Question 154
For what values of $k$ will the system below have no solution?
$$
\begin{aligned}
(k-3)x + 2y &= k^2 - 1 \\
x + \left(\frac{k-4}{3}\right)y &= 0
\end{aligned}
$$
$$
\begin{aligned}
(k-3)x + 2y &= k^2 - 1 \\
x + \left(\frac{k-4}{3}\right)y &= 0
\end{aligned}
$$