Q1. 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
Ans: Let us substitute different values for x in the linear equation y = 3x + 5,
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.
Q2. Write four solutions for each of the following equations:
(i) 2x + y = 7
Ans: 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
Ans: 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 = 9π
(1, 9π)
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, 9π), (9/π, 0), (1, 9+π)
(iii) x = 4y
Ans: 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 =
Then,
x = 4y
x = 4×1
x = 4
(4,1)
The solutions are (0,0), (1,1/4), (16,4), (4,1)
Q3. Check which of the following are solutions of the equation x–2y = 4 and which are not:
(i) (0, 2)
Ans: (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
Therefore, (0, 2) is not a solution of the equation x – 2y = 4
(ii) (2, 0)
Ans: (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)
Ans: (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
⟹ 40 = 4
⟹ 4 = 4
(4, 0) is a solution of the equation x – 2y = 4
(iv) (√2, 4√2)
Ans: (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
√28√2 = 4
But, 7√2 ≠ 4
(√2, 4√2) is not a solution of the equation x – 2y = 4
(v) (1, 1)
Ans: (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
Q4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k.
Ans: 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 equation 2x + 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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