Q. 125.0( 3 Votes )

# The system x – 2y = 3 and 3x + ky = 1 has a unique solution only when

A. k = – 6

B. k ≠ – 6

C. k = 0

D. k ≠ 0

Answer :

We have,

x – 2y – 3 = 0

3x + ky – 1 = 0

The given equation is in the form: a_{1}x _{+} b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0

Here, we have:

a_{1} = 1, b_{1} = – 2 and c_{1} = – 3

And, a_{2} = 3, b_{2} = k and c_{2} = – 1

∴ , and

These graph lines will intersect at a unique point when we have:

∴

Hence, k has all real values other than – 6

Thus, option B is correct

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