问题1.在以下每个方程式系统中，确定系统是否具有唯一解，无解或无限解。如果有一个独特的解决方案：
x-3y-3 = 0，
3x-9y-2 = 0
解决方案：
Given that,
x − 3y − 3 = 0 …(1)
3x − 9y − 2 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y – c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 1, b1 = −3, c1 = −3
a2 = 3, b2 = −9, c2 = −2
Let’s check the equation’s,
a1/a2 = 1/3
b1/b2 = -3/-9 = 1/3
c1/c2 = -3/-9 = 3/2
a1/a2 = b1/b2 ≠ c1/c2
Hence, the given set of equations has no solution.
问题2.在以下每个方程式系统中，确定系统是否具有唯一解，无解或无穷大。如果有一个独特的解决方案：
2x + y-5 = 0，
4x + 2y-10 = 0
解决方案：
Given that,
2x + y − 5 = 0 …(1)
4x + 2y − 10 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = 1, c1 = −5 and
a2 = 4, b2 = 2, c2 = −10
Lets check the equation’s,
a1/a2 = 2/4 = 1/2
b1/b2 = 1/2
and c1/c2 = -5/-10 = 1/2
Therefore, a1/a2 = b1/b2 = c1/c2
Hence, the given set of equations has infinitely many solutions.
问题3.在以下每个方程式系统中，确定系统是否具有唯一解，无解或无穷大。如果有一个独特的解决方案：
3x-5y = 20，
6x-10y = 40
解决方案：
Given that,
3x − 5y = 20 …(1)
6x − 10y = 40 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 3, b1 = −5, c1 = − 20
a2 = 6, b2 = −10, c2 = − 40
Lets check the equation’s,
a1/a2 = 3/6 = 1/2
b1/b2 = -5/-10 – 1/2 and
c1/c2 = -20/-40 = 1/2
Therefore, a1/a2 = b1/b2 = c1/c2
Hence, the given set of equations has infinitely many solutions.
问题4.在以下每个方程式系统中，确定系统是否具有唯一解，无解或无穷大。如果有一个独特的解决方案：
x-2y-8 = 0，
5x-10y-10 = 0
解决方案：
Given that,
x − 2y − 8 = 0 …(1)
5x − 10y − 10 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 1, b1 = −2, c1 = −8
a2 = 5, b2 = −10, c2 = −10
Lets check the equation’s,
a1/a2 = 1/5
b1/b2 = -2/-10 and
c1/c2 = -8/-10
Therefore, a1/a2 = b1/b2 ≠ c1/c2
Hence, the given set of equations has no solution.
问题5.为以下每个具有唯一解的方程组找到k的值：
kx + 2y-5 = 0，
3x + y − 1 = 0
解决方案：
Given that,
kx + 2y − 5 = 0 …(1)
3x + y − 1 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = k, b1 = 2, c1 = −5
a2 = 3, b2 = 1, c2 = −1
For unique solution,
a1/a2 ≠ b1/b2
k/3 ≠ 2/1
k ≠ 6
So, the given set of equations will have unique solution for all real values of k other than 6.
问题6.为以下每个具有唯一解的方程组找到k的值：
4x + ky + 8 = 0，
2x + 2y + 2 = 0
解决方案：
Given that,
4x + ky + 8 = 0 …(1)
2x + 2y + 2 = 0 …(2)
So, the given equations are in the form of:
a1x +b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 4, b1 = k, c1 = 8
a2 = 2, b2 = 2, c2 = 2
For unique solution,
a1/a2 ≠ b1/b2
4/2 ≠ k/2
k ≠ 4
So, the given set of equations will have unique solution for all real values of k other than 4.
问题7.为以下每个具有唯一解的方程组找到k的值：
4x-5y = k，
2x-3y = 12
解决方案：
Given that,
4x − 5y − k = 0 …(1)
2x − 3y − 12 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 4, b1 = −5, c1 = −k
a2 = 2, b2 = -3, c2 = -12
For unique solution,
a1/a2 ≠ b1/b2
4/2 ≠ -5/-3
Here, k can have any real values.
Hence, the given set of equations will have unique solution for all real values of k.
问题8.为以下每个具有唯一解的方程组找到k的值：
x + 2y = 3，
5x + ky + 7 = 0
解决方案：
Given that,
x + 2y = 3 …(1)
5x + ky + 7 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 1, b1 = 2, c1 = −3
a2 = 5, b2 = k, c2 = 7
For unique solution,
a1/a2 ≠ b1/b2
1/5 ≠ 2/k
k ≠ 10
So, the given set of equations will have unique solution for all real values of k other than 10.
问题9.求k的值，以下每个方程组都有无限多个解：
2x + 3y-5 = 0，
6x-ky-15 = 0
解决方案：
Given that,
2x + 3y − 5 = 0 …(1)
6x − ky − 15 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = 3, c1 = −5
a2 = 6, b2 = k, c2 = −15
For unique solution,
We have
a1/a2 = b1/b2 = c1/c2
2/6 = 3/k
k = 9
Hence, when k = 9 the given set of equations will have infinitely many solutions.
问题10。找到以下每个方程组具有无限多个解的k的值：
4x + 5y = 3，
x + 15y = 9
解决方案：
Given that,
4x + 5y = 3 …(1)
kx +15y = 9 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 4, b1 = 5, c1 = 3
a2 = k, b2 = 15, c2 = 9
For unique solution,
We have
a1/a2 = b1/b2 = c1/c2
4/k = 5/15 = -3/-9
4/k = 1/3
k = 12
Hence, when k = 12 the given set of equations will have infinitely many solutions.
问题11。找到k的值，下面的每个方程组都有无限多个解：
kx-2y + 6 = 0，
4x + 3y + 9 = 0
解决方案：
Given that,
kx − 2y + 6 = 0 …(1)
4x + 3y + 9 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = k, b1 = −2, c1 = 6
a2 = 4, b2 = −3, c2 = 9
For unique solution
We have
a1/a2 = b1/b2 = c1/c2
k/4 = -2/-3 = 2/3
k = 8/3
Hence, when k = 8/3 the given set of equations will have infinitely many solutions.
问题12。找到k的值，下面的每个方程组都具有无限多个解：
8x + 5y = 9
kx + 10y = 19
解决方案：
Given that,
8x + 5y = 9 …(1)
kx + 10y = 19 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 8, b1 = 5, c1 = −9
a2 = k, b2 = 10, c2 = −19
For unique solution
We have
a1/a2 = b1/b2 = c1/c2
8/k = 5/10 = k = 16
Hence, when k = 16 the given set of equations will have infinitely many solutions.
问题13。找到以下每个方程组具有无限多个解的k的值：
2x-3y = 7
(k + 2)x −(2k +1)y = 3(2k − 1)
解决方案：
Given that,
2x − 3y = 7 …(1)
(k + 2)x − (2k + 1)y = 3(2k − 1) …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = −3, c1 = −7
a2 = k, b2 = − (2k + 1), c2 = −3(2k − 1)
Now, for unique solution
We have
a1/a2 = b1/b2 = c1/c2
= 2/(k + 2) = -3/-(2k + 1) = -7/-3(2k – 1)
= 2/(k + 2) = -3/-(2k + 1) and -3/-(2k + 1) = -7/-3(2k – 1)
= 2(2k + 1) = 3(k + 2) and 3 × 3(2k − 1) = 7(2k + 1)
= 4k + 2 = 3k + 6 and 18k − 9 = 14k + 7
= k = 4 and 4k = 16
= k = 4
Hence, when k = 4 the given set of equations will have infinitely many solutions.