Question

(x/3)-(y/4)=3,(x/2)+(y/3)=13二元一次方程.

还有x-(y-1/2)=3,2(x-y)+3(2y-x)=-6。
(2x-1/5)+(3y-2/4)=2,(3x+1/5)-(3y+2/4)=0.
3(x-1)=y+5,5(y-1)=3(x+5).
(2u/3)+(3u/4)=(1/2),(4u/5)+5u/6)=(7/12)这四个也解下

Answer 1

{(x/3)-(y/4)==3, (x/2)+(y/3)==13} 得 {x=18, y=12}
{x-(y-1/2)==3, 2(x-y)+3(2y-x)==-6} 得 {x=4/3, y=-7/6}
{(2x-1/5)+(3y-2/4)==2, (3x+1/5)-(3y+2/4)==0} 得 {x=3/5, y=1/2}
{3(x-1)==y+5, 5(y-1)==3(x+5)} 得 {x=5, y=7}
{(2u/3)+(3u/4)==(1/2), (4u/5)+(5u/6)==(7/12)} 无解

PS:应是{(2u/3)+(3v/4)==(1/2), (4u/5)+(5v/6)==(7/12)}吧
解得{u=15/32, v=1/4}

需要过程请追问

Answer 2

降级处理，