Here is a tricky one to keep you busy while waiting for the AP to announce the election results...
Can you solve the following system of equations for real variables
x, y and z\begin{cases}
\frac{x}{1 + a} + \frac{y}{2 + a} + \frac{z}{3 + a} = 1 \\
\frac{x}{1 + b} + \frac{y}{2 + b} + \frac{z}{3 + b} = 1 \\
\frac{x}{1 + c} + \frac{y}{2 + c} + \frac{z}{3 + c} = 1
\end{cases}where
a, b and c are distinct real numbers?
x = A(1+a)(1+b)(1+c)y = B(2+a)(2+b)(2+c)z = C(3+a)(3+b)(3+c)\frac{x}{1+a} + \frac{y}{2+a} + \frac{z}{3+a}=A(1+b)(1+c) + B(2+b)(2+c) + C(3+b)(3+c)=A(1+b+c+bc) + B(4+2b+2c+bc) + C(9+3b+3c+bc)1, so we get a system of equations:A + 4B + 9C = 1A + 2B + 3C = 0A + B + C = 0A=0.5,B=-1,C=0.5.