x=(2y+4)/3 y=(3x-4)/3 这个二元一次方程怎么解

(x-y) (x-2y)+(x+2y) (x-3y)-2(x-3y) (x-4y)= 2（x+y)-3(x-y)=3 ,4(x+y)+3(x-y)=15 {3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1 x+y/2+x-y/3=6,4(x+y)-3(x-y)=-20 {(x+y)/2+(x-y)/3=6 4(x+y)-3(x-y)=-20 4(x+y)-3(x-y)=-20,2/x+y+3/x-y=6 3(x+y)-2(x-y)=11 4(x+y)+3(x-y)=9 3(x+y)+2(x-y)=36 4(x+y)-5(x-y)=2 （x+y）/2- x-y/2=6 3(x+y)=4(x-y) x+y/3+x-y/2=6 4(x+y)-2(x-y)=40 4(x-y)-3(x-y)=9 5(x-y)+2(x-y)=11 3(x+y)-4(x-y)=11,2(x-y)+5(x+y)=27 先化简再求值(x-y)(x-2y)+(x-2y)(x-3y)-2(x-3y)(x-4y) x=4 y=5 已知2y-x=-3,求[4y(2x-y)-2x(2x-y)]/(2x-y x+y+2(-x-y+1)=3(1-y-x)-4-(y+x-1),x+y=? 先化简 再求值 (2x-y)(y+x)-(x-2y)(2y+x)-(-3y+x)^2其(√x+1)+y^2+4=-4y 化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1 已知x-y=1,求[(x+2y)^2+(2x+y)(x-4y)-3(x+y)(x-y)]除以y的值大神们帮帮忙