SOLUTION: Four numbers, taken two at a time, give the sums 84, 88, 100, 100, 112 and 116. What are the four numbers?

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Question 999050: Four numbers, taken two at a time, give the sums 84, 88, 100, 100, 112 and 116. What are the four numbers?
Found 2 solutions by macston, ikleyn:
Answer by macston(5194)   (Show Source): You can put this solution on YOUR website!
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A, B, C, and D are the four numbers
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GIVEN:
A+B=84
A+C=88
A+D=100
B+C=100
B+D=112
C+D=116
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SOLVE:
.
A+B=84
A=84-B
.
B+C=100
C=100-B
.
B+D=112
D=112-B
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A+C=88
(84-B)+(100-B)=88
-2B=-96
B=48
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A+B=84
A+48=84
A=36
.
B+C=100
48+C=100
C=52
.
B+D=112
48+D=112
D=64
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ANSWER:
A=36
B=48
C=52
D=64
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CHECK:
A+B=84
36+48=84
84=84
.
A+C=88
36+52=88
88=88
.
A+D=100
36+64=100
100=100
.
B+C=100
48+52=100
100=100
.
B+D=112
48+64=112
112=112
.
C+D=116
52+64=116
116=116
Answer by ikleyn(52824)   (Show Source): You can put this solution on YOUR website!
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Four numbers, taken two at a time, give the sums 84, 88, 100, 100, 112 and 116. What are the four numbers?
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GIVEN:
A+B=84
A+C=88
A+D=100
B+C=100
B+D=112
C+D=116

Let us consider those sums that relate to 3 numbers A, B and C:

A+B=84     (1)
A+C=88     (2)
B+C=100    (3)

Sum up the left and the right sides. You will get

2*(A + B + C) = 84 + 88 + 100 = 272. It gives 

A + B + C =  = 136   (4).

Now, distract (1) from (4). You will get

C = 136 - 84 = 52.

Distract (2) from (4). You will get

B = 136 - 88 = 48.

Distract (3) from (4). You will get

A = 136 - 100 = 36.

Now you can easily determine D from A+D=100, for instance:

D = 100 - 36 = 64.

Answer. A=36, B=48, C=52, D=64.

Notice. We have 4 unknowns, therefore 4 equations is enough to find the unknowns. The rest of equations are excessive.

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