问题1：以下哪个选项是正确的，为什么?
y = 3x + 5有
(i)一个独特的解决方案，(ii)仅两个解决方案，(iii)无限多个解决方案
解决方案：
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.
问题2：为以下每个方程式写四个解决方案：
(i)2x + y = 7
(ii)πx+ y = 9
(iii)x = 4y
解决方案：
(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)
问题3：检查以下哪些是方程x – 2y = 4的解，哪些不是：
(i)(0，2)
(ii)(2，0)
(iii)(4，0)
(iv)(√2，4√2)
(v)(1、1)
解决方案：
(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.
问题4：如果x = 2，则y = 1是方程式2x + 3y = k的解，求出k的值。
解决方案：
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.