Consider the system of equations :

$x + y +5z = 3$

$x + 2y + 4z = k$

$x + 2y + mz = 5$ then this system of equation is consistent if

(a) $m \ne 4$

(b) $k \ne 5$

(c) $m = 4$

(d) $k =5$

Reducing, this system by subtracting equations (2) and (3) I get :

$x + y + 5z = 3$

$x + 2y + 4z = k$

$(m -4)z = 5 -k$

Now, the given system is consistent if $m \ne 4$,so only condition I require for consistency is $m \ne 4$

but if I take $k = 5$, then again the system is consistent

It will have unique solutions if $m \ne 4$

and infinite solution if $m = 4$

So, my question is are both (a) and (d) the correct choices for this question ?

Or is only (a) correct ?

Thank you.


If $k=5$ the system is consistent for all values of $m$



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