數學的一點問題

1)
{3x+2y=13---(1)
{2x+3y=12---(2)
(1): 2y=13-3x
y=(13-3x)/2----(3)
Sub. (3) in (2)
2x+3(13-3x)/2=12
4x+39-9x=24
-5x=24-39=-15
x=3
Sub. x=3 in (3)
y=(13-3x3)/2=4/2=2
So, the roots of equation is (3, 2)
(2) {3(x+1)+4(y-1)=24----(1)
(2) {(3x-1)-(y-2)=6-----(2)
(2):
3x-1-y+2=6
3x-y=6+1-2=5
y=3x-5-----(3)
Sub. (3) into(1)
3(x+1)+4(3x-5-1)=24
3x+3+12x-20-4=24
15x=24+24-3=45
x=3
Sub. x=3 in (3)
y=3x3-5=9-5=4
The roots of equation is (3, 4)

2008-02-12 18:03:44 補充：
(1){3x+2y=13------1)(1){2x+3y=12-------2)1)=6x+4y=132)=6x+9y=12下面兄弟，1)現在叫你用代入法，唔係消元法2)即使用消元法，你的13和12都唔記得乘2

2008-02-12 18:04:38 補充：
"即使用消元法，你的13和12都唔記得乘2 "唔好意思，應分別乘2和3

1.{3x+2y=13 ----------(1)
{2x+3y=12 ----------(2)
(1)*2, 6x+4y=26 ----------(3)
(2)*3, 6x+9y=36 ----------(4)
(4)-(3), (6x+9y)-(6x+4y)=36-26
6x+9y-6x-4y=10
5y=10
y=2
substitute y=2 into (1)
3x+2(2)=13
3x+4=13
3x=9
x=3
Therefore, the solution is (3,2)

2.{3(x+1)+4(y-1)=24 ----------(1)
{(3x-1)-(y-2)=6 ----------(2)
From(1), 3x+3+4y-4=24
3x+4y=24-3+4
3x+4y=25 ----------(3)
From(2) 3x-1-y+2=6
3x-y=6+1-2
3x-y=5 ----------(4)
(1)-(2), (3x4yy)-(3x-y)=25-5
3x+4y-3x+y=25-5
5y=20
y=4
substitute y=4 into (3)
3x+4(4)=25
3x+16=25
3x=25-16
3x=9
x=3
Therefore, the solution is (3,4)

I hope my solution can help you! ^_^"

(1){3x+2y=13------1)
(1){2x+3y=12-------2)

1)=6x+4y=13
2)=6x+9y=12

1)-2)=(6x+4y)-(6x+9y)=13-12
-5y=1
-1=5y
5y=-1
y=-0.2

If y=-0.2,
1)=3x-0.4=13
3x=13.4
x=4.466666...


x=4.466666...(134/30),y=-0.2


(2) {3(x+1)+4(y-1)=24----1)
(2) {(3x-1)-(y-2)=6------2)

1)-2)=3(x+1)+4(y-1)-[(3x-1)-(y-2)]=24-6
3x+3+4y-4-(3x-1-y+2)=18
3x+3+4y-4-(3x+1-y)=18
3x+3+4y-4-3x-1+y=18
3x-3x+3-4-1+4y+y=18
5y=18+2
5y=20
y=4


If y=4,1)=3(x+1)+4(4-1)=24
3x+3+12=24
3x=24-3-12
3x=9
x=3


x=3,y=4

1) 10 59
y = - ----- (負10 over 3) ， x = ------- (59 over 9)
3 9


2) x= 3 ， y = 4

2008-02-12 17:50:24 補充：
1) y= -10/3 ， x=59/9

2008-02-12 17:51:49 補充：
sor~ 1) 係 y=2, x=3

1)
2x+3y=12
2x=12-3y
x=12-3y/2

3x+2y=13
3(12-3y/2)+2y=13
36-9y/2+2y=13
36-9y/2=13-2y
36-9y=2(13-2y)
36-26=-4y+9y
10=5y
y=2

X=3(自己計)