I interpret this as a system of equations.
Since 3x+m-1=y-3, y=3x+m+2.
Substitute for y in the other equation:
(m+1)x+3x+m+2=m+3,
(m+4)x=1, x=1/(m+4).
y=3/(m+4)+m+2, y=(3+m²+6m+8)/(m+4)=(m²+6m+11)/(m+4).
The role of m is as a parameter, where x and y are in terms of m. Note that m≠-4 because that would cause division by zero for x and y. As m gets large and positive or negative, x→0, y→m, so y→±∞ as m→±∞. By eliminating m we get y=3x-2+(1/x).
[Note that the title of the question gives a different system of equations from the additional details section. “=y” has been switched round to “y=” in the second equation.
If the title carries the correct form of the system, the answers are different. x=-m(m+2)/(m²+2) and y=6/(m²+2). There are no restrictions on the value of m because the denominator cannot be zero. I have assumed the additional details section to be correct, hence the restrictions on the value of m.]