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

NCERT Solutions Class 9 Maths 
Chapter – 4 (Linear Equations in Two Variables) 

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 4 Linear Equations in Two Variables Exercise 4.2 has been provided here to help the students in solving the questions from this exercise. 

Chapter 4: Linear Equations in Two Variables

• NCERT Solution Class 9 Maths Ex – 4.1
• NCERT Solution Class 9 Maths Ex – 4.3
• NCERT Solution Class 9 Maths Ex – 4.4

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
Answer – 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
(ii) πx + y = 9
(iii) x = 4y

Answer –
(i) 2x + y = 7
y = 7 – 2x
Then,

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

(ii) πx + y = 9
y = 9 – πx

The solutions are (0, 9), (1, 9 – π), (2, 9 – π), (-1, 9 + π)

(iii) x = 4y
y = x/4

The solutions are (0, 0), (4, 1), (8, 2), (12, 3)

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)

Answer –
(i) (0, 2)
(x, y) = (0, 2)
Puffing x = 0 and y = 2 in x – 2y = 4, we get
L.H.S.
⇒ 0 – 2(2)
⇒ -4
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)
(x, y) = (2, 0)
Putting x = 2 and y = 0 in x – 2y = 4, we get
L.H.S.
⇒ 2 – 2(0)
⇒ 2 – 0
⇒ 2
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)
(x, y) = (4, 0)
Putting x = 4 and y = o in x – 2y = 4, we get
L.H.S.
⇒ 4 – 2(0)
⇒ 4 – 0
⇒ 4
R.H.S. = 4
∴ L.H.S. = R.H.S.
(4, 0) is a solution of the equation x – 2y = 4

(iv) (√2, 4√2)
(x, y) = (√2, 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
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)
(x, y) = (1, 1)
Putting x = 1 and y = 1 in x – 2y = 4, we get
LH.S.
⇒ 1 – 2(1)
⇒ 1 – 2
⇒ -1
R.H.S = 4
∴ LH.S. ≠ R.H.S.
(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.
Answer – We have 2x + 3y = k
putting x = 2 and y = 1 in 2x + 3y = k, we get
2(2) + 3(1)
⇒ k = 4 + 3 – k
⇒ 7 = k
Thus, the required value of k is 7.