问题14。找到k的值，下面的方程组每个方程式都有无限多个解：
2x + 3y = 2
(k + 2)x +(2k +1)y = 2(k − 1)
解决方案：
Given that,
2x + 3y = 2 …(1)
(k + 2)x + (2k + 1)y = 2(k − 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 = −2
a2 = (k + 2), b2 = (2k + 1), c2 = −2(k − 1)
For unique solution, we have
a1/a2 = b1/b2 = c1/c2
= 2/(k + 2) = 3/(2k + 1) = -2/-2(k – 1)
= 2/(k + 2) = 3/(2k + 1) and 3/(2k + 1) = 2/2(k – 1)
= 2(2k + 1) = 3(k + 2) and 3(k − 1) = (2k + 1)
= 4k + 2 = 3k + 6 and 3k − 3 = 2k + 1
= k = 4 and k = 4
Hence, when k = 4 the given set of equations will have infinitely many solutions.
问题15：找到k的值，下面的每个方程组都有无限多个解：
x +(k + 1)y = 4，
(k + 1)x + 9y =(5k + 2)
解决方案：
Given that,
x + (k + 1)y = 4 …(1)
(k + 1)x + 9y = (5k + 2) …(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 = (k + 1), c1 = −4
a2 = (k + 1), b2 = 9, c2 = − (5k + 2)
For unique solution, we have
a1/a2 = b1/b2 = c1/c2
1/(k + 1) = (k + 1)/9 = -4/-(5k + 2)
1/(k + 1) = (k + 1)/9 and (k + 1)/9 = 4/(5k + 2)
9 = (k + 1)2 and (k + 1)(5k + 2) = 36
9 = k2 + 2k + 1 and 5k2 + 2k + 5k + 2 = 36
k2 + 2k − 8 = 0 and 5k2 + 7k − 34 = 0
k2 + 4k − 2k − 8 = 0 and 5k2 + 17k − 10k − 34 = 0
k(k + 4) −2 (k + 4) = 0 and (5k + 17) − 2 (5k + 17) = 0
(k + 4)(k − 2) = 0 and (5k + 17)(k − 2) = 0
k = – 4 or k = 2 and k = -17/5 or k = 2
Hence, k = 2 satisfies both the condition.
So, when k = 2 the given set of equations will have infinitely many solutions.
问题16。找到k的值，对于该值，以下每个方程组均具有无限多个解：
kx + 3y = 2k + 1
2(k + 1)x + 9y =(7k +1)
解决方案：
Given that,
kx + 3y = 2k + 1 …(1)
2(k + 1)x + 9y = (7k + 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 = k, b1 = 3, c1 = −(2k + 1)
a2 = 2(k + 1), b2 = 9, c2 = −(7k + 1)
For unique solution, we have
a1/a2 = b1/b2 = c1/c2
1/2(k + 2) = 3/9 = -2k + 1/-(7k + 1)
1/2(k + 2) = 3/9 and 3/9 = 2k + 1/(7k + 1)
9k = 3 × 2(k + 1) and 3(7k + 1) = 9(2k + 1)
9k − 6k = 6 and 21k − 18k = 9 − 3
3k = 6 ⇒ k = 2 and k = 2
Hence, when k = 2 the given set of equations will have infinitely many solutions.
问题17。找到以下每个方程组具有无限多个解的k的值：
2x +(k − 2)y = k，
6x +(2k − 1)y =(2k + 5)
解决方案：
Given that,
2x +( k − 2)y = k …(1)
6x + (2k − 1)y = (2k + 5) …(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 = (k − 2), c1 = −k
a2 = 6, b2 = (2k − 1), c2 = −(2k + 5)
For unique solution, we have
a1/a2 = b1/b2 = c1/c2
2/6 = k – 1/(2k – 1) = -k/-2(2k + 5)
2/6 = k – 1/(2k – 1) and (k – 1) / (2k – 1) = k / 2(2k + 5)
2k − 3k = −6 + 1 and k + k = 10
−k = −5 and 2k = 10 = k = 5 and k = 5
Hence, when k = 5 the given set of equations will have infinitely many solutions.
问题18。找到k的值，对于该值，以下每个方程组均具有无限多个解：
2x + 3y = 7
(k + 1)x +(2k − 1)y =(4k +1)
解决方案：
Given that,
2x + 3y = 7 …(1)
(k + 1)x + (2k − 1)y = (4k + 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 + 1, b2 = 2k − 1, c2 = −(4k + 1)
For unique solution, we have
a1/a2 = b1/b2 = c1/c2
2/(k + 1) = 3/(2k – 1) = -7/-(4k + 1)
2/(k + 1) = 3/(2k – 1) and 3/(2k – 1) = -7/-(4k + 1)
Extra close brace or missing open brace
4k − 2 = 3k + 3 and 12k + 3 = 14k − 7
k = 5 and 2k = 10 = k = 5 and k = 5
Hence, when k = 5 the given set of equations will have infinitely many solutions.
问题19.求k的值，以下每个方程组都有无限多个解：
2x + 3y = k，
(k − 1)x +(k + 2)y = 3k
解决方案：
Given that,
2x + 3y = k …(1)
(k − 1)x + (k + 2)y = 3k …(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 = −k
a2 = k − 1, b2 = k + 2, c2 = −3k
For unique solution, we have
a1/a2 = b1/b2 = c1/c2
2/(k – 1) = 3/(k + 2) = -k/-3k
2/(k – 1) = 3/(k + 2) and 3/(k + 2) = -k/-3k
Extra close brace or missing open brace
2k + 4 = 3k − 3 and 9 = k + 2
2k + 4 = 3k − 3 and 9 = k + 2 ⇒ k = 7 and k = 7
Hence, when k = 7 the given set of equations will have infinitely many solutions.
问题20：找到以下方程式无法解决的k值：
kx-5y = 2
6x + 2y = 7
解决方案：
Given that,
kx − 5y = 2 …(1)
6x + 2y = 7 …(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 = −5, c1 = −2
a2 = 6 b2 = 2, c2 = −7
For no solution, we have
a1/a2 = b1/b2 ≠ c1/c2
1/2 = 2/k ≠ 2/7
k = 4
2k = -30
k = -15
Hence, when k = -15 the given set of equations will have no solutions.
问题21：找到以下方程组无解的k的值：
x + 2y = 0，
2x + ky = 5
解决方案：
Given that,
x + 2y = 0 …(1)
2x + ky = 5 …(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 = 0
a2 = 2, b2 = k, c2 = −5
For no solution, we have
a1/a2 = b1/b2 ≠ c1/c2
k/6 = -5/2 ≠ 2/7
k = 4
Hence, when k = 4 the given set of equations will have no solutions.
问题22。找到以下方程组无解的k的值：
3x-4y + 7 = 0，
kx + 3y-5 = 0
解决方案：
Given that,
3x − 4y + 7 = 0 …(1)
kx + 3y − 5 = 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 = 3, b1 = −4, c1 = 7
a2 = k, b2 = 3, c2 = −5
For no solution, we have
a1/a2 = b1/b2 ≠ c1/c2
3/k = -4/3
k = -9/4
Hence, when k = -9/4 the given set of equations will have no solutions.
问题23：找到以下方程式无解的k的值：
2x-ky + 3 = 0，
3x + 2y-1 = 0
解决方案：
Given that,
2x − ky + 3 = 0 …(1)
3x + 2y − 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 = 2, b1 = −k, c1 = 3
a2 = 3, b2 = 2, c2 = −1
For no solution, we have
a1/a2 = b1/b2 ≠ c1/c2
2/3 = -k/2
k = -4/3
Hence, when k = -4/3 the given set of equations will have no solutions.
问题24.找到以下方程组无解的k的值：
2x + ky − 11 = 0，
5x-7y-5 = 0
解决方案：
Given that,
2x + ky − 11 = 0 …(1)
5x − 7y − 5 = 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 = k, c1 = −11
a2 = 5, b2 = −7, c2 = −5
For no solution, we have
a1/a2 = b1/b2 ≠ c1/c2
2/5 = -k/-7
k = -14/5
Hence, when k = -14/5 the given set of equations will have no solutions.
问题25.找到以下方程组无解的k的值：
kx + 3y = 3，
12x + ky = 6
解决方案：
Given that,
kx + 3y = 3 …(1)
12x + ky = 6 …(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 = 3, c1 = −3
a2 = 12, b2 = k, c2 = − 6
For no solution, we have
a1/a2 = b1/b2 ≠ c1/c2
k/12 = 3/k ≠ 3/6 …(5)
k2 = 36 ⇒ k = + 6 or −6
From eq (5), we get
k/12 ≠ 3/6
k ≠ 6
Hence, when k = -6 the given set of equations will have no solutions.
问题26.对于a的哪个值，以下方程组将不一致?
4x + 6y-11 = 0，
2x + ay-7 = 0
解决方案：
Given that,
4x + 6y − 11 = 0 …(1)
2x + ay − 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 = 4, b1 = 6, c1 = −11
a2 = 2, b2 = a, c2 = −7
For inconsistent solution, we have
a1/a2 = b1/b2 ≠ c1/c2
a1/a2 = b1/b2
4/2 = 6/a
a = 3
Hence, when a = 3 the given set of equations will be inconsistent.