# 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

Latika Leekha
8 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.
Thanks & Regards
Latika Leekha
Rishi Sharma
2 years ago
Dear Student,

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