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

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 are the part of NCERT Solutions for Class 9 Maths. Here you can find the 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) πx + 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) πx + y = 9
If x = 0, we have π(0) + y = 9
y = 9 – 0 y = 9
Therefore, one solution is (0, 9).
If x = 1, we have π(1) + y = 9
y = 9 – π
Therefore, the second solution is (1, (9 – π)).
If x = 2, we have π(2) + y = 9
y = 9 – 2π
Therefore, the third solution is (2, (9 – 2π)).
If x = -1, we have π(-1) + y = 9
y = 9 + π
Therefore, the fourth solution is (-1, (9 + π)).

(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.


NCERT Solutions for Maths Class 10

NCERT Solutions for Maths Class 12

Please do not enter any spam link in the comment box.

Post a Comment (0)
Previous Post Next Post