Solution
= Solution
Substitution for $y$ reduces the two <simultaneous equations> to
$$
x^2+(x-1)^4-1=x(x-1)(x^2-3x+4)=0.
$$
The remaining quadratic factor is positive for all real $x$, since $x^2-3x+4=(x-3/2)^2+7/4$. Thus there are exactly $\boxed{2\text{ real solutions}}$.