NCERT Solutions Class 9th Maths Chapter – 4 Linear Equations in Two Variables Exercise – 4.2

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.

Question 2 Write four solutions for each of the following equations:
(i) 2x + y = 7

Solution – 2x + y = 7
Let x = 0, 2(0) + y = 7 ⇒ y = 7
∴ Solution is (0, 7)
Let x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5
∴ Solution is (1, 5)
Let x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3
∴ Solution is (2, 3)
Let x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1
∴ Solution is (3, 1).

(ii) πx + y = 9

Solution – πx + y = 9
Let x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9
∴ Solution is (0, 9)
Let x = 1, π(1) + y = 9 ⇒ y = 9 – π
∴ Solution is (1, (9 – π))
Let x = 2, π(2) + y = 9 ⇒ y = 9 – 2π
∴ Solution is (2, (9 – 2π))
Let x = -1,π(-1) + y = 9 ⇒ y = 9 + π
∴ Solution is (-1, (9 + π))

(iii) x = 4y

Solution – x = 4y
Let x = 0, 4y = 1 ⇒ y = 0
∴ Solution is (0, 0)
Let x = 1, 4y = 1 ⇒ y = 14
∴ Solution is (1,14 )
Let x = 4, 4y = 4 ⇒ y = 1
∴ Solution is (4, 1)
Let x = 4, 4y = 4 ⇒ y = -1
∴ Solution is (-4, -1)

Question 3 Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0,2)

Solution – (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
So, L.H.S. ≠ R.H.S.
Thus, (0, 2) is not a solution.

(ii) (2,0)

Solution – (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
So, L.H.S. ≠ R.H.S.
Thus, So, (2,0) is not a solution.

(iii) (4, 0)

Solution – (4, 0) means x = 4 and y = 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
So, L.H.S. = R.H.S.
Thus, (4, 0) is a solution.

(iv) (√2, 4√2)

Solution – (√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
So, L.H.S. ≠ R.H.S.
Thus, (√2 , 4√2) is not a solution.

(v) (1, 1)

Solution – (1, 1)means x =1 and y = 1
Putting x = 1 and y = 1 in x – 2y = 4, we get
LH.S. = 1 – 2(1)
= 1 – 2 = -1
But R.H.S = 4
So, LH.S. ≠ R.H.S.
Thus, (1, 1) is not a solution.

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
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.