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Solution:
Given linear equation: y = 3x + 5
Let x = 0, Therefore y = 3 × 0 + 5
= 0 + 5 = 5
Hence, (0, 5) is one solution
Now, let x = 1, Therefore y = 3 × 1 + 5
= 3 + 5 = 7
Hence, (1, 8) is another solution
Now, let y = 0, Therefore 0 = 3x + 5
x = 5/3
Hence, (5/3, 0) is one another solution.
This concludes that different values of x and y give the different values of y and x respectively.
As there is also no end to the different types of solution for the linear equation in two variables. Therefore, it can have infinitely many solutions.
Hence, option "(iii) infinitely many solutions" is the correct answer.
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:
(i) 2x + y = 7
Given: 2x + y = 7
To find the four solutions we have to substitute different values for x.
Let x = 0
Then,
2x+y = 7
(2 × 0) + y = 7
y = 7
Therefore, we get (x, y) = (0, 7)
Let x = 1
Then,
2x+y = 7
(2×1)+y = 7
2+y = 7
y = 7-2
y = 5
Therefore, we get (x, y) = (1, 5)
Let x = 2
Then,
2x + y = 7
(2×2) + y = 7
4 + y = 7
y = 7 - 4
y = 3
Therefore, we get (x, y) = (2, 3)
Let x = 3
Then,
2x + y = 7
(2×3) + y = 7
6 + y = 7
y = 7 - 6
y = 1
Therefore, we get (x, y) = (3, 1)
Finally, the four solutions are (0, 7), (1,5), (2,3), (3, 1)
(ii) πx + y = 9
Given: πx+y = 9
To find the four solutions we have to substitute different values for x.
Let x = 0
Then,
πx + y = 9
(π×0) + y = 9
y = 9
Therefore, we get (x, y) = (0, 9)
Let x = 1
Then,
πx +y = 9
(π×1) + y = 9
π + y = 9
y = 9 - π
Therefore, we get (x, y) = (1, 9 - π)
Let x = 2
Then,
πx +y = 9
(π×2) + y = 9
2π + y = 9
y = 9 - 2π
Therefore, we get (x, y) = (1, 9 - 2π)
Let x = 3
Then,
πx +y = 9
(π×3) + y = 9
3π + y = 9
y = 9 - 3π
Therefore, we get (x, y) = (1, 9 - 3π)
Finally, the four solutions are (0, 9), (1, 9 - π), (2, 9 - 2π), (3, 9 - 3π)
(iii) x = 4y
Given: x = 4y
To find the four solutions we have to substitute different values for x.
Let x = 0
Then,
x = 4y
0 = 4y
4y= 0
y = 0/4
y = 0
Therefore, we get (x, y) = (0,0)
Let x = 1
Then,
x = 4y
1 = 4y
4y = 1
y = 1/4
Therefore, we get (x, y) = (1,1/4)
Let x = 2
Then,
x = 4y
2 = 4y
4y = 2
y = 2/4
Therefore, we get (x, y) = (2, 1/2)
Let x = 3
Then,
x = 4y
3 = 4y
4y = 3
y = 3/4
Therefore, we get (x, y) = (2, 3/4)
Finally, the four solutions are (0, 0), (1,1/4), (2, 1/2), (2, 3/4)
(i) (0, 2)
(ii) (2, 0)
(iii) (4, 0)
(iv) (√2 , 4√2)
(v) (1, 1)
Solution:
(i) (0, 2)
Given: x - 2y = 4
As, x=0 and y=2
Hence, substituting the values of x and y in the equation, we get,
x – 2y = 4
0 – (2×2) = 4
-4 ≠ 4
L.H.S ≠ R.H.S
Therefore, (0, 2) is not a solution to the given equation x - 2y = 4.
(ii) (2, 0)
Given: x - 2y = 4
As, x = 2 and y = 0
Hence, substituting the values of x and y in the equation, we get,
x - 2y = 4
2 - (2×0) = 4
2 - 0 = 4
2 ≠ 4
L.H.S ≠ R.H.S
Therefore, (2, 0) is not a solution to the given equation x - 2y = 4.
(iii) (4, 0)
Given: x - 2y = 4
As, x= 4 and y=0
Hence, substituting the values of x and y in the equation, we get,
x - 2y = 4
4 - 2×0 = 4
4 - 0 = 4
4 = 4
L.H.S = R.H.S
Therefore, (4, 0) is a solution to the given equation x - 2y = 4.
(iv) (√2, 4√2)
Given: x - 2y = 4
As, x = √2 and y = 4√2
Hence, substituting the values of x and y in the equation, we get,
x - 2y = 4
√2 - (2×4√2) = 4
√2 - 8√2 = 4
-7√2 ≠ 4
L.H.S ≠ R.H.S
Therefore, (√2, 4√2) is not a solution to the given equation x - 2y = 4.
(v) (1, 1)
Given: x - 2y = 4
As, x= 1 and y= 1
Hence, substituting the values of x and y in the equation, we get,
x - 2y = 4
1 - (2×1) = 4
1 - 2 = 4
-1 ≠ 4
L.H.S ≠ R.H.S
Therefore, (1, 1) is not a solution to the given equation x - 2y = 4.
Solution:
Given: 2x + 3y = k
According to the question, x = 2 and y = 1 is solution of the given equation.
Hence, substituting the values of x and y in the equation 2x+3y = k, we get,
2x + 3y = k
(2×2) + (3×1) = k
4 + 3 = k
7 = k
k = 7
Therefore, the value of k is 7.
