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

k = 4

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k = 1

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k = 2

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k = 5

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Solution

The correct option is C $k = 2$

The given pair of equations are:
$3 x + y = 1... (1)$
$(2 k - 1) x + (k - 1) y = 2 k + 1.. (2)$
Now rearranging eq1 and eq2 will get
$3 x + y - 1 = 0... (3)$
$(2 k - 1) x + (k - 1) y - (2 k + 1) = 0.. (4)$
Now compare with
$a_{1} = 3, b_{1} = 1, c_{1} = - 1$
$a_{2} = 2 k - 1, b_{2} = k - 1, c_{2} = - (2 k + 1)$
Now we get
$\frac{a_{1}}{a_{2}} = \frac{3}{2 k - 1}, \frac{b_{1}}{b_{2}} = \frac{1}{k - 1}, \frac{c_{1}}{c_{2}} = \frac{- 1}{- (2 k + 1)}$
Now will take
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$
$\Rightarrow \frac{3}{2 k - 1} = \frac{1}{k - 1}$
$\Rightarrow 3 k - 3 = 2 k - 1$
$\Rightarrow 3 k - 2 k = - 1 + 3$
$\Rightarrow k = 2$
Hence $k = 2$ is the value.

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