A1xyz5 + B1x + C1y + D1 =0
A2xyz5 + B2x + C2y + D2 =0
A3xyz5 + B3x + C3y + D3 =0
is equivalent to
z5 = a,     x = b,     y = c,         (some a, b, c)

which has 5 solutions.















       
A1x + B1y + C1(z) =0
A2x + B2y + C2(z) =0         (deg(Ci(z))=5)
A3x + B3y + C3(z) =0
is equivalent to a system
f(z) = 0,   x = g(z),   y = h(z),       (deg(f)=5)

which has 5 solutions.