﻿ Class 9 NCERT Math Solution
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TOPICS
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 :-
```  y = 3x + 5 is a linear equation in two variables and it has infinite possible solutions.
As for every value of x, there will be a value of y satisfying the above equation and vice-versa.
Hence, the correct answer is (iii).
```

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
For x = 0,
2(0) + y = 7
y = 7
Therefore, (0, 7) is a solution of this equation.

For x = 1,
2(1) + y = 7
y = 5
Therefore, (1, 5) is a solution of this equation.

For x = −1,
2(−1) + y = 7
y = 9
Therefore, (−1, 9) is a solution of this equation.

For x = 2,
2(2) + y = 7
y = 3
Therefore, (2, 3) is a solution of this equation.
```
```
(ii) πx + y = 9
For x = 0, π(0)
0 + y = 9
y = 9
Therefore, (0, 9) is a solution of this equation.

For x = 1,
π(1) + y = 9
y = 9 − π
Therefore, (1, 9 − π) is a solution of this equation.

For x = 2,
π(2) + y = 9
y = 9 − 2π
Therefore, (2, 9 −2π) is a solution of this equation.

For x = −1,
π(−1) + y = 9
y = 9 + π
(−1, 9 + π) is a solution of this equation.
```
```
(iii) x = 4y
For x = 0,
0 = 4y
y = 0
Therefore, (0, 0) is a solution of this equation.

For y = 1,
x = 4(1) = 4
Therefore, (4, 1) is a solution of this equation.

For y = −1,
x = 4(−1)
x = −4
Therefore, (−4, −1) is a solution of this equation.

For x = 2,
2 = 4y
y = 2/4
y = 1/2
Therefore, (2, 1/2) is a solution of this equation.
```

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)
Putting x = 0 and y = 2 in the L.H.S of the given equation,
x − 2y = 0 − 2 x 2 = − 4 ≠ 4
L.H.S ≠ R.H.S
Therefore, (0, 2) is not a solution of this equation.
```
```(ii) (2, 0)
Putting x = 2 and y = 0 in the L.H.S of the given equation,
x − 2y = 2 − 2 × 0 = 2 ≠ 4
L.H.S ≠ R.H.S
Therefore, (2, 0) is not a solution of this equation.
```
```(iii) (4, 0)
Putting x = 4 and y = 0 in the L.H.S of the given equation, x
− 2y = 4 − 2(0)
= 4 = R.H.S
L.H.S. = R.H.S.
Therefore, (4, 0) is a solution of this equation.
```
```
(iv) (√2, 4√2)
Putting x = √2 and y = 4√2 in the L.H.S. of the given equation,
x - 2y = √2 - 2(4√2)
= √2 - 8√2 = -7√2 ≠ 4
L.H.S. ≠ R.H.S.
Therefore, (√2, 4√2) is not a solution of given equation.
```
```
(v) (1, 1) is not a solution of this equation.
Putting x = 1 and y = 1 in the L.H.S of the given equation,
x − 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4
L.H.S ≠ R.H.S
Therefore, (1, 1) is not a solution of this equation.
```

Question-4 :-  Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Solution :-
```  Putting x = 2 and y = 1 in the given equation,
2x + 3y = k
2(2) + 3(1) = k
4 + 3 = k
k = 7
Therefore, the value of k is 7.
```
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