Using only Point 1 and 2 you have:
$$
\begin{align}
\alpha&=50-\beta\\
\alpha+5\beta&=45\\
\Rightarrow 50+4\beta&=45\\
\beta&=-1.25
\end{align}
$$
and then using the first line you get $\alpha=50-(-1.25)=51.25$. Using MATLAB, it's a simple linear system which you can solve using linsolve.
A = [1 1; 1 5]
B = [50 ; 45]
linsolve(A, B)
What this command does is that it solves the system but written in matrix form:
$$
\begin{align}
\mathbf{Ax}&=\mathbf{B}\\
\begin{pmatrix}1 &1\\1 & 5\end{pmatrix}\begin{pmatrix}\alpha \\ \beta\end{pmatrix}&=\begin{pmatrix}50 \\ 45\end{pmatrix}
\end{align}
$$
I think MATLAB for computational reasons uses LU decomposition, but you could also premultiply by $\mathbf{A}^{-1}$:
$$
\begin{align}
\mathbf{x}&=\mathbf{A}^{-1}\mathbf{B}\\
\begin{pmatrix}\alpha \\ \beta\end{pmatrix}&=\frac{1}{4}\begin{pmatrix}5 &-1\\-1 & 1\end{pmatrix}\begin{pmatrix}50 \\ 45\end{pmatrix}=\begin{pmatrix}\frac{205}{4}\\-\frac{5}{4}\end{pmatrix}=\begin{pmatrix}51.25\\-1.25\end{pmatrix}
\end{align}
$$
Using this approach, you could run the following in MATLAB instead of the linsolve command:
A = [1 1; 1 5]
B = [50 ; 45]
inv(A)*B
EDIT: If you were to do the same for Point 2 and Point 3, the system is instead:
$$
\begin{align}\alpha+5\beta&=45\\
\alpha+12\beta&=38
\end{align}
$$
Rewriting this in matrix notation gives us:
$$
\begin{pmatrix}\alpha+5\beta\\\alpha+12\beta\end{pmatrix}=\begin{pmatrix}45\\38\end{pmatrix}\Leftrightarrow \begin{pmatrix}1 & 5 \\ 1 & 12 \end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}=\begin{pmatrix}45\\38\end{pmatrix}
$$
This means that
$$
\mathbf{A}=\begin{pmatrix}1 & 5 \\ 1 & 12 \end{pmatrix}, \quad \mathbf{B}=\begin{pmatrix}45\\38\end{pmatrix}
$$
which in MATLAB then is
A = [1 5; 1 12]
B = [45 ; 38]
linsolve(A, B)
