I'm trying to solve the following exercise:
Given the following system of linear equations: $\begin{cases}-4x + 3y = 2\\ 5x - 4y = 0 \\ 2x-y = k\end{cases}$
Find the values for k that allow for a unique solution, no solutions, infinitely many solutions.
So I tried to find the values of X and Y:
$\begin{cases}-4x + 3y = 2\\ 5x - 4y = 0 \end{cases}$
I managed to arrive at $x = - 8$ and $y = -10$
But I'm stuck now, how do I find these conditions on $k$?
$x=-8$ and $y=-10$ is a solution of the first two equations. The system has a solution
$\iff 2(-8)-(-10)=k \iff k=-6$.
Conclusion: the system has a unique solution $ \iff k=-6$, it has no solution $ \iff k \ne -6$, infinitely many solutions are not possible.