If the pair of linear equations 13x+15y=3 and 65x+ky=6 have no solution then the value of k is ✅ Chi Tiết
Thủ Thuật Hướng dẫn If the pair of linear equations 13x+15y=3 and 65x+ky=6 have no solution then the value of k is Chi Tiết
Hoàng Đức Anh đang tìm kiếm từ khóa If the pair of linear equations 13x+15y=3 and 65x+ky=6 have no solution then the value of k is được Update vào lúc : 2022-10-04 14:50:36 . Với phương châm chia sẻ Thủ Thuật về trong nội dung bài viết một cách Chi Tiết 2022. Nếu sau khi Read tài liệu vẫn ko hiểu thì hoàn toàn có thể lại phản hồi ở cuối bài để Admin lý giải và hướng dẫn lại nha.Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.5
Question: 1
In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,
Nội dung chính- Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.5Question: 1Question: 2Question: 3Question: 4Question: 5Question: 6Question: 7Question: 8Question: 9Question: 10Question: 11Question: 12Question: 13Question: 14Question: 15Question: 16Question: 17Question: 18Question: 19Question: 20Question: 21Question: 22Question: 23Question: 24Question: 25Question: 26Question: 27Question: 28Question: 29Question: 30Question: 31Question: 32Question: 33Question: 34Question: 35Question: 36For what value of k the pair of linear equations 3x 5y 3 and 6x KY 8 has no solution?How do you find the value of k in a linear equation?What is the condition for pair of equation has no solution?For what values of k do the equations KX 4y K 4 16x Ky K represent coincident lines?
x − 3y − 3 = 0, x − 3y − 3 = 0
3x − 9y − 2 = 0, 3x − 9y − 2 = 0
Solution:
The given system may be written as
x − 3y − 3 = 0
3x − 9y − 2 = 0
The given system of equation is of the form
a2x + b2y − c2 = 0 a2x + b2y − c2 = 0
Where, a1 = 1, b1 = −3, c1 = −3
a2 = 3, b2 = −9, c2 = −2
We have,
Therefore, the given equation has no solution.
Question: 2
In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,
2x + y − 5 = 0
4x + 2y − 10 = 0
Solution:
The given system may be written as
2x + y − 5 = 0 4x + 2y − 10 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = 1, c1 = −5
a2 = 4, b2 = 2, c2 = −10
We have,
Therefore, the given equation has infinitely many solution.
Question: 3
In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,
3x − 5y = 20
6x − 10y = 40
Solution:
The given system may be written as
3x − 5y = 20 6x − 10y = 40
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 3, b1 = −5, c1 = − 20
a2 = 6, b2 = −10, c2 = − 40
We have,
Therefore, the given equation has infinitely many solution.
Question: 4
In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,
x − 2y − 8 = 0
5x − 10y − 10 = 0
Solution:
The given system may be written as
x − 2y − 8 = 0 5x − 10y − 10 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 1, b1 = −2, c1 = −8
a2 = 5, b2 = −10, c2 = −10
We have,
Therefore, the given equation has no solution.
Question: 5
Find the value of k for each of the following system of equations which have a unique solution
kx + 2y − 5 = 0
3x + y − 1 = 0
Solution:
The given system may be written as
kx + 2y − 5 = 0
3x + y − 1 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0
a2x + b2y − c2 = 0
Where, a1 = k, b1 = 2, c1 = −5
a2 = 3, b2 = 1, c2 = −1
For unique solution, we have
Therefore, the given system will have unique solution for all real values of k other than 6.
Question: 6
Find the value of k for each of the following system of equations which have a unique solution
4x + ky + 8 = 0
2x + 2y + 2 = 0
Solution:
The given system may be written as
4x + ky + 8 = 0 2x + 2y + 2 = 0
The given system of equation is of the form
a1x +b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 4, b1 = k, c1 = 8
a2 = 2, b2 = 2, c2 = 2
For unique solution, we have
Therefore, the given system will have unique solution for all real values of k other than 4.
Question: 7
Find the value of k for each of the following system of equations which have a unique solution
4x − 5y = k
2x − 3y = 12
Solution:
The given system may be written as
4x − 5y − k = 0
2x − 3y − 12 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 4, b1 = −5, c1 = −k
a2 = 2, b2 = -3, c2 = -12
For unique solution, we have
⇒ k can have any real values.
Therefore, the given system will have unique solution for all real values of k.
Question: 8
Find the value of k for each of the following system of equations which have a unique solution
x + 2y = 3
5x + ky + 7 = 0
Solution:
The given system may be written as
x + 2y = 3
5x + ky + 7 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where a1 = 1, b1 = 2, c1 = −3
a2 = 5, b2 = k, c2 = 7
For unique solution, we have
Therefore, the given system will have unique solution for all real values of k other than 10.
Question: 9
Find the value of k for which each of the following system of equations having infinitely many solution:
2x + 3y − 5 = 0
6x − ky − 15 = 0
Solution:
The given system may be written as
2x + 3y − 5 = 0 6x − ky − 15 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = 3, c1 = −5
a2 = 6, b2 = k, c2 = −15
For unique solution, we have
Therefore, the given system of equation will have infinitely many solutions, if k = 9.
Question: 10
Find the value of k for which each of the following system of equations having infinitely many solution:
4x + 5y = 3
x + 15y = 9
Solution:
The given system may be written as
4x + 5y = 3
kx +15y = 9
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 4, b1 = 5, c1 = 3
a2 = k, b2 = 15, c2 = 9
For unique solution, we have
Therefore, the given system will have infinitely many solutions if k = 12.
Question: 11
Find the value of k for which each of the following system of equations having infinitely many solution:
kx − 2y + 6 = 0
4x + 3y + 9 = 0
Solution:
The given system may be written as
kx − 2y + 6 = 0 4x + 3y + 9 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = k, b1 = −2, c1 = 6
a2 = 4, b2 = −3, c2 = 9
For unique solution, we have
Therefore, the given system of equations will have infinitely many solutions, if k = 8/3.
Question: 12
Find the value of k for which each of the following system of equations having infinitely many solution:
8x + 5y = 9
kx + 10y = 19
Solution:
The given system may be written as
8x + 5y = 9 kx + 10y = 19
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 =0
Where, a1 = 8, b1 = 5, c1 = −9
a2 = k, b2 = 10, c2 = −18 a2 = k, b2 = 10,c2 = −18
For unique solution, we have
Therefore, the given system of equations will have infinitely many solutions, if k = 16.
Question: 13
Find the value of k for which each of the following system of equations having infinitely many solution:
2x − 3y = 7
(k + 2)x − (2k + 1)y = 3(2k − 1)
Solution:
The given system may be written as
2x − 3y = 7 (k + 2)x − (2k + 1)y = 3(2k − 1)
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = −3, c1 = −7
a2 = k, b2 = − (2k + 1), c2 = −3(2k − 1)
For unique solution, we have
⇒ 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
Therefore, the given system of equations will have infinitely many solutions, if k = 4.
Question: 14
Find the value of k for which each of the following system of equations having infinitely many solution:
2x + 3y = 2
(k + 2)x + (2k + 1)y = 2(k − 1)
Solution:
The given system may be written as
2x + 3y = 2 (k + 2)x + (2k + 1)y = 2(k − 1)
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = 3,c1 = −2
a2 = (k + 2), b2 = (2k + 1),c2 = −2(k − 1)
For unique solution, we have
⇒ 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
Therefore, the given system of equations will have infinitely many solutions, if k = 4.
Question: 15
Find the value of k for which each of the following system of equations having infinitely many solution:
x + (k + 1)y = 4
(k + 1)x + 9y = (5k + 2)
Solution:
The given system may be written as
x + (k + 1)y = 4 (k + 1)x + 9y = (5k + 2)
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 1, b1 = (k + 1), c1 = −4
a2 = (k + 1), b2 = 9, c2 = − (5k + 2)
For unique solution, we have
⇒ 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
Thus, k = 2 satisfies both the condition.
Therefore, the given system of equations will have infinitely many solutions, if k = 2.
Question: 16
Find the value of k for which each of the following system of equations having infinitely many solution:
kx + 3y = 2k + 1
2(k + 1)x + 9y = (7k + 1)
Solution:
The given system may be written as
kx + 3y = 2k + 1 2(k + 1)x + 9y = (7k + 1)
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = k, b1 = 3, c1 = −(2k + 1)
a2 = 2(k + 1), b2 = 9, c2 = −(7k + 1)
For unique solution, we have
⇒ 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
Therefore, the given system of equations will have infinitely many solutions, if k = 2.
Question: 17
Find the value of k for which each of the following system of equations having infinitely many solution:
2x + (k − 2)y = k
6x + (2k − 1)y = (2k + 5)
Solution:
The given system may be written as
2x +( k − 2)y = k 6x + (2k − 1)y = (2k + 5)
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2,b1 = (k − 2),c1 = −k
a2 = 6,b2 = (2k − 1),c2 = −(2k + 5)
For unique solution, we have
⇒ 2k − 3k = −6 + 1 and k + k = 10
⇒ −k = −5 and 2k = 10 ⇒ k = 5 and k = 5
Therefore, the given system of equations will have infinitely many solutions, if k = 5.
Question: 18
Find the value of k for which each of the following system of equations having infinitely many solution:
2x + 3y = 72x + 3y = 7
(k + 1)x + (2k − 1)y = (4k + 1)
Solution:
The given system may be written as
2x + 3y = 7 (k + 1)x + (2k − 1)y = (4k + 1)
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = 3, c1 = −7
a2 = k + 1, b2 = 2k − 1, c2 = −(4k + 1)
For unique solution, we have
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
Therefore, the given system of equations will have infinitely many solutions, if k = 5.
Question: 19
Find the value of k for which each of the following system of equations having infinitely many solution:
2x + 3y = k
(k − 1)x + (k + 2)y = 3k
Solution:
The given system may be written as
2x + 3y = k (k − 1)x + (k + 2)y = 3k
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2,b1 = 3, c1 = −k
a2 = k − 1, b2 = k + 2, c2 = −3k
For unique solution, we have
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
Therefore, the given system of equations will have infinitely many solutions, if k = 7.
Question: 20
Find the value of k for which the following system of equation has no solution:
kx − 5y = 2
6x + 2y = 7
Solution:
The given system may be written as
kx − 5y = 2 6x + 2y = 7
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = k, b1 = −5, c1 = −2
a2 = 6 b2 = 2, c2 = −7
For no solution, we have
⇒ 2k = -30 ⇒ k = -15
Therefore, the given system of equations will have no solutions, if k = −15.
Question: 21
Find the value of k for which the following system of equation has no solution:
x + 2y = 0
2x + ky = 5
Solution:
The given system may be written as
x2y = 0 2x + ky = 5
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 1, b1 = 2, c1 = 0
a2 = 2, b2 = k, c2 = −5
For no solution, we have
⇒ k = 4
Therefore, the given system of equations will have no solutions, if k = 4.
Question: 22
Find the value of k for which the following system of equation has no solution:
3x − 4y + 7 = 0
kx + 3y − 5 = 0
Solution:
The given system may be written as
3x − 4y + 7 = 0 kx + 3y − 5 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 3, b1 = −4, c1 = 7
a2 = k, b2 = 3, c2 = −5
For no solution, we have
Therefore, the given system of equations will have no solutions, if k = - 9 /4.
Question: 23
Find the value of k for which the following system of equation has no solution:
2x − ky + 3 = 0
3x + 2y − 1 = 0
Solution:
The given system may be written as
2x − ky + 3 = 0 3x + 2y − 1 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = −k, c1 = 3
a2 = 3, b2 = 2, c2 = −1
For no solution, we have
Therefore, the given system of equations will have no solutions, if k = – 4/3.
Question: 24
Find the value of k for which the following system of equation has no solution:
2x + ky − 11 = 0
5x − 7y − 5 = 0
Solution:
The given system may be written as
2x + ky − 11 = 0 5x − 7y − 5 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = k, c1 = −11
a2 = 5, b2 = −7, c2 = −5a2 = 5, b2 = −7, c2 = − 5
For no solution, we have
Therefore, the given system of equations will have no solutions, if k = -14/5.
Question: 25
Find the value of k for which the following system of equation has no solution:
kx + 3y = 3
12x + ky = 6
Solution:
The given system may be written as
kx + 3y = 3 12x + ky = 6
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = k, b1 = 3, c1 = −3
a2 = 12, b2 = k, c2 = − 6
For no solution, we have
⇒ k2 = 36 ⇒ k = + 6 or −6
From (i)
Therefore, the given system of equations will have no solutions, if k = − 6.
Question: 26
For what value of a, the following system of equation will be inconsistent?
4x + 6y − 11 = 0
2x + ay − 7 = 0
Solution:
The given system may be written as
4x + 6y − 11 = 0 2x + ay − 7 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 4, b1 = 6, c1 = −11
a2 = 2, b2 = a, c2 = −7
For unique solution, we have
Therefore, the given system of equations will be inconsistent, if a = 3.
Question: 27
For what value of a, the following system of equation have no solution?
ax + 3y = a − 3
12x + ay = a
Solution:
The given system may be written as
ax + 3y = a − 3 12x + ay = a
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = a, b1 = 3, c1 = - (a − 3)
a2 = 12, b2 = a, c2 = − a
For unique solution, we have
And,
⇒ a2 = 36
⇒ a = + 6 or – 6?
a ≠ 6 ⇒ a = – 6
Therefore, the given system of equations will have no solution, if a = − 6.
Question: 28
Find the value of a, for which the following system of equation have
(i) Unique solution
(ii) No solution
kx + 2y = 5
3x + y = 1
Solution:
The given system may be written as
kx + 2y − 5 = 0 3x + y − 1 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = k, b1 = 2, c1 = −5
a2 = 3, b2 = 1, c2 = −1
(i) For unique solution, we have
Therefore, the given system of equations will have unique solution, if k ≠ 6 k ≠ 6.
(ii) For no solution, we have
Therefore, the given system of equations will have no solution, if a = 6.
Question: 29
For what value of c, the following system of equation have infinitely many solution (where c ≠ 0 c ≠ 0)?
6x + 3y = c − 3
12x + cy = c
Solution:
The given system may be written as
6x + 3y − (c − 3) = 0 12x + cy − c = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 6, b1 = 3, c1 = −(c − 3)
a2 = 12, b2 = c, c2 = - c
For infinitely many solution, we have
⇒ c = 6 and c – 3 = 3
⇒ c = 6 and c = 6
Therefore, the given system of equations will have infinitely many solution, if c = 6.
Question: 30
Find the value of k, for which the following system of equation have
(i) Unique solution
(ii) No solution
(iii) Infinitely many solution
2x + ky = 1
3x − 5y = 7
Solution:
The given system may be written as
2x + ky = 1 3x − 5y = 7
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = k, c1 = −1a1 = 2, b1 = k, c1 = −1
a2 = 3, b2 = −5, c2 = −7
(i) For unique solution, we have
Therefore, the given system of equations will have unique solution, if k ≠ -10/3.
(ii) For no solution, we have
Therefore, the given system of equations will have no solution, if k = -10)/3.
(iii) For the given system to have infinitely many solution, we have
So there is no value of k for which the given system of equation has infinitely many solution.
Question: 31
For what value of k, the following system of equation will represent the coincident lines?
x + 2y + 7 = 0
2x + ky + 14 = 0
Solution:
The given system may be written as
x + 2y + 7 = 0 2x + ky + 14 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, 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 solution.
Therefore, the given system of equations will have infinitely many solution, if k = 4.
Question: 32
Find the value of k, for which the following system of equation have unique solution.
ax + by = c
lx + my = n
Solution:
The given system may be written as
ax + by − c = 0 lx + my − n = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = a, b1 = b, c1 = − c
a2 = l, b2 = m, c2 = − n
For unique solution, we have
Therefore, the given system of equations will have unique solution, if am ≠ bl.
Question: 33
Find the value of a and b such that the following system of linear equation have infinitely many solution:
(2a − 1)x + 3y − 5 = 0
3x + (b − 1)y − 2 = 0
Solution:
The given system of equation may be written as,
(2a − 1)x + 3y − 5 = 0 3x + (b − 1)y − 2 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = (2a − 1), b1 = 3, c1 = −5
a2 = 3, b2 = b − 1, c2 = −2
The given system of equation will have infinitely many solution, if
⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)
⇒ 4a − 2 = 15 and 6 = 5b − 5 ⇒ 4a = 17 and 5b = 11
Question: 34
Find the value of a and b such that the following system of linear equation have infinitely many solution:
2x − 3y = 7
(a + b)x − (a + b − 3)y = 4a + b
Solution:
The given system of equation may be written as,
2x − 3y − 7 = 0 (a + b)x − (a + b − 3)y − (4a + b) = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = −3, c1 = −7
a2 = (a + b), b2 = −(a + b − 3), c2 = −(4a + b)
The given system of equation will have infinitely many solution, if
⇒ 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 we have
5( – b – 6) – 4b = – 21
⇒ − 5b − 30 − 4b = − 21
⇒ 9b = − 9 ⇒ b = −1
As a = – 6 – b
⇒ a = − 6 + 1 = − 5
Hence the given system of equation will have infinitely many solution if
a = – 5 and b = –1.
Question: 35
Find the value of p and q such that the following system of linear equation have infinitely many solution:
2x − 3y = 9
(p + q)x + (2p − q)y = 3(p + q + 1)
Solution:
The given system of equation may be written as,
2x − 3y − 9 = 0 (p + q)x + (2p − q)y − 3(p + q + 1) = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 =2, b1 = 3, c1 = −9
a2 = (p + q), b2 = (2p − q), c2 = -3(p + q + 1)
The given system of equation will have infinitely many solution, if
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
Hence the given system of equation will have infinitely many solution if
p = 5/3 and q = 1/3.
Question: 36
Find the values of a and b for which the following system of equation has infinitely many solution:
(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
Solution:
(i) The given system of equation may be written as,
(2a − 1)x + 3y − 5 = 0 3x + (b − 2)y − 3 = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2a − 1, b1 = 3, c1 = −5
a2 = 3, b2 = b − 2, c2 = -3(p + q + 1)
The given system of equation will have infinitely many solution, if
2a - 1 = 5 and - 9 = 5(b - 2)
⇒ a = 3 and -9 = 5b - 10
a = 3 and b = 1/5
Hence the given system of equation will have infinitely many solution if
a = 3 and b = 1/5.
(ii) The given system of equation may be written as,
2x − (2a + 5)y = 5 (2b + 1)x − 9y = 15
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = - (2a + 5), c1 = −5
a2 = (2b + 1), b2 = −9, c2 = −15
The given system of equation will have infinitely many solution, if
Hence the given system of equation will have infinitely many solution if
a = - 1 and b = 5/2.
(iii) The given system of equation may be written as,
(a − 1)x + 3y = 2 6x + (1 − 2b)y = 6
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = a-1, b1 = 3, c1 = −2
a2 = 6, b2 = 1 − 2b, c2 = −6
The given system of equation will have infinitely many solution, if
⇒ a – 1 = 2 and 1 – 2b = 9
⇒ a - 1 = 2 and 1 - 2b = 9
⇒ a = 3 and b = -4
⇒ a = 3 and b = -4
Hence the given system of equation will have infinitely many solution if
a = 3 and b = −4.
(iv) The given system of equation may be written as,
3x + 4y − 12 = 0 (a + b)x + 2(a − b)y − (5a − 1) = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 3, b1 = 4, c1 = −12
a2 = (a + b), b2 = 2(a − b), c2 = – (5a − 1)
The given system of equation will have infinitely many solution, if
⇒ 3(a - b) = 2a + 2b and 2(5a - 1) = 12(a - b)
⇒ a = 5b and -2a = -12b + 2
Substituting a = 5b in -2a = -12b + 2, we have
-2(5b) = -12b + 2
⇒ −10b = −12b + 2 ⇒ b = 1
Thus a = 5
Hence the given system of equation will have infinitely many solution if
a = 5 and b = 1.
(v) The given system of equation may be written as,
2x + 3y − 7 = 0 (a − 1)x + (a + 1)y − (3a − 1) = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = 3, c1 = −7
a2 = (a − 1), b2 = (a + 1), c2 = - (3a − 1)
The given system of equation will have infinitely many solution, if
⇒ 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
Hence the given system of equation will have infinitely many solution if
a = 5 and b = 1.
(vi) The given system of equation may be written as,
2x + 3y − 7 = 0 (a − 1)x + (a + 2)y − 3a = 0
The given system of equation is of the form
a1x + b1y − c1 = 0 a2x + b2y − c2 = 0
Where, a1 = 2, b1 = 3, c1 = −7
a2 = (a − 1), b2 = (a + 2), c2 = −3a
The given system of equation will have infinitely many solution, if
⇒ 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
Hence the given system of equation will have infinitely many solution if
a = 7and b = 1.
For what value of k the pair of linear equations 3x 5y 3 and 6x KY 8 has no solution?
⇒k=10.How do you find the value of k in a linear equation?
Given: Linear equation 2x + 3y = k. We can find the value of k by substituting the values of x and y in the given equation. Therefore, the value of k is 7.What is the condition for pair of equation has no solution?
Inconsistent Pair of Linear Equations If (a1/a2) = (b1/b2) ≠ (c1/c2), then there will be no solution. This type of system of equations is called an inconsistent pair of linear equations. If we plot the graph, the lines will be parallel and system of equations have no solution.For what values of k do the equations KX 4y K 4 16x Ky K represent coincident lines?
⇒ k=8 is correct value for infinite solution. Tải thêm tài liệu liên quan đến nội dung bài viết If the pair of linear equations 13x+15y=3 and 65x+ky=6 have no solution then the value of k is