Suppose a, b, c, and d are real numbers which satisfy the system of equations
a + 2b + 3c + 4d = 10
4a + b + 2c + 3d = 4
3a + 4b + c + 2d = 10
2a + 3b + 4c + d = 4.
Find a + b + c + d.
+2668 Add all the equations:
\(10a+10b+10c+10d=28\)
Divide by 10
\(a + b+c+d=\color{brown}\boxed{2.8}\)
