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Find the value of k for which the system of equations $k x − y = 2, 6 x − 2 y = 3$ , has a unique solution.

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k is any real number except 7

k is any real number except 6

k is any real number except 5

k is any real number except 3

answer is D.

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Detailed Solution

Given that the system of equations

k x − y = 2, 6 x − 2 y = 3

, has a unique solution.
We know that any of two linear equations of the form

a 1 x + b 1 y + c 1 = 0

and

a 2 x + b 2 y + c 2 = 0

are having unique solutions, if

a 1 a 2 ≠ b 1 b 2

.
Let us convert the given linear equations to the form

a 1 x + b 1 y + c 1 = 0

k x − y − 2 = 0

……………….(1)

6 x − 2 y − 3 = 0

…………….(2)
After converting, we can observe that,

a 1 = k

;

b 1

=-1;

c 1 = − 2

from equation (1) and

a 2 = 6; b 2 = − 2; c 2 = − 3

from equation (2).
Let us substitute the above values in the condition for infinitely many solutions.

a 1 a 2 ≠ b 1 b 2

k 6 ≠ − 1 − 2

Perform cross multiplication.

k × − 2 ≠ − 1 × 6

k ≠ − 1 × 6 − 2

k ≠ 3

As a result, k can be any real number other than 3.
The correct option is (4).

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