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The number of values of k, for which the system of equations (k + 1)x + 8y = 4k kx + (k + 3)y = 3k - 1 has no solution, is (1) 1 (2) 2 (3) 3 (4) infinite

The number of values of k, for which the system of equations (k + 1)x + 8y = 4k kx + (k + 3)y = 3k - 1 has no solution, is (1) 1 (2) 2 (3) 3 (4) infinite

Grade:upto college level

2 Answers

Latika Leekha
askIITians Faculty 165 Points
7 years ago
The given system of equations is
(k + 1)x + 8y = 4k
kx + (k + 3)y = 3k - 1
The system has no solution, hence, we must have the determinant equal to zero.
Hence, [(k+1)(k+3) – 8k = 0
Simplifying we get,
k2 + 4k + 3 – 8k = 0
This gives k2 - 4k + 3 = 0
(k-3)(k-1) = 0
This gives the values of k as 1 and 3. But for no solution, we need to satisfy another condition.
Atleast one of ∆x or ∆y must be non-zero.
We try to find ∆x corresponding to k =1.
x = 4k(k+3) – 8(3k-1)
= 4k2 -12k + 8
For k = 1, we have ∆x = 0.
Similarly ∆y is also 0 for k = 1. Hence, k = 1 is ot possible if we need a no solution.
(You may proceed to test k = 3 and check we get a non-zero determinant corresponding to this value).
Hence, k = 3 is the only possible solution.
Hence the sysytem has no solution for one value of k.
The correct answer is (1).
Thanks & Regards
Latika Leekha
askIITians Faculty
Rishi Sharma
askIITians Faculty 646 Points
11 months ago
Dear Student,
Please find below the solution to your problem.

In the given system of equations
(k + 1)x + 8y = 4k
kx + (k + 3)y = 3k - 1
For system having no solution determinant must be equal to zero.
Hence, [(k+1)(k+3) – 8k = 0
Simplifying we get,
k^2 + 4k + 3 – 8k = 0
This gives k^2 - 4k + 3 = 0
(k-3)(k-1) = 0
k=1,3But for no solution, we need to satisfy another condition.
Atleast one of ∆x or ∆y must be non-zero.
We try to find ∆x corresponding to k =1.
∆x = 4k(k+3) – 8(3k-1)
= 4k2 -12k + 8
For k = 1, we have ∆x = 0.
Similarly ∆y is also 0 for k = 1. Hence, k = 1 is ot possible if we need a no solution.
when we test k = 3 and check we get a non-zero determinant corresponding to this value.
Hence, k = 3 is the only possible solution.
Hence the sytem has no solution for 1 value of k.

Thanks and Regards

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