5X+3Y+Z/3= 100 A
X+Y+Z= 100 B
3a-b de
7X+4Y= 100
This equation is a multi-solution equation. From the equation, we know that x must be an even number Let x = 2, 4, 6, 8, 10, 12 and 14 be substituted respectively. It is known that x = 4,8, 12 conforms to the equation and the actual situation, so there are three solutions to this problem, namely.
X=4，Y= 18，Z=78
X=8，Y= 1 1，Z=8 1
X= 12，Y=4，Z=84