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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.

 Aug 3, 2022
 #1
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Add all the equations and we get

 

10a + 10b + 10c + 10d =  0

 

So   dividing through by 10

 

a + b + c + d  =  0

 

 

 

cool cool cool

 Aug 3, 2022

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