Question 1068184
USING DETERMINANTS:
Of you have learned about using matrices and determinants
to solve systems of equations,
you would know that the matrix associated with that system is
{{{(matrix(3,3,1,1,3,-4,2,5,k,0,1))}}} ,
and that the system has no solution if the determinant for that matrix is zero:
{{{abs(matrix(3,3,1,1,3,-4,2,5,k,0,1))=0}}} <--> {{{2+5k-6k+4=0}}} <--> {{{6-k=0}}} <--> {{{highlight(k=0)}}} .
 
WITHOUT DETERMINANTS:
One way to solve that system would be
to make a linear combination of the first two equations that would have no term with {{{y}}} ,
and then try to solve the system made from the resulting equation and
{{{kx+z=3}}}
Adding the first equation times {{{2}}}
plus the second one times {{{(-1)}}} , we get
{{{6x+z=13}}} .
Then you would solve the system
{{{system(6x+z=13,kx+z=3)}}}
Obviously, if {{{highlight(k=6)}}} ,
that system (and the original system)
Would have no solution.