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.
- 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 4.3
- NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4
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)
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.
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