Can anyone confirm if I am correct for this question, thank you.
There are positive real numbers $x$ and $y$ which solve the equations $2x + ky = 4, \;x + y = k$,for
(a) all values of $k$
(b) no values of $k$
(c) $k = 2$ only
(d) only $k > −2$.
My attempt: $$k-y=x\\2(k-y)+ky=4\\2k+ky-2y=4\\k(2+y)=4+2y\\2+y=4+2y\implies y=-2\\k=4+2(-2)\implies k=0$$ There is a problem with my solution. $y$ is a positive real value, and my answer is not correct. I suspect it is line 4 - can I not treat $k$ as a factor similar to factorising quadratics?
Hints for solving the problem:
You have $2k+ky-2y=4$, which means $y(k-2)=-2k+4$.
Solve for $y$ if $k\ne 2$.
And can you see that, if $k=2$, then all real numbers $x$ and $y$ solve the equations?