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Please help and take me through the steps. I have been told the answer is (1,2,3,4) but I need to know how to go through the steps to be able to apply to other problems.

x + y + z + w = 10

x + 2y + 3z + 6w = 22

x + 3y + 7z + 7w = 26

x + 4y + 7z + 7w = 37

Guest Feb 18, 2019

#1**+2 **

\(\text{subtract eq 3 from eq 4}\\ y = 11\\ \text{subtract eq 2 from eq 3}\\ y + 4z + w = 4\\ 11 + 4z + w = 4\\ 4z + w = -7\)

\(\text{subtract eq 1 from eq 2}\\ y + 2z + 5w = 12\\ 11 + 2z + 5w = 12\\ 2z + 5w = 1\)

\(\text{now we have 2 equations in 2 variables}\\ w + 4z = -7\\ 5w+2z = 1\\ \text{subtract twice eq 2 from eq 1}\\ -9w = -9\\ w = 1\\ \text{and from eq 1}\\ 1 + 4z = -7\\ z = -2\)

\(\text{now dump everything we know into eq 1}\\ x + 11 + -2 + 1 = 10\\ x + 10 = 10\\ x = 0\)

\((w, x, y, z) = (1, 0, 11, -2)\)

\(\text{So someone was pulling your leg}\)

.Rom Feb 18, 2019

#1**+2 **

Best Answer

\(\text{subtract eq 3 from eq 4}\\ y = 11\\ \text{subtract eq 2 from eq 3}\\ y + 4z + w = 4\\ 11 + 4z + w = 4\\ 4z + w = -7\)

\(\text{subtract eq 1 from eq 2}\\ y + 2z + 5w = 12\\ 11 + 2z + 5w = 12\\ 2z + 5w = 1\)

\(\text{now we have 2 equations in 2 variables}\\ w + 4z = -7\\ 5w+2z = 1\\ \text{subtract twice eq 2 from eq 1}\\ -9w = -9\\ w = 1\\ \text{and from eq 1}\\ 1 + 4z = -7\\ z = -2\)

\(\text{now dump everything we know into eq 1}\\ x + 11 + -2 + 1 = 10\\ x + 10 = 10\\ x = 0\)

\((w, x, y, z) = (1, 0, 11, -2)\)

\(\text{So someone was pulling your leg}\)

Rom Feb 18, 2019