If the system of equations is inconsistent, then k =
Question:

If the system of equations is inconsistent, then k =

$3 x+y=1$

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

(a) 1

(b) 0

(c) $-1$

(d) 2

Solution:

The given system of equations is inconsistent,

$3 x+y=1$

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

If the system of equations is in consistent, we have

$a_{1} b_{2}-a_{2} b_{1}=0$

$3 \times(k-1)-1(2 k-1)=0$

$3 k-3-2 k+1=0$

$1 k-2=0$

$1 k=2$

Therefore, the value of $k$ is 2 .

Hence, the correct choice is $d$.

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