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.

Add all the equations and we get

10a + 10b + 10c + 10d = 0

So dividing through by 10

a + b + c + d = 0