Solve the Mystery of Four Numbers with Unusual Sums | Homework Help

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The discussion focuses on solving a system of equations involving four numbers A, B, C, and D, where the sums of three numbers equal 20, 22, 24, and 27. The equations derived from the problem are A + B + C = 20, A + C + D = 22, A + B + D = 24, and B + C + D = 27. By substituting A in terms of B and C into the other equations, the problem can be simplified to three equations with three unknowns, allowing for a straightforward solution.

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Okay so there are four numbers represented by A,B,C,D. Their sums when adding three of the four equal the following sums.. 20,22,24, and 27. This is what I have discovered so far but am really stuck.

a+b+c+d=x 20= b+c+d or 20= x-a 22=x-b 24= x-c 27=x-d

Thanks in advance to anyone who can solve this, I really appreciate it!
 
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First, you should be aware that there is a homework forum and I am going to move this thread there.

Second, the 4 3-member subsets of {A, B, C, D} are {A, B, C}, {A, C, D}, {A, B, D}, and {B, C, D}. You have 4 equations: A+ B+ C= 20, A+ C+ D= 22, A+ B+ D= 24, and B+ C+ D= 27. You should be able to solve 4 equations for the 4 unknown numbers. There is no need to introduce a fifth, "x".
Those should be easy to solve. For example, from the first equation, A= 20- B- C. Replace A in the other 3 equations and you have reduced form 4 to 3 equations. Keep doing that.
 

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