聯立方程問題2(要步驟)

2007-12-04 8:09 am
求未知數,,,
1.2x+5y+5=0
4x+3y+17=0

2.2x+3y=18
3x+2y=12

3.2x+5y-3=0
3x-2y+24=0

4.(x+4)-2(y-7)=15
4(2+x)-3(7-5y)=21

5.3x-(y-10)=0
x-2y=0

要步驟,,,thx a lot=]

回答 (2)

2007-12-04 8:39 am
✔ 最佳答案
1.2x+5y+5=0 - (1)
4x+3y+17=0 - (2)

由 (1) : 2x+5y = -5 - (3)
由 (2) : 4x+3y = -17 - (4)
(3) x 2 : 4x+10y= -10 - (5)
(5) - (4) : 7y = 7

把 y = 1 代入 (3)
2x+5 = -5
2x= -10
x= -5

2.2x+3y=18 - (1)
3x+2y=12 - (2)

(1)x2 : 4x+6y=36 - (3)
(2)x3 : 9x+6y =36 - (4)
(4)-(3) : 5x = 0
x=0
把x=0 代入 (3)
y=6

3. 2x+5y-3=0 - (1)
3x-2y+24=0 - (2)

(1) x 3 : 6x+15y-9=0 - (3)
(2) x 2 : 6x-4y+48=0 - (4)
(3)-(4) : 19y-57=0
y = 3
把 y = 3 代入 (4)
6x- 12 + 48 = 0
x = -6

4.(x+4)-2(y-7)=15 - (1)
4(2+x)-3(7-5y)=21 - (2)

由(1) : x+4-2y+14 = 15
x-2y= -3 - (3)
由(2) : 8+4x-21+15y = 21
4x+15y = 34 - (4)
(3)x4 : 4x-8y = -12 - (5)
(4)-(5) : 23y = 46
y = 2
把 y = 2 代入 (3)
x-4= -3
x = 1

5. 3x-(y-10)=0 - (1)
x-2y=0 -(2)

由 (1) : 3x-y = -10 - (3)
(2) x 3 : 3x-6y = 0 - (4)
(3)-(4) = 5y = -10
y = -2
把y = -2 代入 (3)
3x+2= -10
x = -4
2007-12-04 9:03 am
1.
2x+5y+5=0 <---------(1)
4x+3y+17=0 <----------(2)

In (1)
X=-5/2(y+1) <------------ (3)

Put (3) into (2)
-4{5/2(y+1)} + 3y +17=0
-10y - 10 + 3y + 17 = 0
7y = 7
y = 1

put y=1 into (2)
4x + 3(1) + 17=0
4x + 20 = 0
x = -5

so X=-5 and Y=1




2.
2x+3y=18 <----------- (1)
3x+2y=12 <----------- (2)

In (1)
x = 3/2(6-y) <------------- (3)

put (3) into (2)
3{3/2(6-y)} + 2y = 12
9(6-y) + 4y = 24
54 - 5y = 24
y = 6

put y=6 into (1)
2x+3(6)=18
x=0

so X=0 and Y=6




3.
2x+5y-3=0 <------------(1)
3x-2y+24=0 <------------(2)

In (2)
x = 2/3(y-12) <------------(3)

put (3) into (1)
2{2/3(y-12)} + 5y - 3 = 0
4(y-12) + 15y - 9 = 0
19y = 57
y = 3

put y=3 into (2)
3x - 2(3) + 24 = 0
x = -6

so X=-6 and Y=3




4.
(x+4)-2(y-7)=15 <--------------- (1)
4(2+x)-3(7-5y)=21 <---------------- (2)

In (1)
(x+4)-2(y-7)=15
x + 4 - 2y + 14 = 15
x = 2y - 3 <----------- (3)

In (2)
4(2+x)-3(7-5y)=21
8 + 4x - 21 + 15y = 21
4x + 15y = 34
4(2y - 3) + 15y = 34 (by (3))
8y - 12 + 15y = 34
23y = 46
y = 2

put y = 2 into (1)
(x + 4) - 2(2 - 7) = 15
(x + 4) + 10 = 15
x = 1

so X=1 and Y=2




5.
3x-(y-10)=0 <----------- (1)
x-2y=0 <------------ (2)

In (2)
x = 2y

put x = 2y in to (1)
3(2y) - (y - 10) = 0
6y - y + 10 = 0
y = -2

put y = -2 into (2)
x = -4

so X=-4 and Y=-2




好野
做完功課啦


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