數學知識交流---解方程

3x+7y+2z=109 .....(1)
8x+5y+3z=149 .....(2)(1)*3 - (2)*2 :9x + 21y - 16x - 10y = 109*3 - 149*2
11y - 7x = 29
11y - 7x = 22 + 7
11(y - 2) = 7(x + 1) .... (*)因 11 和 7 互質 , 故可令y - 2 = 7k , (k 為整數)y = 7k + 2 , 代回 (*) :77k = 7x + 7x = 11k - 1把 x , y 代回 (1) :3x + 7y + 2z = 109
(33k - 3) + (49k + 14) + 2z = 10982k + 11 + 2z = 109z = 49 - 41k故方程組的通解為 :x = 11k - 1
y = 7k + 2
z = 49 - 41k因正整數解 :x = 11k - 1 > 0 ==> k > 1/11
y = 7k + 2 > 0 ==> k > - 2/7
z = 49 - 41k > 0 ==> k < 49/41綜合得整數 k = 1故方程組的正整數解為 :x = 11 - 1 = 10
y = 7 + 2 = 9
z = 49 - 41 = 8