Linear Equations In Two Variables | Exercise 4.2 | NCERT Solution Maths 9 | Chapter 4
Exercise 4.2 Page: 70
1. Which one of the following options is true, and why?
y = 3x+5 has
- A unique solution
- Only two solutions
- Infinitely many solutions
Solution:
Let us substitute different values for x in the linear equation y = 3x+5,
| x | 0 | 1 | 2 | …. | 100 |
| y, where y=3x+5 | 5 | 8 | 11 | …. | 305 |
From the table, it is clear that x can have infinite values, and for all the infinite values of x, there are infinite values of y as well.
Hence, (iii) infinitely many solutions is the only option true.
2. Write four solutions for each of the following equations:
(i) 2x+y = 7
Solution:
To find the four solutions of 2x+y =7 we substitute different values for x and y
Let x = 0
Then,
2x+y = 7
(2×0)+y = 7
y = 7
(0,7)
Let x = 1
Then,
2x+y = 7
(2×1)+y = 7
2+y = 7
y = 7-2
y = 5
(1,5)
Let y = 1
Then,
2x+y = 7
(2x)+1 = 7
2x = 7-1
2x = 6
x = 6/2
x = 3
(3,1)
Let x = 2
Then,
2x+y = 7
(2×2)+y = 7
4+y = 7
y =7-4
y = 3
(2,3)
The solutions are (0, 7), (1,5), (3,1), (2,3)
(ii) πx+y = 9
Solution:
To find the four solutions of πx+y = 9 we substitute different values for x and y
Let x = 0
Then,
πx+y = 9
(π×0)+y = 9
y = 9
(0,9)
Let x = 1
Then,
πx +y = 9
(π×1)+y = 9
π+y = 9
y = 9-π
(1, 9-π)
Let y = 0
Then,
πx+y = 9
πx+0 = 9
πx = 9
x = 9/π
(9/π,0)
Let x = -1
Then,
πx + y = 9
(π×-1) + y = 9
-π+y = 9
y = 9+π
(-1,9+π)
The solutions are (0,9), (1,9-π), (9/π,0), (-1,9+π)
(iii) x = 4y
Solution:
To find the four solutions of x = 4y we substitute different values for x and y
Let x = 0
Then,
x = 4y
0 = 4y
4y= 0
y = 0/4
y = 0
(0,0)
Let x = 1
Then,
x = 4y
1 = 4y
4y = 1
y = 1/4
(1,1/4)
Let y = 4
Then,
x = 4y
x= 4×4
x = 16
(16,4)
Let y = 1
Then,
x = 4y
x = 4×1
x = 4
(4,1)
The solutions are (0,0), (1,1/4), (16,4), (4,1)
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)
Solutions:
(i) (0, 2)
(x,y) = (0,2)
Here, x=0 and y=2
Substituting the values of x and y in the equation x–2y = 4, we get,
x–2y = 4
⟹ 0 – (2×2) = 4
But, -4 ≠ 4
(0, 2) is not a solution of the equation x–2y = 4
(ii) (2, 0)
(x,y) = (2, 0)
Here, x = 2 and y = 0
Substituting the values of x and y in the equation x -2y = 4, we get,
x -2y = 4
⟹ 2-(2×0) = 4
⟹ 2 -0 = 4
But, 2 ≠ 4
(2, 0) is not a solution of the equation x-2y = 4
(iii) (4, 0)
Solution:
(x,y) = (4, 0)
Here, x= 4 and y=0
Substituting the values of x and y in the equation x -2y = 4, we get,
x–2y = 4
⟹ 4 – 2×0 = 4
⟹ 4-0 = 4
⟹ 4 = 4
(4, 0) is a solution of the equation x–2y = 4
(iv) (√2,4√2)
Solution:
(x,y) = (√2,4√2)
Here, x = √2 and y = 4√2
Substituting the values of x and y in the equation x–2y = 4, we get,
x –2y = 4
⟹ √2-(2×4√2) = 4
√2-8√2 = 4
But, -7√2 ≠ 4
(√2,4√2) is not a solution of the equation x–2y = 4
(v) (1, 1)
Solution:
(x,y) = (1, 1)
Here, x= 1 and y= 1
Substituting the values of x and y in the equation x–2y = 4, we get,
x –2y = 4
⟹ 1 -(2×1) = 4
⟹ 1-2 = 4
But, -1 ≠ 4
(1, 1) is not a solution of the equation x–2y = 4
4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k.
Solution:
The given equation is
2x+3y = k
According to the question, x = 2 and y = 1.
Now, Substituting the values of x and y in the equation2x+3y = k,
We get,
(2×2)+(3×1) = k
⟹ 4+3 = k
⟹ 7 = k
k = 7
The value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k, is 7.
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