Given the equation $x+y =1$, $x^2+y^2 =2$, $x^5 +y^5 =A$. Let N be the number of solution pairs (x,y) to this system of equations. Then AN is equal to

From \(x+y=1\) and \(x^2+y^2=2\): $(x+y)^2=x^2+y^2+2xy$

\[ \;\Rightarrow\; 1=2+2xy \;\Rightarrow\; xy=-\tfrac12. \] Thus \(x,y\) are roots of \(t^2 - t - \tfrac12=0\), giving two ordered pairs \((x,y)\). Hence \(N=2\). Let \(p_n=x^n+y^n\). For quadratic roots with \(s_1=x+y=1\) and \(s_2=xy=-\tfrac12\), \[ p_n = s_1 p_{n-1} - s_2 p_{n-2} \quad (n\ge2), \] with \(p_0=2,\; p_1=1\). \[ p_2=1\cdot1-(-\tfrac12)\cdot2=2,\quad\] $$ p_3=1\cdot2-(-\tfrac12)\cdot1=\tfrac52,$$ \[ p_4=\tfrac72,\quad p_5=1\cdot\tfrac72-(-\tfrac12)\cdot\tfrac52=\tfrac{19}{4}. \] Therefore \(A=p_5=\tfrac{19}{4}\). With \(N=2\), \[ AN=\frac{19}{4}\cdot 2=\boxed{\tfrac{19}{2}}. \]