關於a, b, c, d, e的方程組

a + b + c = m …… (1)
a + c + e = n …… (2)
a + c + d = o …… (3)
b + d + e = p …… (4)

(1) + (2) + (3) + (4)x2:
3a + 3b + 3c + 3d + 3e = m + n + o + 2p
3(a + b + c + d + e) = m + n + o + 2p
Hence, a + b + c + d + e = (m + n + o + 2p)/3

2010-10-24 17:46:43 補充：
樓下 003 lai 的答案錯誤：

(1) - (2) - (3) - (4) 應為：
-2a - 2c - 4d - 4e = 2(m - n - o - p) ...... (6)

All mutiply by 2,
2a+2b+2c=2m.............(1)
2a+2c+2e=2n..............(2)
2a+2c+2d=2o..............(3)
2b+2d+2e=2p..............(4)

(1)+(2)++(3)+(4),
6a+4b+6c+4d+4e=2(m+n+o+p)...........(5)
(1)-(2)-(3)-(4),
-2a-2c=2(m-n-o-p)...............................(6)

(5)+(6),
(6a+4b+6c+4d+4e)+(-2a-2c)=2(m+n+o+p)+2(m-n-o-p)
(6a-2a)+4b+(6c-2c)+4d+4e = 2(m+n+o+p+m-n-o-p)
4a+4b+4c+4d+4e=2(2m)
4(a+b+c+d+e)=4m
a+b+c+d+e=4m/4
a+b+c+d+e=m

a+b+c=m
a+c+e=n
a+c+d=o
b+d+e=p so
3a+2b+2c+2d+2e=m+n+o+p
a+b+c+d+e=m+n+o+p-2a-b-c-d-e