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# help with system

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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

#1
+524
+3

We have the following system of  equations -

\(a+2b+3c+4d=10\)

\(4a+b+2c+3d=4\)

\(3a+4b+c+2d=-10\)

\(2a+3b+4c+d=-4\)

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

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

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

∴ 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.

May 1, 2021
edited by amygdaleon305  May 1, 2021

#1
+524
+3

We have the following system of  equations -

\(a+2b+3c+4d=10\)

\(4a+b+2c+3d=4\)

\(3a+4b+c+2d=-10\)

\(2a+3b+4c+d=-4\)

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

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

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

∴ 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.

amygdaleon305 May 1, 2021
edited by amygdaleon305  May 1, 2021
#2
+121065
+3

Very  nice, Amy  !!!!

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

CPhill  May 1, 2021
#3
+524
+1

You flatter me Phill!! Thank you!!

amygdaleon305  May 1, 2021