Question 1206698
{{{matrix(2,3,k, 1, -2,
4, -1, 2)}}}

system is:

{{{kx+y=-2}}}....eq.1
{{{4x-y=2}}}.....eq.2
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{{{kx+y=-2}}}...eq.1, solve for {{{x}}}

{{{x=(-2-y)/k}}}....eq.1a


{{{4x-y=2}}}.....eq.2, solve for {{{x}}}
{{{x=(2+y)/4}}}.....eq.2a

from eq.1a and eq.2a we have

{{{(-2-y)/k=(2+y)/4}}}....., solve for {{{k}}}

{{{4(-2-y)=k(2+y)}}}

{{{-4(2+y)=k(2+y)}}}....both sides divide by {{{2+y}}}

{{{-4=k}}}

{{{k=-4}}}