If the system of equations is inconsistent, then k =

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Solution

Q.
If the system of equations $3 x + y = 1 and (2 k - 1) x + (k - 1) y = 2 k + 1$ is inconsistent, then k =

1 -1

2 0

3 1

4 2

Ans.
(4)

$3 x + y = 1$ ...(i)

and $(2 k - 1) x + (k - 1) y = 2 k + 1$ ...(ii)

Comparing eq. (i) with $a_{1} x + b_{1} y + c_{1} = 0$ and eq. (ii)

with $a_{2} x + b_{2} y + c_{2} = 0,$ we get

$a_{1} = 3, a_{2} = 2 k - 1, b_{1} = 1, b_{2} = k - 1, c_{1} = - 1$ and $c_{2} = - (2 k + 1)$

Since, system is inconsistent, then $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{3}{2 k - 1} = \frac{1}{k - 1} \neq \frac{- 1}{- (2 k + 1)} \Rightarrow \frac{3}{2 k - 1} = \frac{1}{k - 1} or \frac{1}{k - 1} \neq \frac{1}{2 k + 1}$

$\Rightarrow 3 k - 3 = 2 k - 1 or 2 k + 1 \neq k - 1 \Rightarrow k = 2 or k \neq - 2$

Hence, the value of $k$ is 2.

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