**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 4.2**

**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 in the given equation,
for every value of x, we get a corresponding value of y and vice-versa.

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)**Ï€x + y = 9

**(iii)**x = 4y

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

If x = 0, we have 2(0) + y = 7 ⇒ y = 7

∴ Therefore, one solution is (0, 7).

If x = 1, we have 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5

∴ Therefore,
the second solution is (1, 5).

If x = 2, we have 2(2) + y = 7 ⇒ y = 7 – 4 ⇒ y = 3

∴ Therefore, the third solution is
(2, 3).

If x = 3, we have 2(3) + y = 7 ⇒ y = 7 – 6 ⇒ y = 1

∴ Therefore, the fourth solution is
(3, 1).

**(ii)**
Ï€x + y = 9

If x = 0, we have Ï€(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9

∴ Therefore, one solution is (0, 9).

If x = 1, we have Ï€(1) + y = 9 ⇒ y = 9 – Ï€

∴ Therefore, the second solution is (1, (9 – Ï€)).

If x = 2, we have Ï€(2) + y = 9 ⇒ y = 9 – 2Ï€

∴ Therefore, the third solution is (2, (9 – 2Ï€)).

If x = -1, we have Ï€(-1) + y = 9 ⇒ y = 9 + Ï€

∴ Therefore, the fourth solution is (-1, (9 + Ï€)).

**(iii)** x = 4y

If x = 0, we have 4y = 0 ⇒ y = 0

∴ Therefore, one solution is (0, 0).

If x = 1, we have 4y = 1 ⇒ y = 1/4

∴ Therefore, the second solution is (1, 1/4).

If x = 4, we have 4y = 4 ⇒ y = 1

∴ Therefore, the third solution is (4, 1).

If x = -4, we have 4y = -4 ⇒ y = -1

∴ Therefore, the fourth solution is (-4, -1).

**Ex 4.2 Class 9 Maths Question
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)

**Solution:****
(i)** (0, 2) means x = 0 and y = 2

Putting x = 0 and y = 2 in x – 2y = 4, we get

L.H.S. = 0 – 2(2) = -4

But R.H.S. = 4

∴ L.H.S. ≠ R.H.S.

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

**(ii)**
(2, 0) means x = 2 and y = 0

Putting x = 2 and y = 0 in x – 2y = 4, we get

L.H:S. = 2 – 2(0) = 2 – 0 = 2

But R.H.S. = 4

∴ L.H.S. ≠ R.H.S.

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

**(iii)**
(4, 0) means x = 4 and y = 0

Putting x = 4 and y = 0 in x – 2y = 4, we get

L.H.S. = 4 – 2(0) = 4 – 0 = 4 = R.H.S.

∴ L.H.S. = R.H.S.

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

**(iv)**
(√2, 4√2) means x = √2 and y = 4√2

Putting x = √2 and y = 4√2 in x – 2y = 4, we get

L.H.S. = √2 – 2(4√2) = √2 – 8√2 = -7√2

But R.H.S. = 4

∴ L.H.S. ≠ R.H.S.

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

**(v)**
(1, 1) means x = 1 and y = 1

Putting x = 1 and y = 1 in x – 2y = 4, we get

L.H.S. = 1 – 2(1) = 1 – 2 = -1.

But R.H.S = 4

∴ L.H.S. ≠ R.H.S.

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

**Ex 4.2 Class
9 Maths Question
4.**

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

**Solution:
**We have 2x + 3y = k

x = 2, y = 1 is a solution of the given equation.

Putting x = 2 and y = 1 in 2x + 3y = k, we get

2(2) + 3(1) = k

⇒ k = 4 + 3

⇒ k = 7

Thus, the required value of k is 7.

**Related Links:**

**NCERT Solutions for Maths Class 10**

**NCERT Solutions for Maths Class 11**