For 3 variables x, y, z, you need 3 equations to get a unique solution. As it stands we can only find two variable in terms of the third.
Multiply x2-xy+yz=3 by 3:
3x2-3xy+3yz=9 and substitute 3y=8-2x:
3x2-8x+2x2+8z-2xz=9,
5x2-2x(4+z)+8z-9=0,
x=(2(4+z)±√(4(4+z)2-160z+180))/10,
x=(4+z±√(16+8z+z2-40z+45))/5,
x=(4+z±√(z2-32z+61))/5. We don't know z so we can't find x or y. The quadratic in z has real solutions only when z2-32z+61≥0.
One set of solutions is when z=2, so x=7/5 or 1, making y=26/15 or 2.