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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.

 
 May 1, 2021

Best Answer 

 #1
avatar+274 
+3

We have the following system of  equations - 

  \(a+2b+3c+4d=10\)

  \(4a+b+2c+3d=4\)

  \(3a+4b+c+2d=-10\)

  \(2a+3b+4c+d=-4\)

 

Adding all of them, 

\(10a+10b+10c+10d=10-10+4-4\)

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

\(a+b+c+d=0\)

 

∴ The value of a+b+c+d is 0. 

 

P.S. This maybe due to the reason that either pair of a,b,c,d amount to 0 due to summing of negative and positive integers. 

 

smiley

 
 May 1, 2021
edited by amygdaleon305  May 1, 2021
 #1
avatar+274 
+3
Best Answer

We have the following system of  equations - 

  \(a+2b+3c+4d=10\)

  \(4a+b+2c+3d=4\)

  \(3a+4b+c+2d=-10\)

  \(2a+3b+4c+d=-4\)

 

Adding all of them, 

\(10a+10b+10c+10d=10-10+4-4\)

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

\(a+b+c+d=0\)

 

∴ The value of a+b+c+d is 0. 

 

P.S. This maybe due to the reason that either pair of a,b,c,d amount to 0 due to summing of negative and positive integers. 

 

smiley

 
amygdaleon305 May 1, 2021
edited by amygdaleon305  May 1, 2021
 #2
avatar+118504 
+3

Very  nice, Amy  !!!!

 

I wouldn't  have seen  that  summing  " trick"....clever  .....!!!

 

 

cool cool cool

 
CPhill  May 1, 2021
 #3
avatar+274 
+1

You flatter me Phill!! Thank you!! laugh

 
amygdaleon305  May 1, 2021

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