Determine, in accordance to $k$, how many solutions does the given system of equations have: $ \begin{cases}kx+(k+1)y=k-1\\4x+(k+4)y=k\end{cases} $ And check, for which values of $k$ this system has exactly one solution lying within the boundaries of a triangle of vertexes: $A=(0, 0), B=(\frac{2}{3}, 0), C=(0, 2)$.
Can I just solve it using determinants? I mean: $D=k(k+4)-4(k+1)=k^2-4=(k+2)(k-2)$ $D_x=(k-1)(k+4)-(k+1)k=2k-4$ $D_y=k^2-4(k-1)=k^2-4k+4$
Our system has one solution if $D$ is not $0$. This happens when $k$ is neither $2$ nor $-2$.
It has no solutions when $D=0$ and ($D_x \neq 0$ or $D_y \neq 0$). $D=0$ for $k=2$ or $k=-2$, $D_x \neq 0$ when $k \neq 2$ and $D_y \neq 0$ when $k \neq 2$. Though for $k=2$ our $D$ equals $0$, neither $D_x$ nor $D_y$ is $\neq 0$ so it has no solutions only when $k=-2$.
It has infinitely many solutions for $D=0$ (which means $k=2$ or $k=-2$), $D_x = D_y = 0$. This happens when $k=2$.
Summing up: $ \begin{cases} \text{no solution}, &\text{for } k=-2, \\ \text{infinitely many solutions}, &\text{for } k=2, \\ \text{one solution}, &\text{for every other } k. \end{cases} $
Is this all right? I need to know that before proceeding to the triangle-thing.