(2y-1)/3-(10y+1)/6=(2y+1),5x/12=(2x-1)/4-1 1/3[3y-(10-7y)/2]-1/2[2y-(2y+2)/3]=y/2-1

问题描述:

(2y-1)/3-(10y+1)/6=(2y+1),5x/12=(2x-1)/4-1 1/3[3y-(10-7y)/2]-1/2[2y-(2y+2)/3]=y/2-1

(2y-1)/3-(10y+1)/6=(2y+1)2(2y-1)-10-1=12y+6-8y=19y=-19/85x/12=(2x-1)/4-15x=4(2x-1)-3x=-4x=4/31/3[3y-(10-7y)/2]-1/2[2y-(2y+2)/3]=y/2-1y-(10-7y)/6-y+(2y+2)/6=y-(10-7y)+2y+2=6y3y=8y=8/3