PrecalculusChallengeSystems of Equations

Systems of Equations · Challenge problem

A system like this always has the solution $x = y = 0$ . Give the two values of $k$ for which it has OTHER solutions as well (smaller first).

\begin{aligned} \left(k - 1\right)x + 2 y = 0 \\ 2 x + \left(k - 4\right)y = 0 \end{aligned}