(3x-y)2+(x-5)2=0 is only valid when x=5 and y=3x=15, because the sum of two squares is always positive and is only zero when each square itself is zero.
SOLUTION: x=5, y=15
This is the only real solution but there are complex solutions:
3x-y=(x-5)i,
3x-y=ix-5i,
y=3x-ix+5i,
y=x(3-i)+5i. This linear complex equation represents many solutions for x and y. Graphically, it's a line in 3-dimensional space: x=y/3=5-z which intersects the x-y plane (the "real plane") at x=5, y=15.