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)=16
1个回答

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