The system of simultaneous equations $3 m + n = 1$ and $(2 k - 1) m + (k - 1) n = 2 k + 1$ , is inconsistent. What is the value of 'k'?

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Solution

The correct option is B $2$
Comparing $3 m + n = 1$ with $a_{1} x + b_{1} y + c_{1} = 0$ , we get, $a_{1} = 3, b_{1} = 1, c_{1} = - 1$ and $(2 k - 1) m + (k - 1) n = 2 k + 1$ with $a_{2} x + b_{2} y + c_{2}$ , we get $a_{2} = 2 k - 1, b_{2} = k - 1, c_{2} = - (2 k + 1)$
Given: the equations are inconsistent.
$∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\frac{3}{2 k - 1} = \frac{1}{k - 1} \neq \frac{- 1}{- (2 k + 1)}$
When,
$\frac{3}{2 k - 1} = \frac{1}{k - 1}$
$⟹ 3 (k - 1) = 2 k - 1$
$⟹ 3 k - 3 = 2 k - 1$
$⟹ 3 k - 2 k = 3 - 1$
$⟹ k = 2$
Answer : C

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For what value of 'K' will the system of equations 3x + y = 1,
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3 x + y = 1

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

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