Question

# 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

A
infinite
B
1
C
2
D
3

Solution

## The correct option is A 1(k+1)x+8y=4k...(1) kx+(k+3)y=3k−1...(2) k+1k=8k+3≠4k3k−1 Now, k+1k=8k+3 ⇒k2−4k+3=0 ⇒k=1,3 If k=1 81+3=43−1 ⇒2=2 ∴k=1 is not the correct value So, if k=3 83+3≠129−1 ⇒43≠32 ∴k=3 is the only value for which system of equations has no solution Hence, number of values of k is 1.Mathematics

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