2元一次方程

2007-09-01 1:21 am
第一題 (用加減消元法)

2x + 3y = 9
3x - 2y = 7


第二題 (用加減消元法)

( x/4 ) + ( y/3 ) = 4
3y - 2( x - 1)=4


第三題 (隨意)

( x/2 ) - ( y/3 ) = 4
( x/4 ) + ( y/2 ) = 8

回答 (3)

2007-09-01 1:30 am
✔ 最佳答案
第一題2x + 3y = 9..(1)
3x - 2y = 7..(2)
(1)X2+(2)X3
13x=39
x=3
Sub x=3 into(1)
y=1
第二題
( x/4 ) + ( y/3 ) = 4
3x+4y=48..(1)
3y - 2( x - 1)=4
3y-2x=2..(2)
(1)X2+(2)X3
17y=102
y=6
Sub y=6 into(1)
x=8
第三題
( x/2 ) - ( y/3 ) = 4
3x-2y=24..(1)
( x/4 ) + ( y/2 ) = 8
x+2y=32..(2)
(1) +(2)
4x=56
x=14
Sub x=14 into (1)
y=9
2007-09-01 3:54 am
1)2x+3y=9
3(2x+3y)=3x9
6x+9y=27..............1
3x-2y=7
2(3x-2y)=2x7
6x-4y=14...............2
將1-2
13y=13
y=1
2x+3(1)=9
2x=9-3
x=3,y=1#
2)(x/4)+(y/3)=4
8(x/4)+8(y/3)=8x4
2x+8/3y=32...................1
3y-2(x-1)=4
3y-2x=4-2
2x-3y=-2.......................2
將1-2
8/3y+3y=34
8y+9y=102
17y=102
y=6
2x-3(6)=-2
2x=16
x=8,y=6#
3)(x/2)-(y/3)=4
2(x/2)-2(y/3)=2x4
x-2/3y=8................1
(x/4)+(y/2)=8
4(x/4)+4(y/2)=4x8
x+2y=32
x=32-2y..................2
將2代入1
32-2y-2/3y=8
3x32-3(2y)-3(2/3)y=3x8
96-6y-2y=24
8y=72
y=9
x=32-2x9
x=14,y=9#
2007-09-01 1:44 am
1.2x + 3y = 9(一式)
3x - 2y = 7(二式)
用一式*2加二式*3
4x+6y=18
+ 9x-6y=21
13x=39
x=3
用x=3代入一式
2*3+3y=9
3y=3
y=1

2. ( x/4 ) + ( y/3 ) = 4(一式)
12( x/4 ) +12 ( y/3 )=48
3x+4y=48(三式)
3y - 2( x - 1)=4(二式)
3y-2x+2=4
3y-2x=2(四式)
用三式*2加四式*3
6x+8y=96
+ 9y-6x=6
17y=102
y=6
用y=6代入四式
3*6-2x=2
16=2x
x=8

3.( x/2 ) - ( y/3 ) = 4(一式)
6( x/2 ) -6 ( y/3 ) = 24
3x-2y=24(三式)
( x/4 ) + ( y/2 ) = 8(二式)
4( x/4 ) +4 ( y/2 ) = 32
x+2y=32(四式)
用三式加四式
3x-2y=24
+ x+2y=32
4x=56
x=14
用x=14代入四式
14+2y=32
2y=18
y=9


收錄日期: 2021-04-29 21:44:43
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