F3 Maths

2007-09-16 12:54 am
請用代入消元法解下列二元一次方程
{ 2(x+1)=5(y-1)+11
{ 3(x+5)-2y=-1
列出步驟

回答 (4)

2007-09-16 1:07 am
✔ 最佳答案
{ 2(x+1)=5(y-1)+11----(1)
{ 3(x+5)-2y=-1------(2)
從(1), 得:
2x+2=5y-5+11
2x+2=5y+6
2x=5y+4
x=2.5y+2------(3)
代(3)入(2):
3(2.5y+2+5)-2y=-1
3(2.5y+7)-2y=-1
7.5y+21-2y=-1
5.5y+21=-1
5.5y=-22
y=-4
代y=-4入(3):
x=2.5(-4)+2=-8
要求的解為:x=-8, y=-4.
2007-09-16 1:10 am
(1)2(x+1)=5(y-1)+11
2x + 2 = 5y-5 +11
2x +2= 5y +6
2x= 5y+4
x= (5/2)y +(4/2)
x= (5/2)y +2

(2) 3(x+5)-2y=-1
3x +15 -2y = -1
3x-2y = -16
代入x= (5/2)y +2
3((5/2)y +2) -2y =-16
(15/2)y + 6 -2y=-16
(15/2)y -2y= -16-6
15y -4y = 2( - 22)
11y = -44
y=-4
代入y=4
x= (5/2)y +2
x= (5/2) ( -4) +2
x= -10+2
x= -8
2007-09-16 1:05 am
{ 2(x+1)=5(y-1)+11
{ 3(x+5)-2y=-1

Solution:
{ 2(x+1)=5(y-1)+11 -1
{ 3(x+5)-2y=-1 -2

From 1,
2x+2=5y-5+11
2x-5y=4
6x-15y=12 -3

From2,
3x+15-2y=-1
3x-2y=-16
6x-4y=-32 -4

3-4,
-11y=44
y=-4

Put y=-4 into 3,
6x-15*-4=12
6x+60=12
6x=-48
x=-8

The solution is x= -8, y= -4.
2007-09-16 1:02 am
2(x+1)=5(y-1)+11.................(1)
3(x+5)-2y=-1.........................(2)
From (1)
2x+2-5y+5-11=0
2x-5y=4.................................(3)
From(2)
3x-2y = -16............................(4)
(3) * 3 , 6x-15y=12......(5)
(4) * 2 6x-4y= -32......(6)
(6)- (5)
11y = -44
y = -4
put in (3)
2x - 5(-4) = 4
x= -8
參考: me


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