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For which value of k will the following pair of linear equations have no solution? $3 x + y = 1 (2 k - 1) x + (k - 1) y = 2 k + 1$ .

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k = 4

k = 1

k = 2

k = 5

answer is C.

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Detailed Solution

Given pair of equations are:

3 x + y = 1

…………(1)

(2 k - 1) x + (k - 1) y = 2 k + 1

………….(2)
Now rearranging equation (1) and equation (2) we get,

3 x + y - 1 = 0

……………..(3)

(2 k - 1) x + (k - 1) y - (2 k + 1) = 0

…………….(4)
Now compare with general equation

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

.
We get,

a_{1} = 3

a_{2} = (2 k - 1)

b_{1} = 1

b_{2} = (k - 1)

c_{1} = - 1

and

c_{2} = - (2 k + 1)

Now we get,

\frac{a_{1}}{a_{2}} = \frac{3}{(2 k - 1)}

\frac{b_{1}}{b_{2}} = \frac{1}{(k - 1)}

and

\frac{c_{1}}{c_{2}} = \frac{- 1}{- (2 k + 1)}

Now will take,

a 1 a 2 = b 1 b 2 ⇒ 3 2 k − 1 = 1 k − 1 ⇒ 3 k − 3 = 2 k − 1 ⇒ 3 k − 2 k = − 1 + 3 ⇒ k = 2

Therefore, option (3) is correct.

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