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For what value of k, the pair of linear equations 3x+y=3 and 6x+ky=8 does not have a solution.

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k = 1

k = 3

k = 4

k = 2

answer is D.

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Detailed Solution

Given pair of linear equations are,

3 x + y = 3 … … .. (i)

6 x + k y = 8 … … .. (i i)

We know that, the general form for a pair of linear equations in two variables x and y is

a 1 x + b 1 y + c 1 = 0 a 2 x + b 2 y + c 2 = 0.

Where

a 1, a 2, b 1, b 2, c 1, c 2

are all real numbers.
If

a 1 a 2 = b 1 b 2 ≠ c 1 c 2

then the pair of linear equations has no solution and their graph represents parallel lines.
On comparing

3 x + y − 3 = 0

with

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

, we can observe

a 1 = 3 b 1 = 1 c 1 = − 3

By comparing

6 x + k y − 8 = 0

with

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

, we get

a 2 = 6 b 2 = k c 2 = − 8

Here,

a 1 a 2 = 36 = 12 b 1 b 2 = 1 k c 1 c 2 = − 3 − 8 = 38

The condition for two linear equations with no solution is

a 1 a 2 = b 1 b 2 ≠ c 1 c 2

.
Substitute the values.

a 1 a 2 = b 1 b 2 ≠ c 1 c 2 12 = 1 k ≠ 38

Thus,

12 = 1 k k = 2

Therefore, for the value

k = 2,

the given pair of linear equations

3 x + y = 3

and

6 x + k y = 8

does not have a solution.
The correct option is (4).

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