2y-3x-1=0 (1)
2y²-2xy-3x²=1 (2)
From (1), y=(3x+1)/2. Substitute for y:
2(3x+1)²/4-x(3x+1)-3x²=1,
(9x²+6x+1)/2-3x²-x-3x²=1,
9x²+6x+1-12x²-2x-2=0,
-3x²+4x-1=0,
3x²-4x+1=0=(3x-1)(x-1).
So x=⅓ or 1 and 2y-1-1=0⇒y=1 or 2y-3-1=0⇒y=2 respectively.
Solution is (x,y)=(⅓,1) or (1,2).