數學難題,幫幫手!

let x y z be the no of stamps that 甲、乙和丙 have

so x=2y y=3z

2y =6z...(1)
then x =6z...(2)

x + y + z = 120
6z + 3z + z = 120
10z = 120
z = 12

Sub z = 12 into (2)
x = 6z
x = 6x12
x = 72

所以甲有72個郵票

設A,B,C 分別為甲,乙,丙所得郵票數量. 按題目設立以下方程式:
A = 2B
B = 3C
A + B + C = 120

=> 2(3C) + (3C) + C = 120
=> 10C = 120
=> C = 12
=> B = 3C = 36
=> A = 2B = 72

Let x be the no. of stamp that 丙 has
2(3x)+3x+x=120
10x=120
x=12
therefore, no of stamp that 甲 has =2(3X12)=72