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# Ex 4.2 NCERT Solutions - Linear Equations in Two Variables Class 9 Notes | EduRev

## Class 9 Mathematics by Full Circle

Created by: Full Circle

## Class 9 : Ex 4.2 NCERT Solutions - Linear Equations in Two Variables Class 9 Notes | EduRev

The document Ex 4.2 NCERT Solutions - Linear Equations in Two Variables Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.
All you need of Class 9 at this link: Class 9

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 a linear equation has an infinitely many solutions.

Question 2: Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) p x + y = 9
(iii) x = 4y
Solution:
(i) 2x + y = 7
When x = 0, 2(0) + y = 7
⇒ 0 + y = 7
⇒ y= 7
∴ Solution is (0, 7).
When x = 1, 2(1) + y = 7
⇒ y = 7 – 2
⇒ y= 5
∴ Solution is (1, 5).
When x = 2, 2(2) + y = 7
⇒ y = 7 – 4
⇒ y = 3
∴ Solution is (2, 3).
When x = 3 2(3) + y = 7 ⇒ y = 7 – 6
⇒ y = 1
∴ Solution is (3, 1).

(ii) πx + y = 9
When x = 0 π(0) + y = 9
⇒ y = 9 – 0
⇒ y= 9
∴ Solution is (0, 9).
When x = 1, π(1) + y = 9
⇒ y = 9 – π
∴ Solution is {1, (9 – π)}.
When x = 2 π(2) + y = 9
⇒ y = 9 – 2π
∴ Solution is {2, (9 – 2π)}.
When x = –1, p(–1) + y = 9
⇒ –π + y = 9
⇒ y = 9 + π
∴ Solution is {–1, (9 + π)}.

(iii) x = 4y When x = 0, 4y = 0
⇒ y = 0
∴ Solution is (0, 0).
When x = 1, 4y = 1
⇒ y = (1/4)

∴ Solution is (1, (1/4))
When x = 4, 4y = 4
⇒            y= (4/4) = 1

∴ Solution is (4, 1).
When x = –4, 4y = –4
⇒ y = (-4/4) = –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) (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 have L.H.S. = 0 – 2(2) = –4
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ x = 0, y = 0 is not a solution.

(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.

(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
But R.H.S. = 4
∴ L.H.S. = R.H.S.
∴ (4, 0) is a solution.

(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
= √2(1 – 8) = –7√2

But R.H.S. = 4 ∴ L.H.S. ≠ R.H.S.
∴ (√2, 4√2) is not a solution.

(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.

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 = 7.

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