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.
+36916 Add all of the equations together to get
10a + 10b + 10 c + 10 d = 20 now divide by 10
a + b + c + d = 2
+2666 You made a mistake Pavlov, in adding \(-4 -10+10 +4 = 0\).
Thus, \(10a+10b+10c+10d = 0\)
So, \(\color{brown}\boxed{a + b + c+d=0}\)
+36916 Yep, I sure did ! 
I seem to miss those tiny ' - ' signs pretty regularly
Thanx for checking my math, BuilderBoi !
