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.

Guest Feb 1, 2022

#1**+1 **

Add all of the equations together to get

10a + 10b + 10 c + 10 d = 20 now divide by 10

a + b + c + d = 2

ElectricPavlov Feb 1, 2022

#2**+1 **

You made a mistake Pavlov, in adding \(-4 -10+10 +4 = 0\).

Thus, \(10a+10b+10c+10d = 0\)

So, \(\color{brown}\boxed{a + b + c+d=0}\)

BuilderBoi
Feb 2, 2022

#3**0 **

Yep, I sure did ! I seem to miss those tiny ' - ' signs pretty regularly

Thanx for checking my math, BuilderBoi !

ElectricPavlov
Feb 2, 2022