求解開 加減消元法

{ a/2 + b/3 = 13 ........1
{ a/3 - b/4 = 3 .........2

第一式 ×24
24 × (a/2 + b/3) = 13 × 24
12a+8b = 312 .......3

第二式 ×36
36 × (a/3 - b/4) = 3 × 36
12a-9b = 108 ........4

之後變成這條式
{ 12a+8b = 312 .......3
{ 12a-9b = 108 ........4

3式-4式
12a+8b - (12a-9b) = 312-108
12a+8b-12a+9b =204
17b =204
b=12

把b=12代入3:
12a+8(12)=312
12a+96=312
12a=216
a=18

∴ a=18 b=12

{ a/2 + b/3 =13 i
{ a/3 - b/4 =3 ii
6*i, 3a+2b=78
b=(78-3a)/2 iii
代iii入ii
a/3 -(78-3a)/12=3
4a-(78-3a)=36
4a+3a-78=36
a=114/7
代a=114/7入ii
b=(78-3*114/7)/2
b=102/7
(a,b)=(114/7,102/7)

{ a/2 + b/3 = 13
{ a/3 - b/4 = 3

第一條式全式 × 6
6 × (a/2 + b/3) = 13 × 6
6 × a/2 + 6 × b/3 = 13 × 6
3a + 2b = 78 ...①

第二條式全式 × 12
12 × (a/3 - b/4) = 12 × 3
12 × a/3 - 12 × b/4 = 12 × 3
4a - 3b = 36 ...②

{ 3a + 2b = 78 ...①
{ 4a - 3b = 36 ...②

① × 3: 9a + 6b = 234
② × 2: 8a - 6b = 72

2015-07-28 15:23:13 補充：
相加可得
17a = 306
a = 18

代入②:
4(18) - 3b = 36
72 - 3b = 36
3b = 36
b = 12

{ a = 18
{ b = 12