Compute the values of $x$ and $y$ dependent on $m$ for the following system, then solve $2x + my = m + 1$ (the last equation) to find the values of parameter $m$ for $x$ and $y$:
\begin{cases}
mx + y = m\\
mx + 2y = 1\\
\end{cases}
So,
\begin{cases}
2mx + 2y =2 m\\
mx + 2y = 1\\
\end{cases}
Subtracting two equations, will have:
$$mx=2m-1$$
If $m \neq 0$, we may divide by $m$ and get $x = (2m-1)/m$ and $y =
1-m$.
If $m = 0$, the system has no solution.
Putting $x$ and $y$ in the last equation ($m\neq 0$), we'll have:
$$m^3-3m+2=0 $$
$$(m^3-1)-3m+3=0$$
$$(m-1)(m^2+m+1)-3(m-1)=0$$
$$(m-1)(m^2+m-2)=0$$
Thus the values of parameter $m$ are $m=1$ or $m=-2$.