

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

**(ii) πx + y = 9**

**(iii) x = 4y**

** ****Solution :**

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

Putting x = 0 in the given equation we get,

2(0) + y = 7

y = 7

Hence (0 , 7) is a solution.

Putting y = 0 in the given equation we get,

2x + 0 = 7

x =

Hence ( , 0) is a solution.

Putting x = 1 in the given equation we get,

2(1) + y = 7

y = 7 – 2

y = 5

Hence (1 , 5) is a solution.

Putting y = 1 in the given equation we get,

2x + 1 = 7

2x = 7 – 1

x = = 3

Hence (3 , 1) is our fourth solution.

Thus (0 , 7), ( , 0), (1 , 5) and (3 , 1) are four solutions of the given equation.

**(ii)** We have πx + y = 9

Putting x = 0 in the given equation we get,

π(0) + y = 9

y = 9

Hence (0 , 9) is a solution.

Putting y = 0 in the given equation we get,

πx + 0 = 9

x =

Hence ( , 0) is a solution.

Putting x = 1 in the given equation we get,

π(1) + y = 9

y = 9 – π

Hence (1 , 9 – π) is a solution.

Putting y = 1 in the given equation we get,

πx + 1 = 9

2π = 9 – 1

x =

Hence ( , 1) is a solution.

Thus (0 , 9), ( , 0), (1 , 9-π) and ( , 1) are four solutions of the given equation.

**(iii)** We have x = 4y

Or, x – 4y = 0

Putting x = 0 in the given equation we get,

0 – 4y = 0

y = 0

Hence (0 , 0) is a solution.

Putting y = 1 in the given equation we get,

x – 4(1) = 0

x = 4

Hence (4 , 1) is a solution.

Putting x = 1 in the given equation we get,

(1) – 4y = 0

y =

Hence (1 , ) is a solution.

Putting y = 2 in the given equation we get,

x – 4(2) = 0

x = 8

Hence (8 , 2) is a solution.

Thus (0 , 0), (4 , 1), (1 , ) and (8 , 2) are four solutions of the given equation.

**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)

(v) (1, 1)

**Solution :**

Given Equation:

x – 2y = 4

x – 2y – 4 = 0

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

Putting x = 0 and y = 2 in L.H.S. we get,

L.H.S. = 0 – 2(2) – 4 = -8 R.H.S.

Therefore, (0,2) is not a solution of the given equation.

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

Putting x = 2 and y = 0 in L.H.S. we get,

L.H.S. = 2 – 2(0) – 4 = -2 R.H.S.

Therefore, (2,0) is not a solution of the given equation.

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

Putting x = 4 and y = 0 in L.H.S. we get,

L.H.S. = 4 – 2(0) – 4 = 0 R.H.S.

Therefore, (4,0) is a solution of the given equation.

**(iv) ( √2 , 4√2 )**

Putting x = √2 and y = 4√2 in L.H.S. we get,

L.H.S. = √2 – 2(4√2) – 4 = -7√2 – 4 R.H.S.

Therefore, ( √2, 4√2 ) is not a solution of the given equation.

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

Putting x = 1 and y = 1 in L.H.S. we get,

L.H.S. = 1 – 2(1) – 4 = -5 R.H.S.

Therefore, (1,1) is not a solution of the given equation.

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

**Solution :**

Putting x = 2, y = 1 in the equation 2x + 3y = k

We get,

2(2) + 3(1) = k

4 + 3 = k

k = 7

Hence the value of k = 7.

**Download NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.2**

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