# The system x + 2y = 3 and 5x + ky + 7 = 0 has no solution, whenA. k = 10B. k ≠ 10C. k = –7/3D. k = – 21

We have,

x + 2y – 3 = 0

And, 5x + ky + 7 = 0

Here, a1 = 1, b1 = 2 and c1 = – 3

a2 = 5, b2 = k and c2 = 7

And,

We know that, for the system having no solution we must have:

k = 10

Hence, option A is correct

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