问题27.对于a的哪个值，方程的以下系统无解：
ax + 3y = a − 3
12x + ay = a
解决方案：
Given that,
ax + 3y = a − 3 …(1)
12x + ay = a …(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 = a, b1 = 3, c1 = – (a − 3)
a2 = 12, b2 = a, c2 = − a
For unique solution, we have
a1/a2 = b1/b2 ≠ c1/c2
a/12 = 3/a ≠ -(a – 3)/-a
a-3 ≠ 3
a ≠ 6
And,
a/12 = 3/a
a2 = 36
a = + 6 or – 6?
a ≠ 6, a = – 6
Hence, when a = -6 the given set of equations will have no solution.
问题28.找到a的值，下面的方程组具有
(i)独特的解决方案
(ii)无解
kx + 2y = 5
3x + y = 1
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
(i) For unique solution, we have
a1/a2 ≠ b1/b2
k/3 ≠ 2
k ≠ 6
Hence, when k ≠ 6 the given set of equations will have unique solution.
(ii) For no solution, we have
a1/a2 = b1/b2 ≠ c1/c2
k/3 = 2/1 ≠ -5/-1
k/3 = 2/1
k = 6
Hence, when k = 6 the given set of equations will have no solution.
问题29.对于c的多少值，以下方程组具有无限多个解(其中c≠0 c≠0)：
6x + 3y = c − 3
12x + cy = c
解决方案：
Given that,
6x + 3y − (c − 3) = 0 …(1)
12x + cy − c = 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 = 6, b1 = 3, c1 = −(c − 3)
a2 = 12, b2 = c, c2 = – c
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
6/12 = 3/c = -(c + 3)/-c
6/12 = 3/c and 3/c = -(c + 3)/-c
c = 6 and c – 3 = 3
c = 6 and c = 6
Hence, when c = 6 the given set of equations will have infinitely many solution.
问题30.求出k的值，以下方程式对此具有
(i)独特的解决方案
(ii)无解
(iii)无限多的解决方案
2x + ky = 1
3x-5y = 7
解决方案：
Given that,
2x + ky = 1 …(1)
3x − 5y = 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 = 2, b1 = k, c1 = −1
a2 = 3, b2 = −5, c2 = −7
(i) For unique solution, we have
a1/a2 ≠ b1/b2
2/3 ≠ -k/-5 ≠ -10/3
Hence, when k ≠ -10/3 the given set of equations will have unique solution.
(ii) For no solution, we have
a1/a2 = b1/b2 ≠ c1/c2
2/3 = k/-5 ≠ -1/-7
2/3 = k/-5 and k/-5 ≠ -1/-7
k = -10/3 and k ≠ -5/7
k = -10/3
Hence, when k = -10/3 the given set of equations will have no solution.
(iii) For the given system to have infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/3 = k/-5 = -1/-7
Clearly a1/a2 ≠ c1/c2
Hence. there is no value of k for which the given set of equation has infinitely many solution.
问题31.对于k的值是多少，以下方程组将表示重合线：
x + 2y + 7 = 0
2x + ky + 14 = 0
解决方案：
Given that,
x + 2y + 7 = 0 …(1)
2x + ky + 14 = 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 = 7
a2 = 2, b2 = k, c2 = 14
The given system of equation will represent the coincident
lines if they have infinitely many solutions
So,
a1/a2 = b1/b2 = c1/c2
1/2 = 2/k = 7/14
1/2 = 2/k = 7/14
k = 4
Hence, when k = 4 the given set of equations will have infinitely many solution.
问题32.求k的值，以下方程组具有唯一的解：
ax + by = c，
lx +我的= n
解决方案：
Given that,
ax + by − c = 0 …(1)
lx + my − n = 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 = a, b1 = b, c1 = − c
a2 = l, b2 = m, c2 = − n
For unique solution, we have
a1/a2 ≠ b1/b2
a/l ≠ b/m
am ≠ bl
Hence, when am ≠ bl the given set of equations will have unique solution.
问题33.找到a和b的值，以使以下线性方程组具有无限多个解：
(2a-1)x + 3y-5 = 0，
3x +(b − 1)y − 2 = 0
解决方案：
Given that,
(2a − 1)x + 3y − 5 = 0
3x + (b − 1)y − 2 = 0
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
a2 = 3, b2 = b − 1, c2 = −2
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
(2a – 1)/3 = 3/(b – 1) = -5/-2
(2a – 1)/3 = 3/(b – 1) and 3/(b – 1) = -5/-2
2(2a − 1) = 15 and 6 = 5(b − 1)
4a − 2 = 15 and 6 = 5b − 5
4a = 17 and 5b = 11
So, a = 17/4 and b = 11/5
问题34.找到a和b的值，使得以下线性方程组具有无限多个解：
2x-3y = 7
(a + b)x −(a + b − 3)y = 4a + b
解决方案：
Given that,
2x − 3y − 7 = 0 …(1)
(a + b)x − (a + b − 3)y − (4a + b) = 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 = −7
a2 = (a + b), b2 = −(a + b − 3), c2 = −(4a + b)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(a + b) = -3/-(a + b – 3) = -7/-(4a + b)
2/(a + b) = -3/-(a + b – 3) and -3/-(a + b – 3) = -7/-(4a + b)
2(a + b − 3) = 3(a + b) and 3(4a + b) = 7(a + b − 3)
2a + 2b − 6 = 3a + 3b and 12a + 3b = 7a + 7b − 21
a + b = −6 and 5a − 4b = −21
a = − 6 − b
Substituting the value of a in 5a − 4b = −21, and we will get,
5( -b – 6) – 4b = -21
− 5b − 30 − 4b = − 21
9b = − 9 ⇒ b = −1
As a = – 6 – b
a = − 6 + 1 = − 5
So, a = – 5 and b = –1.
问题35.找到p和q的值，以使以下线性方程组具有无限多个解：
2x-3y = 9
(p + q)x +(2p − q)y = 3(p + q + 1)
解决方案：
Given that,
2x − 3y − 9 = 0 …(1)
(p + q)x + (2p − q)y − 3(p + q + 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 = 3, c1 = −9
a2 = (p + q), b2 = (2p − q), c2 = -3(p + q + 1)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(p + q) = 3/(2p – q) = -9/-3(p + q + 1)
2/(p + q) = 3/(2p – q) and 3/(2p – q) = -9/-3(p + q + 1)
2(2p – q) = 3(p + q) and (p + q + 1) = 2p – q
4p – 2q = 3p + 3q and -p + 2q = -1
p = 5q and p – 2q = 1
Substituting the value of p in p – 2q = 1, we have
3q = 1
q = 1/3
Substituting the value of p in p = 5q we have
p = 5/3
So, p = 5/3 and q = 1/3.
问题36.找到a和b的值，下面的方程组可以无限地求解这些问题：
(i)(2a − 1)x + 3y = 5，
3x +(b − 2)y = 3
(ii)2x-(2a + 5)y = 5，
(2b +1)x-9y = 15
(iii)(a − 1)x + 3y = 2，
6x +(1-2b)y = 6
(iv)3x + 4y = 12
(a + b)x + 2(a-b)y = 5a – 1
(v)2x + 3y = 7
(a-1)x +(a + 1)y = 3a-1
(vi)2x + 3y = 7
(a − 1)x +(a + 2)y = 3a
(vii)2x + 3y = 7
(a – b)x +(a + b)y = 3a + b + 2
(viii)x + 2y = 1
(a − b)x +(a + b)y = a + b – 2
(ix)2x + 3y = 7
2ax + ay = 28 –
解决方案：
(i) Given that,
(2a − 1)x + 3y − 5 = 0 …(1)
3x + (b − 2)y − 3 = 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 = 2a − 1, b1 = 3, c1 = −5
a2 = 3, b2 = b − 2, c2 = -3(p + q + 1)
For infinitely many solution, we have
a1/a2 = b1/b2 = c1/c2
(2a – 1)/3 = -3/(b – 2) = -5/-3
(2a – 1)/3 = -3/(b – 2) and -3/(b – 2) = -5/-3
2a – 1 = 5 and – 9 = 5(b – 2)
a = 3 and -9 = 5b – 10
a = 3 and b = 1/5
So, a = 3 and b = 1/5.
(ii) Given that,
2x − (2a + 5)y = 5 …(1)
(2b + 1)x − 9y = 15 …(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 = – (2a + 5), c1 = −5
a2 = (2b + 1), b2 = −9, c2 = −15
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(2b + 1) = (-2a + 5)/-9 = -5/-15
2/(2b + 1) = (-2a + 5)/-9 and (-2a + 5)/-9 = -5/-15
6 = 2b + 1 and 2a + 5
3 = b = 5/2 and a = -1
So, a = – 1 and b = 5/2.
(iii) Given that,
(a − 1)x + 3y = 2 …(1)
6x + (1 − 2b)y = 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 = a – 1, b1 = 3, c1 = −2
a2 = 6, b2 = 1 − 2b, c2 = −6
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
(a – 1)/6 = 3/(1 – 2b) = 2/6
(a – 1)/6 = 3/(1 – 2b) and 3/(1 – 2b) = 2/6
a – 1 = 2 and 1 – 2b = 9
a – 1 = 2 and 1 – 2b = 9
a = 3 and b = -4
a = 3 and b = -4
So, a = 3 and b = −4.
(iv) Given that,
3x + 4y − 12 = 0 …(1)
(a + b)x + 2(a − b)y − (5a − 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 = 3, b1 = 4, c1 = −12
a2 = (a + b), b2 = 2(a − b), c2 = – (5a − 1)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
3/(a + b) = 4/2(a – b) = 12/(5a – 1)
3/(a + b) = 4/2(a – b) and 4/2(a – b) = 12/(5a – 1)
3(a – b) = 2a + 2b and 2(5a – 1) = 12(a – b)
a = 5b and -2a = -12b + 2
On substituting a = 5b in -2a = -12b + 2, we have
-2(5b) = -12b + 2
−10b = −12b + 2 ⇒ b = 1
Thus a = 5
So, a = 5 and b = 1.
(v) Given that,
2x + 3y − 7 = 0 …(1)
(a − 1)x + (a + 1)y − (3a − 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 = 3, c1 = −7
a2 = (a − 1), b2 = (a + 1), c2 = – (3a − 1)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(a – b) = 3/(a + 1) = -7/(3a – 1)
2/(a – b) = 3/(a + 1) and 3/(a + 1) = -7/(3a – 1)
2(a + 1) = 3(a – 1) and 3(3a – 1) = 7(a + 1)
2a – 3a = -3 – 2 and 9a – 3 = 7a + 7
a = 5 and a = 5
So, a = 5 and b = 1.
(vi) Given that,
2x + 3y − 7 = 0 …(1)
(a − 1)x + (a + 2)y − 3a = 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 = −7
a2 = (a − 1), b2 = (a + 2), c2 = −3a
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(a – b) = 3/(a + 2) = -7/-3a
2/(a – b) = 3/(a + 2) and 3/(a + 2) = -7/-3a
2(a + 2) = 3(a – 1) and 3(3a) = 7(a + 2)
2a + 4 = 3a – 3 and 9a = 7a + 14
a = 7 and a = 7
So, a = 7 and b = 1.
(vii) 2x + 3y – 7 = 0, …(1)
(a – b)x + (a + b)y – 3a – b + 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 = 2, b1 = 3, c1 = −7
a2 = (a − b), b2 = (a + b), c2 = −(3a + b – 2)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(a – b) = 3/(a + b) = -7/−(3a + b – 2)
2/(a – b) = 3/(a + b) and 3/(a + b) = -7/−(3a + b – 2)
2(a + b) = 3(a – b) and 3(3a + b – 2) = 7(a + b)
2a + 2b = 3a – 3b and 9a + 3b – 6 = 7a + 7b
So, a = 5 and b = 1.
(viii) x + 2y – 1 = 0 …(1)
(a − b)x + (a + b)y – a – b + 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 = 2, c1 = −1
a2 = (a − b), b2 = (a + b), c2 = −(a + b – 2)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
1/(a – b) = 2/(a + b) = -1/−(a + b – 2)
1/(a – b) = 2/(a + b) and 2/(a + b) = -1/−(a + b – 2)
(a + b) = 2(a – b) and 2(a + b – 2) = (a + b)
a + b = 2a – 2b and 2a + 2b – 4 = a + b
So, a = 3 and b = 1.
(ix) 2x + 3y – 7 = 0 …(1)
2ax + ay – 28 + by = 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 = −7
a2 = 2a, b2 = (a + b), c2 = −28
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/2a = 3/(a + b) = -7/−28
2/2a = 3/(a + b) and 3/(a + b) = 7/28
2(a + b) = 6a and 84 = 7(a + b)
So, a = 4 and b = 8.
问题37.其中值(S)λ的，执行对线性方程λx+ Y =λ2，并且x +λy= 1具有
(i)没有解决办法?
(ii)无限多个解决方案?
(iii)独特的解决方案?
解决方案：
Given that,
λx + y = λ2 …(1)
x + λy = 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 = λ, b1 = 1, c1 = -λ2
a2 = 1, b2 = λ, c2 = -λ2
(i) For no solution,
a1/a2 = b1/b2 ≠ c1/c2
λ = 1/λ ≠ -λ2/-1
λ2 – 1 = 0
λ = 1, -1
Here we will take only λ = -1 because at λ = 1 the linear
equation will have infinite many solutions.
(ii) For infinite many solutions,
a1/a2 = b1/b2 = c1/c2
λ = 1/λ = λ2 /1
λ(λ – 1) = 0
When λ ≠ 0 then λ = 1
(iii) For Unique solution,
a1/a2 ≠ b1/b2
λ ≠ 1/λ