2x=4y, so x=2y.
3xy+8=4y.
Substitute for x: 3y×2y+8=4y,
6y²-4y+8=0,
3y²-2y+4=0, which has only a complex solution:
y²-2y/3+4/3=0, y²-2y/3=-4/3,
y²-2y/3+1/9=1/9-4/3=-11/3,
(y-⅓)²=-11/3. We cannot take the square root of each side because √(-11/3) is an imaginary number.
In complex numbers √(-11/3)=i√(11/3) and the solutions would be:
y=⅓+i√(11/3) and y=⅓-i√(11/3).
However, if your equations were 2x=4y and 3x+8=4y, then 3x+8=2x, x+8=0, x=-8 and -16=4y, so y=-4.