Number of solutions - please check my solution? 
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.
 A: First, let's get back to the mainstream terminology: We have a linear system $Ax=b$ with the data 
$$
\pmatrix{k &k+1\\4 &k+4}\pmatrix{x\\y}=\pmatrix{k-1\\k}
$$
Now if $A$ is invertible then there exists a unique solution which is given by
$$
\pmatrix{x\\y} = \pmatrix{k &k+1\\4 &k+4}^{-1}\pmatrix{k-1\\k} = \frac{1}{k^2-4}\pmatrix{k+4 &-(k+1)\\-4&k}\pmatrix{k-1\\k} = \pmatrix{\frac{2}{k+2}\\ 1-\frac{4}{k+2}}
$$
As you have computed, the easiest way to conclude invertibility is having the determinant nonzero. Computing the determinant gives: $\det A = k^2-4 = (k-2)(k+2)$, Thus, we have a unique solution for all $|k|\neq 2$.
Regarding the zero determinant cases, let's look at the system at those points: 
($k=2$)
$$
\pmatrix{2&3\\4&6}\pmatrix{x\\y} = \pmatrix{1\\2}
$$
We can clearly see that the second equation is the 2-multiple of the first one. Hence we actually only have $2x+3y = 1$
($k=-2$)
$$
\pmatrix{-2&-1\\4&2}\pmatrix{x\\y} = \pmatrix{-3\\-2}
$$
We can see that the right hand side is not in the range of $A$, hence there is no solution. Another way to look at it is rewriting as
$$
\pmatrix{-2\\4}x+\pmatrix{-1\\2}y = \pmatrix{-1\\2}(2x+y) = \pmatrix{-1\\2}\alpha = \pmatrix{-3\\-2}
$$
It is obvious that no alpha can satisfy that equality hence no solution. 
Regarding the triangle question, you can rewrite the constraints as
$$
\pmatrix{-3 &-1\\1 &0\\0&1}\pmatrix{x\\y}\geq \pmatrix{-2\\0\\0}
$$
Now plug in the $k$ dependent solution into this and we get
$$
\pmatrix{1-\frac{2}{(k + 2)}\\
       \frac{2}{(k + 2)}\\
   1 - \frac{4}{(k + 2)}}\geq 0
$$
This gives us the condition on $k$:
$$
0\leq \frac{2}{k+2} \leq \frac{1}{2} \implies k>2
$$
Note that we ruled out $k=2$ case since there are infinitely many solutions in that case violating the requirement given in the question. 
A: I don't know your terminology, but the answer is correct.  For $k=2$ the equations are multiples of each other.  For $k=-2$ the left sides are multiples but the rights do not respect that, so the equations are inconsistent.
For the triangle part, in the case there is only one solution, you should be able to express the point as a function of $k$.  In other words, the solution will be $(f(k),g(k))$  This will be a continuous function of $k$, so you can find the values of $k$ where it hits the edges of the triangle.  For $k=2$ you have a line of solutions-does it hit the triangle?  It can only have a single solution in the triangle if it hits one of the vertices and you consider that "within" the triangle.
