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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.

 Feb 1, 2022
 #1
avatar+36417 
+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

 Feb 1, 2022
 #2
avatar+1374 
+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
avatar+36417 
0

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

             Thanx for checking my math, BuilderBoi !   

ElectricPavlov  Feb 2, 2022

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