初一的三道二元一次方程!急用!① [ 2(x-y)]/3 - [x+y]/4=-16(x+y) - 4(2x-y)=16② [x+1]/5=[y-3]/23x+4y=32③ [3x+2y]/4=[2x+y]/5=[x-y+1]/6

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初一的三道二元一次方程!急用!①[2(x-y)]/3-[x+y]/4=-16(x+y)-4(2x-y)=16②[x+1]/5=[y-3]/23x+4y=32③[3x+2y]/4=[2x+y]/5=[x

初一的三道二元一次方程!急用!① [ 2(x-y)]/3 - [x+y]/4=-16(x+y) - 4(2x-y)=16② [x+1]/5=[y-3]/23x+4y=32③ [3x+2y]/4=[2x+y]/5=[x-y+1]/6
初一的三道二元一次方程!急用!
① [ 2(x-y)]/3 - [x+y]/4=-1
6(x+y) - 4(2x-y)=16
② [x+1]/5=[y-3]/2
3x+4y=32
③ [3x+2y]/4=[2x+y]/5=[x-y+1]/6

初一的三道二元一次方程!急用!① [ 2(x-y)]/3 - [x+y]/4=-16(x+y) - 4(2x-y)=16② [x+1]/5=[y-3]/23x+4y=32③ [3x+2y]/4=[2x+y]/5=[x-y+1]/6
① [ 2(x-y)]/3 - [x+y]/4=-1 (1)
6(x+y) - 4(2x-y)=16 (2)
由(1)得
8(x-y)-3(x+y)=-12
5x-11y=-12 (3)
由(2)得
6x+6y-8x+4y=16
-2x+10y=16
x-5y=-8 (4)
解(3)(4)得
x=2
y=2
② [x+1]/5=[y-3]/2 (1)
3x+4y=32 (2)
由(1)得 2x+2=5y-15
2x-5y=-17
6x-15y=-51 (3)
(2)*2-(3)得 23y=115
y=5
把y=5代入(2) 得
x=4
∴x=4
y=5
③ [3x+2y]/4=[2x+y]/5 (1)
[3x+2y]/4=[x-y+1]/6 (2)
由(1)得 15x+10y=8x+4y
7x+6y=0 (3)
由(2)得 9x+6y=2x-2y+2
7x+8y=2 (4)
解(3)(4)得
x=-6/7
y=1