There are *positive* real numbers \(x\) and \(y\) which solve the equations \[2x + ky = 4, \qquad x+y = k\] for

all values of \(k\);

no values of \(k\);

\(k=2\) only;

only \(k>-2\).

The second equation tells us that \(x=k-y\).

We can substitute this into the first equation to give \(2(k-y)+ky=4\), or \(y(k-2) +2k-4=0\).

Note that \(2k-4=2(k-2)\), so the equation factorises to \((k-2)(y+2)=0\).

So either \(k = 2\), or \(y = -2\), or both.

Clearly if \(y = -2\), we will never have a solution to the pair of equations that gives positive values for both \(x\) and \(y\).

Our only hope for such a solution is therefore \(k = 2\).

This gives us \(2x+2y = 4\) and \(x+y=2\). In this case, the lines representing each equation coincide, and we have infinitely many solutions for the pair of equations.

In particular we have the solution \(x = 1, y = 1\), and so a solution where both \(x\) and \(y\) are positive is possible.

Thus the answer is (c).